Algebra 1 · A.SSE.B.3B · F.IF.C.7A

Vertex form: reading, converting and graphing

A parabola’s equation can be written so that its turning point sits in plain sight. How do you read it off, get any quadratic into that shape, and draw the whole curve from that one point?

You need

Completing the square, leading coefficient not oneAlgebra 1 The vertex and the axis of symmetryAlgebra 1

Unlocks

Factored formAlgebra 1 The parabola as a locusPrecalculus Maximum, minimum and projectile problemsAlgebra 1

Where you are 12 in Quadratics

Algebra9

Quadratics12

▴ 6 above Completing the square, leading coefficient not one The quadratic formula4

Vertex form: reading, converting and graphing

Factored form Maximum, minimum and projectile problems ▾ Show all 12 (1 more)

Why this exists

The parabola y = x^2 - 6x + 11 turns somewhere, and nothing in the numbers 1, -6 and 11 tells you where. To find its vertex you need a formula, \displaystyle -\frac{b}{2a}, and then a substitution.

Now look at the same parabola written another way:

y = (x - 3)^2 + 2

Multiply it out and you get x^2 - 6x + 9 + 2 = x^2 - 6x + 11, so it is the same curve. But this time the vertex is printed in the equation: it is (3, 2). No formula, no arithmetic. This way of writing a quadratic is vertex form, and it is the reason anyone bothers to complete the square.

The catch is the sign. The 3 appears as -3 in the bracket, yet the vertex is at x = 3. Reading vertex form means knowing exactly why.

Reading is the first of four things this lesson does with the form. The second is converting: rewriting any quadratic ax^2 + bx + c in vertex form. The third is graphing from vertex form: drawing the whole parabola from its vertex and the number a, without a table of values. The fourth is seeing what the form says about every parabola at once: each one is the single curve y = x^2, moved, flipped and stretched. Those are the transformations of a parabola.

Definition

A quadratic is in vertex form when it is written

y = a(x - h)^2 + k \qquad (a \neq 0)

where a, h and k are numbers. Each of the three does one job:

(h, k) is the vertex, and the line x = h is the axis of symmetry.
The sign of a says which way the parabola opens: upward if a > 0, downward if a < 0.
The size of a says how steep it is: the larger |a|, the narrower the curve.

The form has a minus sign inside the bracket on purpose. With it, the number you read off is the x-coordinate of the vertex, sign and all. An equation with a plus inside has to be read through that minus:

y = (x + 5)^2 - 1 \quad\text{is}\quad y = \big(x - (-5)\big)^2 + (-1)

so h = -5 and k = -1, and the vertex is (-5, -1). Outside the bracket there is no twist: k is added, so +2 means k = 2 and -1 means k = -1.

Vertex form of a parabola beside the other two forms

The same quadratic can be written in three ways, and each makes a different fact easy to see.

Form	Written	Shows at a glance
standard	ax^2 + bx + c	y-intercept c
vertex	a(x - h)^2 + k	vertex (h, k)
factored	a(x - r)(x - s)	real roots r, s

The leading coefficient a is the same number in all three, since each multiplies out to a single x^2 term, ax^2. Standard form hides the vertex, and vertex form hides the roots until you solve for them. Factored form shows the roots the way vertex form shows the turning point, and it exists only when the roots are real; every quadratic has a vertex form.

Transformations of a parabola

The basic parabola is the graph of y = x^2. Its vertex is (0, 0), and it passes through (\pm 1, 1), (\pm 2, 4) and (\pm 3, 9).

A transformation changes every point of a graph by the same rule. Three kinds build every parabola.

A vertical shift, y = x^2 + k, moves every point k units up. A negative k moves it down.
A horizontal shift, y = (x - h)^2, moves every point h units to the right. A negative h moves it left, so y = (x + 2)^2 is a shift of 2 to the left.
A vertical stretch, y = ax^2, multiplies every height by a. When |a| > 1 the curve gets narrower; when 0 < |a| < 1 it gets wider, and this is also called a compression. When a < 0 the curve is also reflected in the x-axis and opens downward.

Doing all three gives vertex form, y = a(x - h)^2 + k: the basic parabola stretched by a, then moved so that its vertex lands on (h, k).

Intuition

The outside of the equation and the inside act on different things. Whatever is done outside the square, after squaring, acts on the output, the height. Adding 3 there adds 3 to every height, and the whole curve rises.

Figure 1Adding 3 to every output lifts every point 3 units.

Whatever is done inside, to x before it is squared, acts on the input. Start from y = x^2, whose lowest output, 0, comes when the thing being squared is 0, at x = 0. In y = (x - 3)^2 the thing being squared is x - 3. That is 0 when x = 3, not when x = 0, so the lowest point has moved to x = 3. Every output of y = x^2 now turns up 3 places later: the height 4 that y = x^2 reached at x = -2 now comes at x = 1. Arriving later on the x-axis means sitting further right, so the graph has slid 3 to the right, even though a 3 was subtracted. That is the whole of why a horizontal shift looks reversed, for this curve and for any graph shifted left or right.

Figure 2Subtracting 3 from the input carries every point 3 units to the right.

Do both at once and the two moves simply add up: the inside moves the curve across, the outside moves it up.

Figure 3y = (x - 3)^2 + 2 is y = x^2 moved 3 right and 2 up. Every point moves the same way, the vertex included.

A number multiplying the outside scales every height by the same factor. Doubling every height makes the curve climb twice as fast, so it looks narrower. Halving makes it wider. Multiplying by -1 turns every height into its opposite, which flips the curve over the x-axis. The vertex, at height 0, stays where it is, because a \cdot 0 = 0.

Figure 4Multiplying the output by a multiplies every height. At x = 2 the height 4 becomes 2 when a = \tfrac{1}{2} and -4 when a = -1.

So vertex form is the plain parabola y = x^2, moved so that its vertex lands on (h, k), and stretched or flipped by a. The a multiplies the vertical distance from the vertex, so it can make the curve narrower, wider or upside down. It cannot move the vertex, because at the vertex the distance it multiplies is 0.

Figure 5Three values of a in y = a(x - 3)^2 + 1. The shape and the direction change; the vertex does not.

Results

Reading vertex form. For y = a(x - h)^2 + k with a \neq 0:

the vertex is (h, k), and the axis of symmetry is the line x = h;
if a > 0 the parabola opens upward, k is its least value, and the range is y \ge k;
if a < 0 it opens downward, k is its greatest value, and the range is y \le k.

Stepping away from the vertex. At a distance d either side of the vertex, at x = h \pm d, the height is k + ad^2. In particular, one unit either side it is k + a.

Missing pieces are zeros. y = ax^2 + k has h = 0, and y = a(x - h)^2 has k = 0.

Moving one point at a time

For y = a(x - h)^2 + k, each point (x, y) of y = x^2 goes to

(x,\ y) \;\mapsto\; (x + h,\ ay + k)

The input only has h added; the output is multiplied by a first and has k added second. The vertex (0, 0) goes to (h, k), and the axis of symmetry x = 0 goes to x = h.

Read as a list of moves, y = a(x - h)^2 + k is y = x^2

stretched vertically by |a| (narrower if |a| > 1, wider if |a| < 1),
reflected in the x-axis if a < 0,
shifted h right and k up,

with the stretch and reflection done before the vertical shift.

Converting to vertex form

Reach for the conversion whenever a question about a quadratic in standard form is really about the turning point: the vertex, the axis of symmetry, the largest or smallest value, the range, or a quick sketch. Converting means finding a, h and k with f(x) = a(x - h)^2 + k for every x. The tool is completing the square (with a leading 1 and without one), used to rewrite an expression rather than to solve an equation. So each step must change how the right side looks without changing its value, and the f(x) on the left is never touched. Take

f(x) = 3x^2 - 12x + 7

Factor the leading coefficient out of the x terms. A square (x - h)^2 starts with a plain x^2, so the 3 has to come out first. Only the two x terms need to go inside the brackets; the constant can stay where it is:

f(x) = 3(x^2 - 4x) + 7

This is the same expression, since multiplying the 3 back in returns 3x^2 - 12x.

Complete the square inside the brackets by adding and subtracting. Half of -4 is -2, and (-2)^2 = 4. There is no other side to add 4 to, so add it and take it away in the same place. The pair adds zero, so the value does not change:

f(x) = 3(x^2 - 4x + 4 - 4) + 7

Write the square, and bring the extra piece out. The first three terms in the brackets are (x - 2)^2. The -4 is still inside the brackets, so it is multiplied by 3 on its way out and leaves as -12:

\begin{aligned} f(x) &= 3(x - 2)^2 - 12 + 7 \\ &= 3(x - 2)^2 - 5 \end{aligned}

Read off a, h and k. Matching a(x - h)^2 + k gives a = 3, h = 2 and k = -5. The vertex is (2, -5), the axis of symmetry is the line x = 2, and since a > 0 the parabola opens upward, so -5 is the smallest value f takes.

Check. Expand the answer: 3(x^2 - 4x + 4) - 5 is 3x^2 - 12x + 12 - 5, which is 3x^2 - 12x + 7, the f(x) we started from.

Two facts come out of these steps every time, and they give a second, faster check.

The a does not change. Expanding a(x - h)^2 + k gives ax^2 as its only x^2 term, so the number in front of the brackets is the leading coefficient of the standard form.

\displaystyle h = -\frac{b}{2a}, and k = f(h). Run the same steps with letters. Factoring a out of ax^2 + bx gives \displaystyle a\left(x^2 + \frac{b}{a}x\right); half of \displaystyle \frac{b}{a} is \displaystyle \frac{b}{2a}, so the square is \displaystyle \left(x + \frac{b}{2a}\right)^2 and \displaystyle h = -\frac{b}{2a}. Then putting x = h into a(x - h)^2 + k makes the bracket zero and leaves k, so k is simply the value of f at h. Here \displaystyle h = -\frac{-12}{2 \cdot 3} = 2 and f(2) = 12 - 24 + 7 = -5, which agrees. The shortcut finds the vertex quickly; completing the square is what produces the form and shows why it is right.

Graphing from vertex form

The vertex is one point and the sign of a is a direction. To draw the curve you also need its shape: how fast it climbs or falls as you move away from the vertex. The same number a answers that, in a pattern simple enough to do in your head. Because the vertex costs nothing, this is faster than graphing from standard form. Work through y = 2(x - 1)^2 - 8.

Read a, h and k. The bracket is x - 1, so h = 1; the number added is -8, so k = -8; the number in front is a = 2.

Plot the vertex and draw the axis. The vertex is (1, -8), and the axis is the dashed line x = 1. Since a = 2 > 0, the parabola opens upward, so (1, -8) is its lowest point.

Step out from the vertex. Take any distance d and move that far to the right of the vertex, to x = h + d. The bracket becomes (h + d) - h = d, so the height there is ad^2 + k. The same happens at x = h - d, where the bracket is -d and (-d)^2 = d^2. So the two points at distance d either side of the axis are at the same height, ad^2 above or below the vertex. Each point you work out gives you its mirror image free. With a = 2:

\begin{aligned} d = 1: &\quad -8 + 2(1) = -6 \\ d = 2: &\quad -8 + 2(4) = 0 \\ d = 3: &\quad -8 + 2(9) = 10 \end{aligned}

That gives (0, -6) and (2, -6), then (-1, 0) and (3, 0), then (-2, 10) and (4, 10).

See the steps as a staircase. On y = x^2, walking right from the vertex one unit at a time, the heights are 0, 1, 4, 9, 16, so each step rises 1, then 3, then 5, then 7: the odd numbers. The reason is that (d + 1)^2 - d^2 = 2d + 1. Multiplying by a multiplies every rise by a, so on y = a(x - h)^2 + k the rises from the vertex are

a,\; 3a,\; 5a,\; 7a,\; \dots

Here a = 2, so the rises are 2, 6, 10. You can plot the right arm by counting: over 1, up 2; over 1, up 6; over 1, up 10. Then copy each point across the axis.

Figure 6Each step is 1 across. The rises are 2, 6, 10, which is a = 2 times 1, 3, 5.

If a is negative, every rise is a fall and the curve opens downward. If a is a fraction, take bigger steps: a step of s units changes the height by as^2, then 3as^2, then 5as^2, so a step size that makes as^2 a whole number keeps every point on the grid.

Find the intercepts if you want them. The y-intercept is where x = 0: y = 2(0 - 1)^2 - 8 = -6, the point (0, -6) found above. The x-intercepts are where y = 0, and vertex form is already the shape for solving with a squared bracket:

\begin{aligned} 2(x - 1)^2 - 8 &= 0 \\ (x - 1)^2 &= 4 \\ x - 1 &= \pm 2 \end{aligned}

so x = 3 or x = -1, the points found at d = 2.

Draw the curve. Join the points with one smooth curve, symmetric about the axis and rounded at the vertex. Near the vertex it is almost flat, because d^2 is tiny when d is small; far from it the curve is steep, because the rises keep growing.

Why it is true

Why the vertex is where the bracket is zero

Everything follows from one fact: a square is never negative, and it is zero only when the thing squared is zero.

So (x - h)^2 \ge 0 for every x, with equality only when x - h = 0, that is, only at x = h. This is why the bracket shows the vertex with its sign flipped: h is the number that makes x - h zero.

Suppose a > 0. A number that is at least 0, multiplied by a positive a, stays at least 0, so a(x - h)^2 \ge 0. Adding k to both sides,

a(x - h)^2 + k \ge k

with equality only at x = h. The output is never below k and reaches k at exactly one input, so (h, k) is the lowest point. If a < 0, multiplying by a reverses the inequality, so a(x - h)^2 \le 0 and y \le k, again with equality only at x = h, and (h, k) is the highest point.

The symmetry comes from the same square. At x = h + d and x = h - d the bracket is d and -d, and both square to d^2, so both heights are k + ad^2. Points at equal distances either side of x = h are at equal heights, which makes x = h the axis of symmetry.

Why every vertex form is the basic parabola moved

Take any point of the basic parabola and call it (p, p^2).

Look at the new curve at the input x = p + h. The bracket is (p + h) - h = p, so the output is

a p^2 + k

So the point (p + h,\ a p^2 + k) is on the new curve. Its input is the old input plus h, and its output is the old output, p^2, times a, plus k. That is the rule (x, y) \mapsto (x + h,\ ay + k).

Every point of the new curve arises this way. Given any input x, set p = x - h; then the point of the new curve above x is the image of (p, p^2). So the rule does not just send points of y = x^2 onto the new curve, it fills the new curve completely.

With a = 1 and h = 0 the rule is (p, p^2) \mapsto (p, p^2 + k), a pure vertical shift. With a = 1 and k = 0 it is (p, p^2) \mapsto (p + h, p^2), a pure horizontal shift. With h = 0 and k = 0 it is (p, p^2) \mapsto (p, a p^2), a pure stretch. The general form is these three done together.

The order inside the output matters. The formula computes a \cdot p^2 first and adds k after, so the stretch comes before the vertical shift. Shifting first and stretching second would give a(p^2 + k) = a p^2 + ak, a different height whenever a \neq 1 and k \neq 0. The horizontal shift acts on the input and the others on the output, so it can be done at any point in the list.

Why converting always works

Multiplying out gives

a(x - h)^2 + k = ax^2 - 2ahx + (ah^2 + k)

so every vertex form is a standard form with b = -2ah and c = ah^2 + k. Going the other way, for any a \neq 0, factoring a out of the x terms and completing the square gives

ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}

which is vertex form with \displaystyle h = -\frac{b}{2a}, the vertex formula, and \displaystyle k = c - \frac{b^2}{4a}. Each step of the conversion added and subtracted the same amount, or factored and multiplied back by the same number, so the value never changed and the two sides agree for every x.

Worked examples

Example 1

Reading the vertex

Find the vertex of y = (x - 4)^2 + 1, say which way the parabola opens, and give its least or greatest value.

Solution

Match it to a(x - h)^2 + k. The bracket is x - 4, so h = 4. The number added is 1, so k = 1. There is no number in front of the bracket, so a = 1.

The vertex is (4, 1). Since a = 1 > 0 the parabola opens upward, and its least value is 1, at x = 4.

Check. At x = 4 the bracket is 0 and y = 1. At x = 3 and x = 5, one unit either side, y = 1 + 1 = 2: higher on both sides, as a lowest point should be.

Example 2

A plus sign inside, and a negative a

Find the vertex of y = -2(x + 3)^2 + 5, and its greatest or least value.

Solution

The form wants a minus inside the bracket, so write the plus as subtracting a negative: x + 3 = x - (-3). So h = -3, not 3. The quicker way to see it is to ask where the bracket is zero: x + 3 = 0 at x = -3. The number outside is +5, so k = 5.

The vertex is (-3, 5). Here a = -2 < 0, so the parabola opens downward and 5 is its greatest value, reached at x = -3. Since |a| = 2, the curve is narrower than y = x^2: one unit either side of the vertex it is already 2 below it, at height 5 - 2 = 3.

Example 3

A missing piece, and fractions

Find the vertex of each parabola.

(a) y = 3x^2 - 7 \qquad (b) y = -(x - 6)^2 \qquad (c) \displaystyle y = \frac{1}{2}\left(x + \frac{3}{2}\right)^2 - 4

Solution

When a piece of the form is missing, its value is 0.

(a) There is no bracket, but x^2 = (x - 0)^2, so h = 0 and k = -7. The vertex is (0, -7), on the y-axis, and it is a lowest point since a = 3 > 0.

(b) Nothing is added after the bracket, so k = 0, and h = 6. The vertex is (6, 0), on the x-axis. The minus in front is a = -1, so the parabola opens downward and touches the x-axis only at its highest point.

(c) The bracket is zero at \displaystyle x = -\frac{3}{2}, so \displaystyle h = -\frac{3}{2}, and k = -4. The vertex is \displaystyle \left(-\frac{3}{2}, -4\right). The \displaystyle \frac{1}{2} in front makes this parabola wider than y = x^2, but a fractional a changes nothing about where the vertex is.

Example 4

The form in disguise

Find the vertex of (a) y = 5 - (x + 1)^2 and (b) y = 2(3 - x)^2 + 1.

Solution

Neither is written in the order a(x - h)^2 + k, so rearrange first.

(a) Addition can be done in either order, so y = -(x + 1)^2 + 5. Now a = -1, h = -1 and k = 5. The vertex is (-1, 5), a highest point.

(b) Here the x in the bracket has a minus in front of it. But 3 - x = -(x - 3), and squaring removes that minus:

(3 - x)^2 = \big(-(x - 3)\big)^2 = (x - 3)^2

So y = 2(x - 3)^2 + 1, and the vertex is (3, 1), a lowest point. A habit of “flip the sign of the number in the bracket” would have given -3, which is wrong. The test that never fails is the one from Example 2: the vertex is where the bracket is zero, and 3 - x = 0 at x = 3.

Example 5

What the vertex tells you about the rest

For y = -3(x - 1)^2 + 12, give the axis of symmetry and the range, and find the x-intercepts.

Solution

The vertex is (1, 12), so the axis of symmetry is the line x = 1. With a = -3 < 0 the parabola opens downward, so 12 is its greatest output and the range is y \le 12.

The top of the curve is above the x-axis and the curve opens downward, so it must cross the x-axis twice, once on each side of x = 1. To find where, set y = 0. Vertex form is already the shape for solving with a squared bracket:

\begin{aligned} -3(x - 1)^2 + 12 &= 0 \\ (x - 1)^2 &= 4 \\ x - 1 &= \pm 2 \end{aligned}

So x = 3 or x = -1. They sit 2 either side of the axis x = 1, as the symmetry says they must.

Example 6

One move at a time

Describe each graph as a transformation of y = x^2, and give its vertex.

(a) y = x^2 - 5 \qquad (b) y = (x + 4)^2

Solution

(a) The -5 is outside the square, so it changes every output: a shift 5 down. The vertex goes from (0, 0) to (0, -5).

(b) The +4 is inside, so it changes the input. Written as x - h, it is x - (-4), so h = -4: a shift 4 to the left. The vertex is (-4, 0), where the bracket is zero.

(c) Every height is multiplied by 3. The curve is stretched vertically by a factor of 3, so it is narrower. The vertex stays at (0, 0).

(d) Here a = -1, so every height becomes its opposite: a reflection in the x-axis. The curve opens downward, and the vertex stays at (0, 0).

Example 7

Writing the equation from the moves

The graph of y = x^2 is reflected in the x-axis, stretched vertically by a factor of 3, moved 2 units left and 5 units up. Write the equation of the new parabola.

Solution

Each move fixes one number. The reflection and the stretch together make a = -3. Moving left 2 means h = -2, so the bracket is x - (-2) = x + 2. Moving up 5 means k = 5:

y = -3(x + 2)^2 + 5

Check. The vertex should be (0, 0) moved to (-2, 5), and it is. The point (1, 1) should go to (1 - 2,\ -3 \cdot 1 + 5) = (-1, 2), and -3(-1 + 2)^2 + 5 = 2.

Example 8

The order of the moves matters

Start from y = x^2. Find the equation and the vertex after (a) stretching by 2 and then moving up 3, and (b) moving up 3 and then stretching by 2.

Solution

(a) Stretching by 2 multiplies every output: y = 2x^2. Moving up 3 then adds 3 to every output: y = 2x^2 + 3. The vertex is (0, 3).

(b) Moving up 3 first gives y = x^2 + 3. Stretching by 2 now multiplies the whole output, the 3 included:

y = 2(x^2 + 3) = 2x^2 + 6

The vertex is (0, 6). The stretch doubled the height of the vertex as well, because by then the vertex was no longer at height 0.

The two answers differ, so a list of moves has to say its order. The form a(x - h)^2 + k always means stretch first, then shift up. A horizontal shift does not interfere: stretching y = (x - 4)^2 by 2 gives 2(x - 4)^2, the same as shifting y = 2x^2 right by 4.

Example 9

Drawing a parabola by moving points

Draw y = -2(x + 1)^2 + 3 by moving points of y = x^2.

Solution

Read the numbers: a = -2, and x + 1 = x - (-1), so h = -1, and k = 3. The rule is

(x,\ y) \;\mapsto\; (x - 1,\ -2y + 3)

Apply it to five points of the basic parabola. Each input loses 1; each output is doubled, negated, then raised by 3:

(0, 0) \mapsto (-1, 3), the new vertex
(1, 1) \mapsto (0, 1) and (-1, 1) \mapsto (-2, 1)
(2, 4) \mapsto (1, -5) and (-2, 4) \mapsto (-3, -5)

Figure 7Five points of y = x^2 carried onto y = -2(x + 1)^2 + 3.

Plot the five new points and join them with a curve opening downward. In words: y = x^2 stretched by 2, flipped over the x-axis, moved 1 left and 3 up.

Check. Put x = 1 into the equation: -2(1 + 1)^2 + 3 = -2(4) + 3 = -5, matching (1, -5).

Example 10

Graphing with a leading coefficient of one

Graph y = (x + 2)^2 - 1.

Solution

Read the form through its minus sign: (x + 2)^2 = \big(x - (-2)\big)^2, so h = -2. The number added is k = -1, and there is no number in front, so a = 1. The vertex is (-2, -1), the axis is x = -2, and the parabola opens upward.

With a = 1 the steps are those of y = x^2: up 1, then up 3. One unit either side of the vertex, the height is -1 + 1 = 0, giving (-3, 0) and (-1, 0). Two units either side it is -1 + 4 = 3, giving (-4, 3) and (0, 3).

The points at d = 1 are on the x-axis, so they are the x-intercepts, and (0, 3) is the y-intercept. Check one: (0 + 2)^2 - 1 = 4 - 1 = 3.

Example 11

Graphing a stretch

Graph y = 3(x - 2)^2 + 1.

Solution

Here h = 2, k = 1 and a = 3. The vertex is (2, 1), the axis is x = 2, and the parabola opens upward.

The rises are 3 and then 9. One unit either side the height is 1 + 3 = 4, at (1, 4) and (3, 4); two units either side it is 1 + 12 = 13, at (0, 13) and (4, 13). The last of these is the y-intercept.

The curve climbs three times as fast as y = x^2, so it is narrower. Its lowest point is 1 unit above the x-axis and it opens upward, so it never reaches the axis: there are no x-intercepts. The algebra agrees, since 3(x - 2)^2 + 1 = 0 would need \displaystyle (x - 2)^2 = -\frac{1}{3}, and no square is negative.

With heights up to 13 from points only 2 apart, a graph on equal scales is tall and thin. A vertical scale of 1 square for 2 units keeps it on the page without changing any point.

Example 12

Graphing with a negative fraction

Graph \displaystyle y = -\frac{1}{2}(x + 1)^2 + 8.

Solution

Here h = -1, k = 8 and \displaystyle a = -\frac{1}{2}. The vertex is (-1, 8) and, since a < 0, it is the highest point. The parabola opens downward and is wider than y = x^2.

Steps of 1 would give heights \displaystyle 8 - \frac{1}{2} = \frac{15}{2}, then 8 - 2 = 6. Steps of 2 avoid the halves, because a \cdot 2^2 = -2 is a whole number. The falls are then 2, then 3 \times 2 = 6:

\begin{aligned} d = 2: &\quad 8 - \tfrac{1}{2}(4) = 6 \\ d = 4: &\quad 8 - \tfrac{1}{2}(16) = 0 \end{aligned}

That gives (-3, 6) and (1, 6), then (-5, 0) and (3, 0), which are the x-intercepts. The y-intercept is the point at d = 1 to the right, \displaystyle \left(0, \frac{15}{2}\right).

Figure 8With a = -\tfrac{1}{2}, steps of 2 keep every height a whole number.

Example 13

Intercepts that miss the steps

Graph y = -2(x - 1)^2 + 6, and find its x-intercepts exactly.

Solution

The vertex is (1, 6), the highest point, since a = -2 < 0. The falls are 2, then 6. One unit either side the height is 6 - 2 = 4, at (0, 4) and (2, 4); two units either side it is 6 - 8 = -2, at (-1, -2) and (3, -2).

Between d = 1 and d = 2 the height drops from 4 to -2, so each arm crosses the x-axis somewhere in between. To find where, set y = 0:

\begin{aligned} -2(x - 1)^2 + 6 &= 0 \\ (x - 1)^2 &= 3 \\ x &= 1 \pm \sqrt{3} \end{aligned}

That is about 2.73 and -0.73, which is 1.73 either side of the axis, between 1 and 2 as the steps predicted.

Example 14

Converting with a leading coefficient of one

Write f(x) = x^2 + 10x + 21 in vertex form, and give its vertex and axis of symmetry.

Solution

The leading coefficient is already 1, so there is nothing to factor out. Half of 10 is 5, and 5^2 = 25. Add and subtract 25 beside the x terms:

\begin{aligned} f(x) &= x^2 + 10x + 25 - 25 + 21 \\ &= (x + 5)^2 - 4 \end{aligned}

To read h, match the form exactly: x + 5 = x - (-5), so h = -5, not 5. With k = -4, the vertex is (-5, -4) and the axis of symmetry is the line x = -5.

Check with k = f(h): f(-5) = 25 - 50 + 21 = -4.

Example 15

Converting with an odd coefficient of x

Write f(x) = x^2 - 3x + 1 in vertex form.

Solution

Half of -3 is \displaystyle -\frac{3}{2}, and \displaystyle \left(-\frac{3}{2}\right)^2 = \frac{9}{4}. Add and subtract it, then combine the two constants over a common denominator, writing 1 as \displaystyle \frac{4}{4}:

\begin{aligned} f(x) &= x^2 - 3x + \frac{9}{4} - \frac{9}{4} + 1 \\[6pt] &= \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} + \frac{4}{4} \\[6pt] &= \left(x - \frac{3}{2}\right)^2 - \frac{5}{4} \end{aligned}

The vertex is \displaystyle \left(\frac{3}{2}, -\frac{5}{4}\right). A fraction here is normal: the vertex sits halfway between any two points of the curve at the same height, and a halfway point need not be a whole number.

Check. \displaystyle f\left(\frac{3}{2}\right) = \frac{9}{4} - \frac{9}{2} + 1 = \frac{9}{4} - \frac{18}{4} + \frac{4}{4} = -\frac{5}{4}.

Example 16

Converting with a negative leading coefficient

Write f(x) = -2x^2 + 8x - 5 in vertex form. Find its vertex, its largest value and its range, and sketch it.

Solution

Factor -2, sign included, out of the x terms. Dividing 8x by -2 gives -4x:

f(x) = -2(x^2 - 4x) - 5

Half of -4 is -2, whose square is 4. Add and subtract 4 inside, then bring the -4 out. It is multiplied by -2 and leaves as +8:

\begin{aligned} f(x) &= -2(x^2 - 4x + 4 - 4) - 5 \\ &= -2(x - 2)^2 + 8 - 5 \\ &= -2(x - 2)^2 + 3 \end{aligned}

The vertex is (2, 3). Now (x - 2)^2 \ge 0, and multiplying by -2 turns that round, so -2(x - 2)^2 \le 0 and f(x) \le 3, with equality only at x = 2. The parabola opens downward, its largest value is 3, and its range is y \le 3.

For the sketch, the y-intercept is f(0) = -5. It equals the constant term c = -5 of the original, as it must, which is a free check on the conversion. The axis x = 2 is a mirror, and x = 0 is 2 to its left, so the point 2 to its right, (4, -5), is on the curve as well.

Figure 9f(x) = -2(x - 2)^2 + 3. The highest point is (2, 3), and (0, -5) has a twin at (4, -5).

Check. -2(x^2 - 4x + 4) + 3 is -2x^2 + 8x - 8 + 3, which is -2x^2 + 8x - 5.

Example 17

When the leading coefficient does not divide b

Write f(x) = 2x^2 + 6x + 1 in vertex form.

Solution

Factoring 2 out of 2x^2 + 6x leaves an odd coefficient inside the brackets, so fractions appear:

f(x) = 2(x^2 + 3x) + 1

Half of 3 is \displaystyle \frac{3}{2}, and \displaystyle \left(\frac{3}{2}\right)^2 = \frac{9}{4}. The subtracted \displaystyle \frac{9}{4} comes out multiplied by 2, as \displaystyle \frac{9}{2}:

\begin{aligned} f(x) &= 2\left(x^2 + 3x + \frac{9}{4} - \frac{9}{4}\right) + 1 \\[6pt] &= 2\left(x + \frac{3}{2}\right)^2 - \frac{9}{2} + 1 \\[6pt] &= 2\left(x + \frac{3}{2}\right)^2 - \frac{7}{2} \end{aligned}

The vertex is \displaystyle \left(-\frac{3}{2}, -\frac{7}{2}\right).

Check with the shortcut:

\begin{aligned} h &= -\frac{6}{2 \cdot 2} = -\frac{3}{2} \\[6pt] f\left(-\frac{3}{2}\right) &= 2 \cdot \frac{9}{4} - 9 + 1 = \frac{9}{2} - 8 = -\frac{7}{2} \end{aligned}

Example 18

Converting from factored form

Write g(x) = -(x + 1)(x - 5) in vertex form.

Solution

Completing the square works on ax^2 + bx + c, so expand first:

g(x) = -(x^2 - 4x - 5) = -x^2 + 4x + 5

Factor -1 out of the x terms, add and subtract 4 inside, and bring the -4 out, where it becomes +4:

\begin{aligned} g(x) &= -(x^2 - 4x) + 5 \\ &= -(x^2 - 4x + 4 - 4) + 5 \\ &= -(x - 2)^2 + 4 + 5 \\ &= -(x - 2)^2 + 9 \end{aligned}

The vertex is (2, 9). The factored form gives a check for free: its roots are -1 and 5, and the axis of symmetry passes halfway between them, at \displaystyle \frac{-1 + 5}{2} = 2.

Example 19

Converting to find the x-intercepts

Find the vertex, the range and the x-intercepts of f(x) = 2x^2 - 8x + 1.

Solution

\begin{aligned} f(x) &= 2(x^2 - 4x) + 1 \\ &= 2(x^2 - 4x + 4 - 4) + 1 \\ &= 2(x - 2)^2 - 7 \end{aligned}

The vertex is (2, -7), and since a = 2 > 0 the range is y \ge -7.

The x-intercepts are where f(x) = 0. In vertex form x appears only once, so the equation can be undone one step at a time:

\begin{aligned} 2(x - 2)^2 - 7 &= 0 \\[6pt] (x - 2)^2 &= \frac{7}{2} \\[6pt] x - 2 &= \pm\frac{\sqrt{14}}{2} \\[6pt] x &= \frac{4 \pm \sqrt{14}}{2} \end{aligned}

The radical was tidied as \displaystyle \sqrt{\frac{7}{2}} = \sqrt{\frac{14}{4}} = \frac{\sqrt{14}}{2}. The two intercepts sit the same distance, \displaystyle \frac{\sqrt{14}}{2}, either side of the axis x = 2, as the mirror says they must.

Example 20

From standard form to moves

Describe \displaystyle y = \frac{1}{2}x^2 - 3x + 1 as a transformation of y = x^2.

Solution

The moves are visible only in vertex form, so convert first. Take the \displaystyle \frac{1}{2} out of the x terms, which leaves -6x inside, since \displaystyle \frac{1}{2} \cdot (-6x) = -3x. Completing the square inside means adding 9 there, and because the bracket is multiplied by \displaystyle \frac{1}{2} that really adds \displaystyle \frac{9}{2}, so subtract \displaystyle \frac{9}{2} outside:

\begin{aligned} y &= \tfrac{1}{2}(x^2 - 6x) + 1 \\ &= \tfrac{1}{2}(x^2 - 6x + 9) - \tfrac{9}{2} + 1 \\ &= \tfrac{1}{2}(x - 3)^2 - \tfrac{7}{2} \end{aligned}

Now read it: compressed vertically by \displaystyle \frac{1}{2}, so wider, then moved 3 right and \displaystyle \frac{7}{2} down. The vertex is \displaystyle \left(3, -\frac{7}{2}\right).

Check. Multiply out: \displaystyle \frac{1}{2}(x^2 - 6x + 9) - \frac{7}{2} is \displaystyle \frac{1}{2}x^2 - 3x + \frac{9}{2} - \frac{7}{2}, which is \displaystyle \frac{1}{2}x^2 - 3x + 1.

Example 21

Writing the equation from the vertex and one point

A parabola has its vertex at (-2, -3) and passes through (0, -1), as drawn below. Write its equation in vertex form.

Figure 10The vertex fixes h and k. One more point fixes a.

Solution

The vertex gives h = -2 and k = -3 at once. Put them in, taking care with the sign inside:

y = a\big(x - (-2)\big)^2 + (-3) = a(x + 2)^2 - 3

Only a is unknown, and one more point is enough to find it. The point (0, -1) is on the curve, so x = 0 and y = -1 must satisfy the equation:

\begin{aligned} -1 &= a(0 + 2)^2 - 3 \\ 2 &= 4a \\ a &= \frac{1}{2} \end{aligned}

The equation is \displaystyle y = \frac{1}{2}(x + 2)^2 - 3.

Check. Multiplied out, \displaystyle \frac{1}{2}(x^2 + 4x + 4) - 3 = \frac{1}{2}x^2 + 2x - 1, and at x = 0 that is -1.

Example 22

Reading the moves off a graph

The parabola below is a transformation of y = x^2. Write its equation.

Figure 11Two steps across from the vertex, the curve has fallen 6. The plain parabola would have risen 4.

Solution

The vertex is (-3, 4), so the curve has been moved 3 left and 4 up: h = -3 and k = 4. The vertex is the highest point, so the curve has also been flipped and a is negative. That leaves the size of a.

Compare one step with the basic parabola. From its vertex, y = x^2 rises 2^2 = 4 when you go 2 across. This curve falls 6 over the same 2 across, to (-1, -2). Every height measured from the vertex has been multiplied by \displaystyle \frac{-6}{4} = -\frac{3}{2}, so \displaystyle a = -\frac{3}{2}:

y = -\frac{3}{2}(x + 3)^2 + 4

In words: y = x^2 stretched by \displaystyle \frac{3}{2}, flipped, moved 3 left and 4 up. This is the previous example’s question answered by comparing steps instead of substituting, and the two methods always agree, because a point d across from the vertex sits ad^2 above or below it.

Check. The other marked point is (-5, -2): \displaystyle -\frac{3}{2}(-5 + 3)^2 + 4 = -\frac{3}{2}(4) + 4 = -2.

Example 23

A squeeze that is really a stretch

Describe y = (2x - 6)^2 + 1 as a transformation of y = x^2.

Solution

The bracket is not x - h, because x is multiplied by 2. Take the 2 out of the bracket first: 2x - 6 = 2(x - 3), and squaring squares the 2 as well:

y = \big(2(x - 3)\big)^2 + 1 = 4(x - 3)^2 + 1

So the graph is y = x^2 stretched vertically by 4, moved 3 right and 1 up, with vertex (3, 1). The 6 in the bracket is not the shift; the 3 is. Asking where the bracket is zero finds it directly: 2x - 6 = 0 at x = 3.

There is a second way to see the same curve. Multiplying the input by 2 squeezes the graph toward the y-axis by a factor of 2, a horizontal compression. For the basic parabola the two descriptions give one curve, since (2x)^2 = 4x^2: squeezing it sideways by 2 and stretching it upward by 4 are the same move.

Check. At x = 4: (8 - 6)^2 + 1 = 5, and 4(4 - 3)^2 + 1 = 5.

Example 24

The highest point of a throw

A ball is thrown straight up from the ground. Its height after t seconds is -5(t - 3)^2 + 45 meters. How high does it go, when, and when does it land?

Solution

The height is in vertex form with t in place of x: a = -5, h = 3 and k = 45. Since a < 0 the vertex is the highest point, so the ball rises to 45 meters, at t = 3 seconds.

It is on the ground when the height is 0:

\begin{aligned} -5(t - 3)^2 + 45 &= 0 \\ (t - 3)^2 &= 9 \\ t - 3 &= \pm 3 \end{aligned}

So t = 0 or t = 6. The t = 0 is the moment of the throw, so the ball lands after 6 seconds. The top of the flight, at t = 3, is exactly halfway through: the symmetry of the parabola, showing up in time.

Example 25

Reading a shape off the steps

The opening under an arch is 12 meters wide at the ground and 9 meters tall at its center. Its height y meters, x meters from one foot, is

y = -\frac{1}{4}(x - 6)^2 + 9 \qquad (0 \le x \le 12)

How wide is the opening at a height of 5 meters, and how tall is it 1 meter in from a foot?

Solution

The vertex is (6, 9): the top of the arch, 6 meters from each foot. With \displaystyle a = -\frac{1}{4}, steps of 2 give whole numbers, since a \cdot 2^2 = -1. The falls from the top are then 1, 3, 5:

9,\quad 8,\quad 5,\quad 0

at 0, 2, 4 and 6 meters from the center. So the arch is 5 meters tall at x = 2 and x = 10, which are 4 meters either side of the center. At a height of 5 meters the opening is 10 - 2 = 8 meters wide.

One meter in from a foot is 5 meters from the center, so d = 5 and the height is

9 - \frac{1}{4}(25) = \frac{11}{4}

that is, 2.75 meters.

Figure 12Stepping out in twos from the top gives heights 9, 8, 5, 0. At height 5 the arch is 8 wide.

Only the part of the parabola with 0 \le x \le 12 is the arch. The rest of the curve dips below the ground and describes nothing.

Example 26

A vertex that must land on the axis

For which values of b does the vertex of y = x^2 + bx + 9 lie on the x-axis?

Solution

Convert with b left as a letter. Half of b is \displaystyle \frac{b}{2}, whose square is \displaystyle \frac{b^2}{4}:

\begin{aligned} y &= x^2 + bx + \frac{b^2}{4} - \frac{b^2}{4} + 9 \\[6pt] &= \left(x + \frac{b}{2}\right)^2 + 9 - \frac{b^2}{4} \end{aligned}

The vertex is \displaystyle \left(-\frac{b}{2},\ 9 - \frac{b^2}{4}\right). It lies on the x-axis when its height is 0:

9 - \frac{b^2}{4} = 0 \quad\Longrightarrow\quad b^2 = 36 \quad\Longrightarrow\quad b = \pm 6

Check. With b = 6 the function is x^2 + 6x + 9 = (x + 3)^2, with vertex (-3, 0); with b = -6 it is (x - 3)^2, with vertex (3, 0). Either way the parabola just touches the x-axis, the case of exactly one root.

Boundaries

The bracket must hold x on its own. y = (2x - 6)^2 + 1 is not in vertex form, and its vertex is not at x = 6. Take the 2 out first, as in Example 23: (2x - 6)^2 = 4(x - 3)^2, so a = 4, not 1, and the vertex is (3, 1). The shift is read from a bare x.

Nothing else may contain x. In y = (x - 1)^2 + 3x the x also appears outside the bracket, so the 1 in the bracket is not the x-coordinate of the vertex. Multiply out and convert before reading anything.

a cannot be 0. With a = 0 the equation is y = k, a horizontal line, which has no vertex. Multiplying every height by 0 flattens the parabola, every step has height 0, and nothing is left to move back.

It is written for parabolas that open up or down. Shifts, vertical stretches and reflections in the x-axis keep the axis vertical; they can never turn a parabola on its side. A parabola opening sideways, such as x = (y - 2)^2 + 1, has the letters in each other’s places. Its vertex is (1, 2), it is the leftmost point of the curve, not a highest or lowest value, and the steps from it are taken up and down and change x, not y. Such a curve is not even a function of x.

Converting rewrites one side only. The conversion changes the look of f(x), never its value, so the completing number is added and subtracted on the same side. Adding it alone gives a different function, whose graph is the old one moved up or down.

Every quadratic has a vertex form, and only one. The steps work for any real a \ne 0, b and c, so the conversion never fails, even when the parabola misses the x-axis: 3x^2 - 6x + 5 = 3(x - 1)^2 + 2 has its vertex at (1, 2) and no real roots. And the form is unique, because expanding a(x - h)^2 + k gives ax^2 - 2ahx + ah^2 + k, so the standard form fixes a, then h, then k.

Converting is a rewrite, not a solution. Vertex form says where the parabola turns. Finding its roots is a separate step, though a short one from there, as Example 19 shows.

Steps give exact points, not the curve. The height is exact at every d, whole or not. The curve drawn between the points comes from knowing the shape: smooth, symmetric, and rounded at the vertex rather than pointed.

The vertex and one other point decide everything. Once (h, k) is fixed, any point off the axis fixes a, because its height is k + ad^2 with d \ne 0. Two parabolas with the same vertex and the same first step are the same parabola.

A left-right flip does nothing to y = x^2, and something to its copies. Reflecting in the y-axis replaces x with -x, and (-x)^2 = x^2, so the basic parabola is its own mirror image. Once it has been moved sideways, the flip shows: y = (x - 2)^2 reflected in the y-axis is y = (-x - 2)^2 = (x + 2)^2, with its vertex moved from (2, 0) to (-2, 0).

Where this goes next

With the vertex in hand, every question about the greatest or least value of a quadratic is answered by reading, and so is its range. The axis splits the graph into the part that falls and the part that rises. In context the same reading answers how high a ball goes or which rectangle holds the most area, which is what maximum and minimum problems are about.

Vertex form is one of three ways to write the same quadratic. Reading a quadratic in its three forms puts standard, vertex and factored side by side, and chooses between them by what a question needs.

The same three moves work on every graph, not only on parabolas. Adding outside and subtracting inside are the shifts, reflections flip across an axis, and stretches scale in either direction. For other curves the horizontal and vertical stretches no longer coincide, and the order they are applied in matters more. The general transformed function writes all of it as one formula. The corner of y = a|x - h| + k is read exactly the same way, and stepping from it works too, except that the rises are all equal, a each time, which is why that graph is a V: see transforming absolute value graphs.

There is a quieter consequence too. Enlarging the whole picture from the origin by a scale factor s > 0 sends each point (p, p^2) of y = x^2 to (sp, sp^2), and \displaystyle sp^2 = \frac{1}{s}(sp)^2, so the image is exactly the curve \displaystyle y = \frac{1}{s}x^2. Taking \displaystyle s = \frac{1}{a} shows that for a > 0 the vertical stretch y = ax^2 is the basic parabola enlarged (or, when a > 1, reduced) by the factor \displaystyle \frac{1}{a}. A negative a adds a reflection, and h and k add a translation. So any two parabolas y = a(x - h)^2 + k are similar, the way any two circles are. They are not congruent unless their values of |a| agree: y = x^2 and y = 2x^2 cannot be laid on each other by sliding and flipping alone. Parabolas differ in size and position, never in shape. This is special to the parabola: stretching a circle vertically gives an ellipse, which is not similar to the circle it came from.

The same conversion, done once in x and once in y, is how the center and radius of a circle are found from an expanded equation. And for the parabola as a curve with a focus, the focus of y = a(x - h)^2 + k sits directly above or below the vertex, at \displaystyle \left(h, k + \frac{1}{4a}\right); see focus and directrix.

Exercises

Reading the vertex

Find the vertex, say whether the parabola opens upward or downward, and give its least or greatest value.

y = (x - 5)^2 + 3
y = (x + 2)^2 - 6
y = -(x - 1)^2 + 4
y = 4(x + 7)^2
\displaystyle y = -\frac{1}{3}x^2 + 9
\displaystyle y = 2\left(x - \frac{5}{2}\right)^2 - \frac{1}{2}

Axis, range and intercepts

Give the axis of symmetry and the range of y = 3(x - 4)^2 - 1.
Give the axis of symmetry and the range of y = -(x + 6)^2 + 10.
For y = 3(x + 2)^2 - 4, find y at x = -1 and at x = -3, and explain why the two answers match.

How many x-intercepts does each parabola have? Find any there are.

y = 2(x - 3)^2 + 5
y = -(x + 1)^2 + 9
y = 7(x - 4)^2

Single moves

Describe each graph as a transformation of y = x^2, and give its vertex.

y = x^2 + 6
y = (x - 5)^2
y = (x + 2)^2
y = 4x^2
\displaystyle y = -\frac{1}{2}x^2

Several moves

Describe each graph as a transformation of y = x^2, and give its vertex.

y = (x + 3)^2 - 4
y = -(x - 1)^2 + 6
y = 3(x + 5)^2 + 2
\displaystyle y = -\frac{1}{4}(x - 2)^2 - 1

Moving points

The point (2, 4) is on y = x^2. Where does it land on y = -3(x - 1)^2 + 5?
Which point of y = x^2 lands on (0, 1) on the graph of y = 2(x + 1)^2 - 1?
Find where (-2, 4), (0, 0) and (2, 4) land on \displaystyle y = \frac{1}{2}(x - 4)^2 - 3.

Step out from the vertex

Give the vertex, say which way the parabola opens, and find the points one and two units either side of the vertex.

y = (x - 3)^2 + 2
y = (x + 1)^2 - 4
y = -(x - 2)^2 + 5
y = 2(x + 3)^2
y = -3(x - 1)^2 + 4
\displaystyle y = \frac{1}{2}x^2 - 2

Graph the parabola

Graph each equation. Mark the vertex, the axis, the y-intercept and its mirror image, and any x-intercepts. Where fewer than five points are marked, add a pair of steps.

y = (x - 2)^2 - 9
y = -(x + 3)^2 + 1
y = 2(x - 1)^2 + 3
\displaystyle y = \frac{1}{4}(x + 2)^2 - 1
y = -2(x - 4)^2 + 8
y = (x + 1)^2 - 3
\displaystyle y = -\frac{1}{3}(x - 3)^2 + 3

Converting with a leading coefficient of one

Write each function in vertex form and give its vertex.

f(x) = x^2 + 6x + 2
f(x) = x^2 - 8x + 19
f(x) = x^2 + 12x + 36
f(x) = x^2 - 5x + 4
f(x) = x^2 + x - 1

Converting with a leading coefficient that is not one

Write each function in vertex form and give its vertex.

f(x) = 2x^2 + 12x + 11
f(x) = 3x^2 - 6x - 2
f(x) = -x^2 - 4x + 1
f(x) = -3x^2 + 18x - 20
f(x) = 4x^2 + 4x + 3
f(x) = 2x^2 - 5x + 1
\displaystyle f(x) = \frac{1}{2}x^2 + 3x + 1

Other starting points

Write f(x) = (x - 2)(x + 6) in vertex form, and check h against the roots.
Write f(x) = 2(x - 1)(x - 4) in vertex form.
Write f(x) = (x + 3)^2 - 2x in vertex form.
Write f(x) = -2(x + 1)^2 + 5 in standard form.

Convert, then graph or describe

Write each equation in vertex form, then give the vertex and the points one and two units either side of it.

y = x^2 + 4x + 1
y = -x^2 + 6x - 5
y = 2x^2 - 4x - 1

Describe each graph as a transformation of y = x^2.

y = x^2 + 8x + 13
y = -2x^2 + 12x - 10

Writing the equation

Write the equation of the parabola in vertex form.

Vertex (3, -2), and a = 4.
Vertex (-5, 1), opening downward, the same shape as y = x^2.
Vertex (1, 2), passing through (3, 10).
Vertex (-4, 6), passing through (-2, 4).
Vertex at the origin, passing through (2, -12).
y = x^2 moved 4 right and 1 down.
y = x^2 stretched vertically by 5, then moved 2 left and 7 down.
y = x^2 reflected in the x-axis, compressed vertically by \displaystyle \frac{1}{3}, then moved 6 right.
y = (x - 2)^2 + 1 moved 3 left and 4 up.

Disguises, order and look-alikes

Find the vertex of y = (4 - x)^2 + 1.
Find the vertex of y = 8 - 2(x + 3)^2, and its greatest value.
Find the vertex of y = (2x + 4)^2 - 1 and the value of a once it is in vertex form, and describe the graph as a vertical stretch and shifts of y = x^2.
Start from y = x^2. Write the equation after (a) moving up 1 and then stretching by 3, and (b) stretching by 3 and then moving up 1. Give both vertices.
Write y = (4x)^2 and \displaystyle y = \left(\frac{x}{3}\right)^2 in the form y = ax^2.

Using vertex form

Find the vertex, the axis of symmetry, the largest or smallest value, and the range of f(x) = -x^2 + 6x - 4.
Find the x-intercepts of f(x) = x^2 - 4x - 1 by first writing it in vertex form.
Show that f(x) = 2x^2 + 4x + 5 has no real x-intercepts.
For which value of c does the vertex of y = x^2 + 8x + c lie on the x-axis?
The parabola y = x^2 + bx + c has its vertex at (3, -2). Find b and c.
A parabola has its vertex at (2, -1) and passes through (3, 2). Without writing its equation, name four more points on it.
For which values of k does the graph of y = -2(x + 1)^2 + k cross the x-axis twice, touch it once, or miss it?
A parabola has its vertex at (-1, 4) and its y-intercept at (0, 2). Find a in y = a(x - h)^2 + k, and find the x-intercepts.

In context

A ball’s height after t seconds is -5(t - 2)^2 + 20 meters. Find its greatest height, when it is reached, and when the ball is next at height 0.
A jet of water leaves a fountain at ground level and follows \displaystyle y = -\frac{1}{2}(x - 4)^2 + 8, where x and y are in meters. How high does it go, where does it land, and how high is it 2 meters from the fountain?

Reasoning

Explain why changing a in y = a(x - 2)^2 + 7 changes the shape of the parabola but never moves its vertex.
Explain why y = (x - 4)^2 is y = x^2 moved 4 to the right and not 4 to the left.
Explain why, on y = x^2, the heights at x = 0, 1, 2, 3, 4 go up by 1, 3, 5, 7, and say what the rises are on y = a(x - h)^2 + k, one unit at a time from the vertex.
Explain why the number in front of the brackets in vertex form is always the leading coefficient of the standard form.
In converting f(x) = x^2 - 6x + 2, explain why x^2 - 6x + 9 + 2 is a different function from f, while x^2 - 6x + 9 - 9 + 2 is the same one.
Show that for any a \ne 0 the vertex of y = ax^2 + bx + c has x-coordinate \displaystyle -\frac{b}{2a}, and explain why its y-coordinate is the value of the function there.
Explain why the vertex and one other point, not on the axis, are enough to fix a parabola of the form y = a(x - h)^2 + k.
Show that y = a(x - h)^2 + k meets the x-axis exactly when k = 0 or a and k have opposite signs, and find the x-intercepts when it does.
Explain why reflecting y = x^2 in the y-axis gives back the same graph, but reflecting y = (x - 2)^2 in the y-axis does not.
The graph of y = a(x - h)^2 + k is reflected in the x-axis. Show that the new equation is y = -a(x - h)^2 - k, and say where the vertex goes.
Explain why no choice of a, h and k makes y = a(x - h)^2 + k a parabola that opens to the right.

Answers

(5, 3), upward, least value 3. The bracket is zero at x = 5, and k = 3.
(-2, -6), upward, least value -6. x + 2 = x - (-2), so h = -2.
(1, 4), downward, greatest value 4. Here a = -1 < 0.
(-7, 0), upward, least value 0. Nothing is added, so k = 0, and the bracket is zero at x = -7.
(0, 9), downward, greatest value 9. x^2 = (x - 0)^2, so h = 0, and \displaystyle a = -\frac{1}{3} < 0.
\displaystyle \left(\frac{5}{2}, -\frac{1}{2}\right), upward, least value \displaystyle -\frac{1}{2}. Here a = 2 > 0.
Axis x = 4; range y \ge -1. The vertex is (4, -1) and a = 3 > 0, so -1 is the least value.
Axis x = -6; range y \le 10. The vertex is (-6, 10) and a = -1 < 0, so 10 is the greatest value.
Both are -1: 3(1)^2 - 4 = -1 and 3(-1)^2 - 4 = -1. The two inputs are each 1 from the axis x = -2, one on each side, so the bracket is 1 and -1 and the squares match.
None. The vertex (3, 5) is above the x-axis and the parabola opens upward, so every output is at least 5.
Two, at x = 2 and x = -4. The vertex (-1, 9) is above the x-axis and the parabola opens downward. Setting y = 0: (x + 1)^2 = 9, so x + 1 = \pm 3.
One, at x = 4. The vertex (4, 0) is on the x-axis, so the parabola touches it there and nowhere else.
Moved 6 up; vertex (0, 6). The 6 is added outside, to every output.
Moved 5 right; vertex (5, 0). The bracket is zero at x = 5.
Moved 2 left; vertex (-2, 0). x + 2 = x - (-2), so h = -2.
Stretched vertically by 4, so narrower; vertex (0, 0).
Reflected in the x-axis and compressed vertically by \displaystyle \frac{1}{2}, so wider and opening downward; vertex (0, 0).
Moved 3 left and 4 down; vertex (-3, -4).
Reflected in the x-axis, moved 1 right and 6 up; vertex (1, 6).
Stretched vertically by 3, moved 5 left and 2 up; vertex (-5, 2).
Reflected in the x-axis, compressed vertically by \displaystyle \frac{1}{4}, moved 2 right and 1 down; vertex (2, -1).
(3, -7). The rule is (x, y) \mapsto (x + 1,\ -3y + 5), so (2, 4) \mapsto (3, -12 + 5). Check: -3(3 - 1)^2 + 5 = -7.
(1, 1). Here h = -1, so the input was 0 - (-1) = 1, and 1^2 = 1. Check: (1, 1) \mapsto (1 - 1,\ 2 \cdot 1 - 1) = (0, 1).
(2, -1), (4, -3) and (6, -1). The rule is \displaystyle (x, y) \mapsto (x + 4,\ \frac{1}{2}y - 3).
Vertex (3, 2), opens upward. One unit either side: (2, 3) and (4, 3). Two units either side: (1, 6) and (5, 6).
Vertex (-1, -4), opens upward. One unit: (-2, -3) and (0, -3). Two units: (-3, 0) and (1, 0).
Vertex (2, 5), opens downward. One unit: (1, 4) and (3, 4). Two units: (0, 1) and (4, 1), since the falls are 1 then 3.
Vertex (-3, 0), opens upward. One unit: (-4, 2) and (-2, 2). Two units: (-5, 8) and (-1, 8). There is no k, so k = 0.
Vertex (1, 4), opens downward. One unit: (0, 1) and (2, 1). Two units: (-1, -8) and (3, -8), since the falls are 3 then 9.
Vertex (0, -2), opens upward. One unit: \displaystyle \left(-1, -\frac{3}{2}\right) and \displaystyle \left(1, -\frac{3}{2}\right). Two units: (-2, 0) and (2, 0). There is no bracket, so h = 0.
Vertex (2, -9) on the axis x = 2; y-intercept (0, -5) and its mirror (4, -5); x-intercepts (-1, 0) and (5, 0), from (x - 2)^2 = 9.
Vertex (-3, 1) on the axis x = -3, opening downward; y-intercept (0, -8) and its mirror (-6, -8); x-intercepts (-4, 0) and (-2, 0), from (x + 3)^2 = 1.
Vertex (1, 3) on the axis x = 1; y-intercept (0, 5) and its mirror (2, 5); no x-intercepts, since the lowest point is above the axis. Extra points: (-1, 11) and (3, 11).
Vertex (-2, -1) on the axis x = -2; y-intercept (0, 0) and its mirror (-4, 0), which are also the x-intercepts. Extra points, four units either side: (-6, 3) and (2, 3).
Vertex (4, 8) on the axis x = 4, opening downward; y-intercept (0, -24) and its mirror (8, -24); x-intercepts (2, 0) and (6, 0), from (x - 4)^2 = 4.
Vertex (-1, -3) on the axis x = -1; y-intercept (0, -2) and its mirror (-2, -2); x-intercepts at \displaystyle x = -1 \pm \sqrt{3}, about 0.73 and -2.73.
Vertex (3, 3) on the axis x = 3, opening downward; y-intercept (0, 0) and its mirror (6, 0), which are also the x-intercepts, since steps of 3 give a fall of 3. The next step falls 9, so extra points six units either side are (-3, -9) and (9, -9).
(x + 3)^2 - 7, vertex (-3, -7). Add and subtract 9: (x^2 + 6x + 9) - 9 + 2.
(x - 4)^2 + 3, vertex (4, 3). Add and subtract 16: (x^2 - 8x + 16) - 16 + 19.
(x + 6)^2, vertex (-6, 0). It is already a perfect square, so k = 0 and the vertex is on the x-axis.
\displaystyle \left(x - \frac{5}{2}\right)^2 - \frac{9}{4}, vertex \displaystyle \left(\frac{5}{2}, -\frac{9}{4}\right). Add and subtract \displaystyle \frac{25}{4}, and \displaystyle 4 - \frac{25}{4} = -\frac{9}{4}.
\displaystyle \left(x + \frac{1}{2}\right)^2 - \frac{5}{4}, vertex \displaystyle \left(-\frac{1}{2}, -\frac{5}{4}\right). Add and subtract \displaystyle \frac{1}{4}, and \displaystyle -1 - \frac{1}{4} = -\frac{5}{4}.
2(x + 3)^2 - 7, vertex (-3, -7). 2(x^2 + 6x + 9 - 9) + 11 is 2(x + 3)^2 - 18 + 11.
3(x - 1)^2 - 5, vertex (1, -5). 3(x^2 - 2x + 1 - 1) - 2 = 3(x - 1)^2 - 3 - 2.
-(x + 2)^2 + 5, vertex (-2, 5). -(x^2 + 4x + 4 - 4) + 1 = -(x + 2)^2 + 4 + 1.
-3(x - 3)^2 + 7, vertex (3, 7). -3(x^2 - 6x + 9 - 9) - 20 is -3(x - 3)^2 + 27 - 20.
\displaystyle 4\left(x + \frac{1}{2}\right)^2 + 2, vertex \displaystyle \left(-\frac{1}{2}, 2\right). \displaystyle 4\left(x^2 + x + \frac{1}{4} - \frac{1}{4}\right) + 3 is \displaystyle 4\left(x + \frac{1}{2}\right)^2 - 1 + 3.
\displaystyle 2\left(x - \frac{5}{4}\right)^2 - \frac{17}{8}, vertex \displaystyle \left(\frac{5}{4}, -\frac{17}{8}\right). Inside \displaystyle 2\left(x^2 - \frac{5}{2}x\right), add and subtract \displaystyle \frac{25}{16}; it leaves the brackets as \displaystyle \frac{25}{8}, and \displaystyle 1 - \frac{25}{8} = -\frac{17}{8}.
\displaystyle \frac{1}{2}(x + 3)^2 - \frac{7}{2}, vertex \displaystyle \left(-3, -\frac{7}{2}\right). \displaystyle \frac{1}{2}(x^2 + 6x + 9 - 9) + 1, and the -9 leaves the brackets as \displaystyle -\frac{9}{2}.
(x + 2)^2 - 16, vertex (-2, -16). Expand: x^2 + 4x - 12. The roots 2 and -6 have midpoint -2, which is h.
\displaystyle 2\left(x - \frac{5}{2}\right)^2 - \frac{9}{2}, vertex \displaystyle \left(\frac{5}{2}, -\frac{9}{2}\right). Expand: \displaystyle 2x^2 - 10x + 8 = 2\left(x^2 - 5x + \frac{25}{4} - \frac{25}{4}\right) + 8, and 8 - \tfrac{25}{2} = -\tfrac{9}{2}. The roots 1 and 4 have midpoint \tfrac{5}{2}.
(x + 2)^2 + 5, vertex (-2, 5). Expand: x^2 + 6x + 9 - 2x = x^2 + 4x + 9. The original is not already in vertex form, because of the -2x outside the square.
-2x^2 - 4x + 3. -2(x^2 + 2x + 1) + 5 = -2x^2 - 4x - 2 + 5.
y = (x + 2)^2 - 3. Vertex (-2, -3); one unit either side, (-3, -2) and (-1, -2); two units, (-4, 1) and (0, 1).
y = -(x - 3)^2 + 4. Vertex (3, 4); one unit either side, (2, 3) and (4, 3); two units, (1, 0) and (5, 0). From -(x^2 - 6x + 9) - 5 + 9.
y = 2(x - 1)^2 - 3. Vertex (1, -3); one unit either side, (0, -1) and (2, -1); two units, (-1, 5) and (3, 5). From 2(x^2 - 2x + 1) - 1 - 2.
Moved 4 left and 3 down. Completing the square, x^2 + 8x + 16 - 16 + 13 gives y = (x + 4)^2 - 3.
Reflected in the x-axis, stretched by 2, moved 3 right and 8 up. -2(x^2 - 6x) - 10 = -2(x^2 - 6x + 9) + 18 - 10 gives y = -2(x - 3)^2 + 8.
y = 4(x - 3)^2 - 2.
y = -(x + 5)^2 + 1. The same shape as y = x^2, opening downward, means a = -1.
y = 2(x - 1)^2 + 2. From 10 = a(3 - 1)^2 + 2, so 4a = 8.
\displaystyle y = -\frac{1}{2}(x + 4)^2 + 6. From 4 = a(-2 + 4)^2 + 6, so 4a = -2.
y = -3x^2. With h = k = 0, -12 = a(2)^2, so a = -3.
y = (x - 4)^2 - 1.
y = 5(x + 2)^2 - 7.
\displaystyle y = -\frac{1}{3}(x - 6)^2.
y = (x + 1)^2 + 5. Moving left 3 takes the vertex from x = 2 to x = -1, and moving up 4 takes it from height 1 to height 5.
(4, 1). (4 - x)^2 = (x - 4)^2; equally, the bracket is zero at x = 4.
(-3, 8), greatest value 8. Rearranged, y = -2(x + 3)^2 + 8, and a = -2 < 0.
(-2, -1), with a = 4. The bracket is zero at x = -2, and (2x + 4)^2 = \big(2(x + 2)\big)^2 = 4(x + 2)^2, so y = 4(x + 2)^2 - 1: y = x^2 stretched vertically by 4, moved 2 left and 1 down.
(a) y = 3x^2 + 3, vertex (0, 3), since 3(x^2 + 1) = 3x^2 + 3. (b) y = 3x^2 + 1, vertex (0, 1).
y = 16x^2 and \displaystyle y = \frac{1}{9}x^2. Squaring squares the coefficient: 4^2 = 16 and \displaystyle \left(\frac{1}{3}\right)^2 = \frac{1}{9}.
f(x) = -(x - 3)^2 + 5: vertex (3, 5), axis x = 3, largest value 5 at x = 3, range y \le 5. -(x^2 - 6x + 9 - 9) - 4 = -(x - 3)^2 + 9 - 4.
\displaystyle x = 2 \pm \sqrt{5}. f(x) = (x - 2)^2 - 5, and (x - 2)^2 = 5 gives \displaystyle x - 2 = \pm\sqrt{5}.
f(x) = 2(x + 1)^2 + 3. Since 2(x + 1)^2 \ge 0, f(x) \ge 3 for every x, so f(x) is never 0.
c = 16. y = (x + 4)^2 + c - 16, and the vertex (-4, c - 16) is on the x-axis when c - 16 = 0.
b = -6, c = 7. The standard form starts with x^2, so a = 1 and the vertex form is y = (x - 3)^2 - 2, which expands to x^2 - 6x + 9 - 2 = x^2 - 6x + 7.
For example (1, 2), (0, 11), (4, 11) and (5, 26). The first step rises 3, so a = 3 and the next rises are 9 and 15; (1, 2) is the mirror of (3, 2).
Twice when k > 0, once when k = 0, never when k < 0. The parabola opens downward from the vertex (-1, k), so it reaches the axis only if its highest point is on or above it.
a = -2, and the x-intercepts are at \displaystyle x = -1 \pm \sqrt{2}. The point (0, 2) is one unit from the vertex and 2 lower, so a = -2; then -2(x + 1)^2 + 4 = 0 gives (x + 1)^2 = 2.
The greatest height is 20 meters, at t = 2 seconds, and the ball is next at height 0 at t = 4 seconds. -5(t - 2)^2 + 20 = 0 gives (t - 2)^2 = 4, so t = 0 or t = 4, and t = 0 is the throw.
It rises to 8 meters, 4 meters from the fountain, and lands 8 meters away, since (x - 4)^2 = 16 gives x = 0 or x = 8. At 2 meters out, d = 2 and the height is 8 - 2 = 6 meters.
At x = 2 the bracket is 0, so a(x - 2)^2 = 0 whatever a is, and y = 7. Every parabola in the family passes through (2, 7) and turns there. Everywhere else a multiplies the distance above or below 7, which changes the width and, when a < 0, the direction.
The height y = x^2 has at an input p, the new curve has at the input p + 4, since (p + 4) - 4 = p. Every height arrives 4 units later along the x-axis, which is 4 to the right. The vertex, for instance, is where the bracket is 0, at x = 4, not at x = -4, where the bracket is -8 and y = 64.
The heights are 0, 1, 4, 9, 16, and (d + 1)^2 - d^2 = 2d + 1, which is 1, 3, 5, 7 for d = 0, 1, 2, 3. On y = a(x - h)^2 + k the height at distance d from the vertex is k + ad^2, so each rise is a times the one for y = x^2: a, 3a, 5a, 7a, \dots
Expanding a(x - h)^2 + k gives a(x^2 - 2hx + h^2) + k, which is ax^2 - 2ahx + ah^2 + k. The only x^2 term is ax^2, and two quadratics that agree for every x have the same coefficients, so a must be the leading coefficient of the standard form.
Adding 9 alone changes the value: for every x the new expression is 9 more than f(x), so its graph is the graph of f moved up 9. Adding 9 and subtracting 9 adds 9 - 9 = 0, so the value is the same for every x, and the expression can then be grouped as (x - 3)^2 - 7.
Factor a from the x terms and complete the square: \displaystyle ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}, so \displaystyle h = -\frac{b}{2a}. The squared term is 0 at x = h and nowhere else, so the turning point is at x = h, and there the function equals 0 + k = k. That is, k = f(h).
The vertex gives h and k. A point (x_1, y_1) with x_1 \ne h must satisfy y_1 = a(x_1 - h)^2 + k, and since (x_1 - h)^2 \ne 0 this has exactly one solution, \displaystyle a = \frac{y_1 - k}{(x_1 - h)^2}. With a, h and k known, the equation is fixed.
Setting y = 0 gives \displaystyle (x - h)^2 = -\frac{k}{a}. A square can equal this only if \displaystyle -\frac{k}{a} \ge 0, which happens when k = 0 or a and k have opposite signs. Then \displaystyle x = h \pm \sqrt{-\frac{k}{a}}, one point if k = 0 and two otherwise. In particular, if a and k are both positive there are no x-intercepts: y \ge k > 0 everywhere.
Reflecting in the y-axis replaces x with -x. For y = x^2 that gives (-x)^2 = x^2, the same equation, because the basic parabola is symmetric about the y-axis. For y = (x - 2)^2 it gives (-x - 2)^2 = (x + 2)^2, a different parabola with vertex (-2, 0) instead of (2, 0).
Reflecting in the x-axis replaces every output y with -y, so the new equation is y = -\big(a(x - h)^2 + k\big) = -a(x - h)^2 - k. The vertex goes from (h, k) to (h, -k), and the parabola opens the other way.
Every such equation gives exactly one output for each input, so each vertical line meets the graph once. A parabola opening to the right is met twice by vertical lines to the right of its vertex. The moves available shift, scale and flip heights but never turn the curve, so the axis of symmetry stays vertical.