Algebra 1 · A.SSE.B.3B · F.IF.C.7A

The vertex and the axis of symmetry

A parabola comes down, turns and goes back up, and its two sides look like mirror images. Where exactly does it turn, where exactly is the mirror, and can you find both from the equation without drawing anything?

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The shape of a parabolaAlgebra 1

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Graphing from standard form The shape of a parabola

The vertex and the axis of symmetry

What makes an expression quadratic

Why this exists

Feed the numbers 0, 1, 2, 3, 4 into x^2 - 4x + 1 and watch what comes out:

\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline x^2 - 4x + 1 & 1 & -2 & -3 & -2 & 1 \end{array}

The outputs fall, stop, and climb again. Somewhere they reach a smallest value, and any question of the form “how high does the ball go”, “what is the largest area”, or “what is the least this can cost” is a question about that one turning point.

Look again and a second pattern shows. The outputs repeat, in pairs: 1 and 3 both give -2, and 0 and 4 both give 1, and every pair is centered on the same number, 2. That is the shape of a parabola showing through: fold the graph along one vertical line and the two halves land on each other. If you know where that line is, you get half of every table for free, you know where the turning point must sit, and you can sketch the whole curve from one side of it.

A table can only guess at either. The smallest output might sit between two of the inputs you happened to try, and the center of the pairs might be a fraction no table lands on. The vertex and the axis of symmetry are the exact answers: the point where the parabola turns and the line it is symmetric about, both found from the coefficients without drawing anything.

Definition

Every parabola y = ax^2 + bx + c, with a \neq 0, has two landmarks: a point where it turns, and a line it is symmetric about. The point lies on the line.

The vertex

The parabola turns around exactly once. The point where it turns is its vertex.

If a > 0 the parabola opens upward, and the vertex is its lowest point.
If a < 0 it opens downward, and the vertex is its highest point.

The vertex is written (h, k). Its first coordinate h is the input where the turn happens. Its second coordinate k is the output there, and that is the least value of the expression when a > 0 and the greatest value when a < 0.

Figure 1When a > 0 the vertex is the lowest point. When a < 0 it is the highest.

The axis of symmetry

The axis of symmetry of a parabola is the line that folds it onto itself: reflect the graph in that line and every point lands on another point of the graph.

For a parabola that is the graph of y = ax^2 + bx + c, with a \ne 0, the axis is a vertical line. A vertical line is named by the one x-coordinate all its points share, so it is written

x = h

and read “the line x equals h”. The axis is a line, not a number: “the axis is x = 3” is the complete statement, and “3” on its own names only where the line crosses the x-axis.

Reflecting in the line x = h sends a point that is d units to the right of the line to the point d units to the left of it at the same height. So the line x = h is an axis of symmetry of y = f(x) exactly when

f(h + d) = f(h - d) \quad\text{for every number } d.

The vertex always lies on the axis of symmetry, and it is the only point of the parabola that does. That is why the same letter h names both the axis x = h and the first coordinate of the vertex (h, k).

Intuition

Go back to the table. The outputs come in matching pairs: x = 1 and x = 3 both give -2, and x = 0 and x = 4 both give 1. Each pair sits the same distance either side of x = 2. That is the axis of symmetry at work: fold the graph along x = 2 and each point lands on its partner.

Figure 2Points at the same height pair up across the line x = 2. The vertex is the one point with no partner.

One point has no partner, because it sits on the fold itself: (2, -3). Every height above -3 is reached twice, once on the way down and once on the way up. The height -3 is reached only once, and nothing lower is reached at all. So the vertex is found in two moves: find the axis, then go up or down it to the curve.

The pairing holds at every height, not only at the ones a table happens to hit. In Figure 3 the parabola is y = x^2 - 10x + 21 and the dashed vertical line is x = 5. Every level segment across the parabola, at whatever height, is cut in half by it.

Figure 3Points at the same height come in pairs, the same distance either side of the line x = 5.

Read the picture as a statement about the equation. Pick any height the parabola reaches above its lowest point, draw the level line there, and it meets the curve twice. Those two inputs give the same output, and they sit the same distance either side of x = 5. Lower the level line and the two points slide inward, always in step, until they meet at the bottom of the curve, which is why the lowest point is on the axis too.

So there are two ways to find the axis. One is a formula in a and b. The other needs no formula: find any two inputs that give the same output and take the number halfway between them. Either way, the vertex is then one substitution away.

Results

The axis of symmetry

The graph of y = ax^2 + bx + c, with a \ne 0, has exactly one vertical axis of symmetry:

x = -\frac{b}{2a}

The constant c plays no part in it.

Equal heights

If two different inputs p and q give the same output, the axis is the line halfway between them:

x = \frac{p + q}{2}

In particular, a parabola with two x-intercepts r and s has its axis at \displaystyle x = \frac{r + s}{2}.

The vertex from standard form

The parabola y = ax^2 + bx + c, with a \neq 0, has its vertex at (h, k), where

h = -\frac{b}{2a} \qquad\text{and}\qquad k = a h^2 + b h + c

That is, h comes from the formula for the axis, and k is what the expression gives at h. Written out in the coefficients, k = c - \dfrac{b^2}{4a}, though substituting h is usually quicker.

Highest or lowest

If a > 0, then k is the least value of ax^2 + bx + c, reached only at x = h. If a < 0, then k is the greatest value, again reached only at x = h.

Other forms

In factored form y = a(x - r)(x - s) the axis is \displaystyle x = \frac{r + s}{2}, so the vertex sits halfway between the roots:

h = \frac{r + s}{2}

In vertex form y = a(x - h)^2 + k the axis is x = h and the vertex is (h, k).

Why it is true

Why the axis is the line \displaystyle x = -\frac{b}{2a}

Write f(x) = ax^2 + bx + c and test a vertical line x = h straight against the definition. Go a distance d either side of h and subtract the two heights. The c cancels at once. The squares differ by (h + d)^2 - (h - d)^2 = 4hd and the inputs by (h + d) - (h - d) = 2d, so

\begin{aligned} f(h + d) - f(h - d) &= a(4hd) + b(2d) \\ &= 2d\,(2ah + b) \end{aligned}

The line x = h is an axis exactly when this is zero for every d. If 2ah + b = 0, it is. If 2ah + b \ne 0, then d = 1 already gives a difference that is not zero. So the line works exactly when 2ah + b = 0, which, since a \ne 0, means

h = -\frac{b}{2a}

That one computation proves both halves of the result: the line \displaystyle x = -\frac{b}{2a} is an axis, and no other vertical line is.

The constant c canceled in the first line. Raising or lowering the whole parabola does not move its axis sideways, which is why c appears nowhere in the formula.

Why two points at the same height are mirror images

Suppose two different inputs p and q give the same output. Then

\begin{aligned} 0 &= f(p) - f(q) \\ &= a(p^2 - q^2) + b(p - q) \\ &= (p - q)\bigl(a(p + q) + b\bigr) \end{aligned}

Since p \ne q, the second factor is zero, so \displaystyle p + q = -\frac{b}{a}, and the midpoint \displaystyle \frac{p + q}{2} is \displaystyle -\frac{b}{2a}, the axis. Two points of a parabola at the same height are always mirror images.

The handiest pair is the y-intercept and its partner. The input 0 gives c, and ax^2 + bx + c = c becomes x(ax + b) = 0, so the input \displaystyle -\frac{b}{a} gives c as well. Halfway between them is \displaystyle -\frac{b}{2a}, a quick way to rebuild the formula if you forget it.

Figure 4The y-intercept always has a partner at the same height. The axis is halfway between them.

Why the vertex is the highest or lowest point

The argument is completing the square with letters in place of numbers. Take a out of the first two terms, then add and subtract the number that completes the square inside the bracket, as in the case with a leading coefficient other than one:

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

The \displaystyle -\frac{b^2}{4a} appears because the \displaystyle \frac{b^2}{4a^2} added inside the bracket is multiplied by a, so \displaystyle a \cdot \frac{b^2}{4a^2} = \frac{b^2}{4a} has to be taken away again.

Now name the two pieces: \displaystyle h = -\frac{b}{2a} and \displaystyle k = c - \frac{b^2}{4a}. The expression is

a(x - h)^2 + k

A square is never negative, so (x - h)^2 \ge 0, and it is 0 only when x = h.

If a > 0, then a(x - h)^2 \ge 0 as well, so the whole expression is at least k, and equals k only at x = h. That makes (h, k) the lowest point. If a < 0, multiplying by a flips the inequality: a(x - h)^2 \le 0, so the expression is at most k, and (h, k) is the highest point.

The same form shows the symmetry a second way. Move a distance d either side of h and the square is d^2 both times, so the outputs at h + d and h - d are equal: x = h is the axis again. The vertex, the one input where the square is 0, sits on it.

The other forms

In factored form a(x - r)(x - s) the inputs r and s both give 0. They are two points at the same height, so they are mirror images across the axis (or the same point, when r = s), and either way the axis, and with it the vertex, is halfway between them.

In vertex form a(x - h)^2 + k the inputs h + d and h - d both give ad^2 + k, because squaring forgets the sign, so x = h is the axis. Since a(x - h)^2 is zero only at x = h, the output there is k, and the turning point is (h, k).

Worked examples

Example 1

The axis from standard form

Find the axis of symmetry of y = x^2 - 8x + 7.

Solution

Read off the coefficients: a = 1 and b = -8. The constant c = 7 plays no part. Substitute into the formula, keeping the sign of b inside the negative:

x = -\frac{b}{2a} = -\frac{-8}{2(1)} = \frac{8}{2} = 4

The axis is the line x = 4.

Check. This one factors, as (x - 1)(x - 7), so its x-intercepts are 1 and 7. The number halfway between them is \displaystyle \frac{1 + 7}{2} = 4, as it should be.

Example 2

A negative leading coefficient and a fraction

Find the axis of symmetry of y = -2x^2 + 6x - 1.

Solution

Here a = -2 and b = 6. Both signs matter:

x = -\frac{6}{2(-2)} = -\frac{6}{-4} = \frac{3}{2}

The axis is \displaystyle x = \frac{3}{2}. A fraction is normal; the axis need not pass through a grid line.

Check. Use the partner of the y-intercept. The input 0 gives -1. The input 3 gives -2(9) + 6(3) - 1 = -18 + 18 - 1 = -1, the same height, and halfway between 0 and 3 is \displaystyle \frac{3}{2}.

Example 3

A missing term, and terms out of order

Find the axis of symmetry of each parabola.

(a) y = 3x^2 - 12 \qquad (b) y = 5x^2 + 20x \qquad (c) y = 4 + x - x^2

Solution

(a) There is no x term, so b = 0 and the axis is \displaystyle x = -\frac{0}{6} = 0, the y-axis. This makes sense directly: 3x^2 - 12 gives the same output for x and -x, since squaring forgets the sign.

(b) Here a = 5 and b = 20, so \displaystyle x = -\frac{20}{10} = -2. With no constant term the x-intercepts are easy to see, since 5x^2 + 20x = 5x(x + 4) is zero at 0 and -4, and halfway between them is -2.

(c) The coefficients are named by the power of x they go with, not by where they stand. The x^2 term is -x^2, so a = -1, and the x term is x, so b = 1:

x = -\frac{1}{2(-1)} = \frac{1}{2}

The axis is \displaystyle x = \frac{1}{2}.

Example 4

A lowest point

Find the axis of symmetry and the vertex of y = x^2 - 6x + 5, and say whether the vertex is a highest or a lowest point.

Solution

Here a = 1, b = -6 and c = 5. The first coordinate comes from the axis:

h = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3

so the axis is the line x = 3. The second coordinate is the value of the expression at x = 3:

k = 3^2 - 6(3) + 5 = 9 - 18 + 5 = -4

The vertex is (3, -4). Since a = 1 > 0, the parabola opens upward, so this is the lowest point: the least value of x^2 - 6x + 5 is -4, at x = 3. It is the left-hand graph in Figure 1.

Check. The expression factors as (x - 1)(x - 5), so its roots are 1 and 5, and halfway between them is \displaystyle \frac{1 + 5}{2} = 3.

Example 5

A highest point, with a not equal to 1

Find the vertex of y = -2x^2 + 8x - 3 and its greatest value.

Solution

Here a = -2, b = 8 and c = -3. The 2a in the formula takes the sign of a with it:

h = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2

Substitute x = 2, squaring before multiplying by -2:

k = -2(2)^2 + 8(2) - 3 = -8 + 16 - 3 = 5

The vertex is (2, 5). Since a = -2 < 0, the parabola opens downward and 5 is the greatest value, reached at x = 2. This is the right-hand graph in Figure 1.

Check. One step either side of the axis should give equal heights: at x = 1 the expression is -2 + 8 - 3 = 3, and at x = 3 it is -18 + 24 - 3 = 3.

Example 6

A vertex at fractions

Find the vertex of y = 2x^2 + 6x - 1.

Solution

h = -\frac{6}{2(2)} = -\frac{6}{4} = -\frac{3}{2}

Substituting a fraction is where arithmetic slips happen, so take it one term at a time. \displaystyle \left(-\frac{3}{2}\right)^2 = \frac{9}{4}, so the first term is \displaystyle 2 \cdot \frac{9}{4} = \frac{9}{2}. The middle term is \displaystyle 6 \cdot \left(-\frac{3}{2}\right) = -9. Then

k = \frac{9}{2} - 9 - 1 = \frac{9}{2} - \frac{20}{2} = -\frac{11}{2}

The vertex is \displaystyle \left(-\frac{3}{2}, -\frac{11}{2}\right), a lowest point since a > 0.

Check. The coefficient form gives the same height: \displaystyle k = c - \frac{b^2}{4a} = -1 - \frac{36}{8} = -1 - \frac{9}{2} = -\frac{11}{2}.

Example 7

Straight from factored form and from vertex form

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

(b) Find the axis of symmetry of y = -(x + 3)(x - 7).

Solution

(a) There is no need to multiply out. The roots are where a factor is zero, x = -3 and x = 5, and the vertex sits halfway between them:

h = \frac{-3 + 5}{2} = 1

Substitute into the factored form, which is quicker than any expanded one:

k = 2(1 + 3)(1 - 5) = 2(4)(-4) = -32

The vertex is (1, -32), and it is the lowest point, since multiplying out would give a leading coefficient of 2. To check, expand: the expression is 2x^2 - 4x - 30, and \displaystyle -\frac{-4}{2(2)} = 1.

(b) The minus sign in front is a leading coefficient of -1. It turns the parabola upside down, but it does not move the intercepts: the product is still zero when x + 3 = 0 or x - 7 = 0, at -3 and 7. The axis is halfway between them:

x = \frac{-3 + 7}{2} = 2

To check, expand: -(x^2 - 4x - 21) = -x^2 + 4x + 21, and \displaystyle -\frac{4}{2(-1)} = 2.

(c) In vertex form a(x - h)^2 + k the axis is x = h. Write x + 5 as x - (-5) to see that h = -5. The axis is x = -5. The sign flips because (x + 5)^2 is smallest where x + 5 = 0, at x = -5. The height there is -1, so the vertex is (-5, -1), a lowest point since a = 2.

Example 8

Choosing a coefficient to place the vertex

Find the value of c for which x^2 - 8x + c has least value 3.

Solution

Changing c slides the parabola up or down but does not move it sideways, because h depends only on a and b:

h = -\frac{-8}{2(1)} = 4

The least value is the height at x = 4:

k = 4^2 - 8(4) + c = 16 - 32 + c = c - 16

Setting c - 16 = 3 gives c = 19. The expression x^2 - 8x + 19 has its vertex at (4, 3).

Check. x^2 - 8x + 19 = (x - 4)^2 + 3, and (x - 4)^2 is never negative, so the least value is 3.

Example 9

Building a parabola from its symmetry

A parabola y = ax^2 + bx + c passes through (-2, 15), (4, 15) and (0, -1). Find its axis of symmetry, name a fourth point on it without any algebra, and then find its equation.

Solution

The points (-2, 15) and (4, 15) are at the same height, so the axis is halfway between them:

x = \frac{-2 + 4}{2} = 1

The point (0, -1) is 1 unit to the left of the axis. Its mirror image is 1 unit to the right, at the same height, so (2, -1) is on the parabola as well.

For the equation, use what each fact says about a, b and c. The point (0, -1) gives c = -1 at once. The axis gives \displaystyle -\frac{b}{2a} = 1, so b = -2a. The point (4, 15) gives

\begin{aligned} 16a + 4b - 1 &= 15 \\ 16a + 4(-2a) &= 16 \\ 8a &= 16 \\ a &= 2 \end{aligned}

and then b = -4. The parabola is y = 2x^2 - 4x - 1.

Check. At x = -2: 8 + 8 - 1 = 15. At x = 4: 32 - 16 - 1 = 15. At x = 2: 8 - 8 - 1 = -1, the point found by symmetry.

Example 10

How long a ball is in the air

A ball is thrown upward from a height of 4 feet. Its height after t seconds is y = -16t^2 + 48t + 4 feet. When is it highest, and how high does it get? It is 36 feet up after 1 second, on the way up. When is it 36 feet up on the way down, and when is it back at 4 feet?

Figure 5The ball’s height over time. The rise and the fall mirror each other about t = 1.5.

Solution

The graph of height against time is a parabola opening downward, since a = -16 < 0, so its highest point is the vertex, on the axis. With a = -16 and b = 48:

t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5

The ball is highest at t = 1.5 seconds. How high it gets is one substitution away:

-16(1.5)^2 + 48(1.5) + 4 = -36 + 72 + 4 = 40

so the vertex is (1.5, 40) and the ball reaches 40 feet. The timing questions need only the axis.

For the second part, use the symmetry instead of solving an equation. The time t = 1 is 0.5 seconds before the axis, so its partner is 0.5 seconds after it, at t = 2. To confirm: -16(4) + 48(2) + 4 = -64 + 96 + 4 = 36.

The same reasoning says the ball is back at its starting height of 4 feet at t = 3, since 0 and 3 are the same distance from 1.5. The rise and the fall take exactly the same time.

Example 11

The largest pen

A farmer has 60 meters of fencing to make a rectangular pen against a long wall, so only three sides need fence (Figure 6). What dimensions give the largest area, and what is that area?

Figure 6Sixty meters of fence make three sides. The wall makes the fourth.

Solution

Call the length of each side meeting the wall x. Those two sides use 2x meters, which leaves 60 - 2x for the side opposite the wall. The area is length times width:

A = x(60 - 2x) = -2x^2 + 60x

This is a quadratic in x with a = -2 < 0, so its graph opens downward and the vertex is the highest point, the largest area:

h = -\frac{60}{2(-2)} = 15

A = -2(15)^2 + 60(15) = -450 + 900 = 450

The sides meeting the wall should be 15 meters, the side opposite the wall 60 - 2(15) = 30 meters, and the area is 450 square meters.

The factored form gives the same h faster: the area is zero when x = 0 or x = 30, and halfway is 15. Note also that x has to lie between 0 and 30 for both lengths to be positive, and 15 does, so the vertex is a pen that can actually be built.

Boundaries

It needs a \neq 0. With a = 0 the expression bx + c is linear, its graph is a straight line, and there is no turning point. Both formulas divide by a and break down in exactly that case. A slanted line has no vertical line of symmetry at all; a level line y = c has every vertical line as one, so no single axis can be named.

Highest or lowest over all inputs, not over some. The vertex gives the greatest or least value when x may be any real number. If x is limited to an interval that does not contain h, the extreme value on that interval sits at one of its ends instead. In Example 11 this did not bite, because 15 lay between 0 and 30; in a problem where the vertex falls outside the allowed inputs, it does.

Collect the terms first. The formulas read a, b and c off ax^2 + bx + c, named by the power of x each goes with, not by where it stands. An expression such as 80x - 2x^2 or (x + 1)^2 + 3x has to be put in that form before a and b can be read, or the numbers you substitute are the wrong ones.

The midpoint of the roots needs real roots, and the formula does not. y = x^2 - 6x + 13 never meets the x-axis, so there is no pair of intercepts to average. It still has an axis, x = 3, and a vertex, (3, 4), because the symmetry has nothing to do with where the x-axis happens to be; \displaystyle h = -\frac{b}{2a} is the way in. Its roots are the complex numbers 3 \pm 2i, whose average is 3 as well. The vertex is on the x-axis exactly when the two roots are the same number.

Equal heights means equal outputs at two different inputs. One point on its own tells you nothing about the axis. Two points at different heights do not either.

Equal heights gives the axis only for a parabola. For other curves two points at the same height need not be mirror images. y = x^3 - x is zero at -1, 0 and 1, and its graph is not symmetric about any vertical line.

Only for parabolas that open up or down. A parabola opening sideways, the graph of x = ay^2 + by + c, has a horizontal axis, \displaystyle y = -\frac{b}{2a}, by the same argument with the letters swapped. It has a vertex, but that vertex is its leftmost or rightmost point, not a highest or lowest value, and the curve is not the graph of a function of x. A tilted parabola has a tilted axis. These are parabolas in other orientations.

Where this goes next

The vertex and the axis are the frame for graphing a quadratic from standard form: mark the axis and the vertex, plot one side, and reflect it; the intercepts finish the sketch.

The proof above rewrote ax^2 + bx + c as a(x - h)^2 + k. That is vertex form, where the vertex and the axis can be read off the equation without any formula, and converting to it is the same completing of the square.

The height k is where the outputs stop, and the axis splits the parabola into the part that falls and the part that rises. Together they settle the range of a quadratic function and where it rises and falls. The vertex is the answer in every maximum and minimum problem, from the highest point of a thrown ball to the largest pen, and to which rectangle of a fixed perimeter has the most area.

Look again at the quadratic formula:

x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}

It is the axis, plus or minus a distance. The two roots are the two points at height zero, one each side of the mirror.

When the parabola is defined by a point and a line, the focus and directrix, the axis is the line through the focus perpendicular to the directrix, and the vertex is the point of the parabola halfway between them.

The same idea, a turning point found exactly, returns in calculus as a critical point, where it works for curves that are not parabolas at all.

Exercises

Opening and turning

Without finding it, say whether the vertex of y = -4x^2 + x + 9 is the highest or the lowest point of the graph, and why.
A parabola passes through (-1, 6) and (5, 6), and its vertex has y-coordinate -3. Find the vertex.
A parabola crosses the x-axis at x = -4 and x = 10. Find the x-coordinate of its vertex.

The axis from standard form

Find the axis of symmetry.

y = x^2 + 6x + 2
y = 3x^2 + 4x - 1
y = -4x^2 - 6x + 7
y = 7 - 5x^2
\displaystyle y = \frac{1}{2}x^2 + 3x - 4

The vertex from standard form

Find the vertex, and say whether it is a highest or a lowest point.

y = x^2 + 8x + 10
y = x^2 - 10x
y = -x^2 + 4x + 1
y = 2x^2 - 12x + 7
y = -3x^2 - 6x + 2
y = 5x^2 - 20
y = x^2 + 3x + 1
y = -2x^2 + 5x
y = 4x^2 - 4x - 3

From factored form and vertex form

Do not multiply out.

Find the vertex of y = (x - 2)(x - 8).
Find the vertex of y = -(x + 1)(x - 5).
Find the vertex of y = 3x(x + 4).
Find the axis of symmetry of y = -3(x + 4)(x - 1).
Find the axis of symmetry of y = x(x + 9).
Find the axis of symmetry of y = 2(x - 5)^2 + 1.
Find the axis of symmetry of y = -(x + 7)^2 - 3.

Using the symmetry

A parabola has axis x = 3 and passes through (-1, 8). Name another point on it.
The parabola y = x^2 - 4x + c has an x-intercept at (-3, 0). Find the other x-intercept, then find c.
Find the value of m for which y = mx^2 + 8x - 1 has axis of symmetry x = 2.
A parabola y = ax^2 + bx + c has axis x = -1 and passes through (0, 3) and (1, 6). Find its equation.
A stone’s height in meters after t seconds is -5t^2 + 30t + 2. At what time is it highest? It is 27 meters up at t = 1. When is it 27 meters up again, and when is it back at its starting height?

Greatest and least values

Find the least value of x^2 - 14x + 50 and the value of x where it occurs.
Find the greatest value of 12 + 4x - x^2 and the value of x where it occurs.
Find c so that x^2 + 6x + c has least value 5.
A stone is thrown upward, and its height after t seconds is -5t^2 + 20t + 2 meters. Find its greatest height and when it is reached.
Two numbers add to 30. Find the largest possible value of their product, and the two numbers that give it.

Reasoning

Points of a parabola at the same height come in pairs, one either side of the axis. Explain why the vertex is the only point without a partner.
A parabola has axis x = h, and (p, q) is a point on it. Show that (2h - p, q) is on it too, and explain what this says when p = h.
A parabola y = ax^2 + bx + c meets the x-axis at only one point, (4, 0). Explain why its axis of symmetry is x = 4.
The parabola y = x^2 + 2x + 5 never meets the x-axis. Explain why it still has an axis of symmetry, and find it.
Use the vertex to explain why y = x^2 - 6x + 10 never meets the x-axis.
Show that the vertex of y = ax^2 + bx + c lies on the y-axis exactly when b = 0.
Show, by substituting \displaystyle h = -\frac{b}{2a}, that the height of the vertex is \displaystyle k = c - \frac{b^2}{4a}.
The parabolas y = x^2 - 4x + c for different values of c all have the vertex at the same x-coordinate. Explain why, and say how the vertex moves as c grows.
Use the quadratic formula to show that when y = ax^2 + bx + c has two real roots, the axis is exactly halfway between them.
Show that y = x^2 - 6x + 1 and y = -2x^2 + 12x + 5 have the same axis of symmetry.

Answers

Highest. The leading coefficient is -4 < 0, so the parabola opens downward and turns at its top.
(2, -3). The two given points are at the same height, so they are mirror images across the axis, which is halfway between them: \displaystyle \frac{-1 + 5}{2} = 2.
x = 3. The vertex is halfway between the roots: \displaystyle \frac{-4 + 10}{2} = 3.
x = -3. Here a = 1, b = 6, so \displaystyle -\frac{6}{2} = -3.
\displaystyle x = -\frac{2}{3}, since \displaystyle -\frac{4}{2(3)} = -\frac{4}{6}.
\displaystyle x = -\frac{3}{4}, since \displaystyle -\frac{-6}{2(-4)} = \frac{6}{-8}.
x = 0. There is no x term, so b = 0.
x = -3, since \displaystyle -\frac{3}{2 \cdot \frac{1}{2}} = -\frac{3}{1}.
(-4, -6), lowest. \displaystyle h = -\frac{8}{2} = -4, and 16 - 32 + 10 = -6.
(5, -25), lowest. \displaystyle h = -\frac{-10}{2} = 5, and 25 - 50 = -25.
(2, 5), highest. \displaystyle h = -\frac{4}{2(-1)} = 2, and -4 + 8 + 1 = 5.
(3, -11), lowest. \displaystyle h = -\frac{-12}{4} = 3, and 18 - 36 + 7 = -11.
(-1, 5), highest. \displaystyle h = -\frac{-6}{2(-3)} = -1, and -3 + 6 + 2 = 5.
(0, -20), lowest. With b = 0, h = 0, and the height there is the constant, -20.
\displaystyle \left(-\frac{3}{2}, -\frac{5}{4}\right), lowest. \displaystyle h = -\frac{3}{2}, and \displaystyle \frac{9}{4} - \frac{9}{2} + 1 = \frac{9 - 18 + 4}{4} = -\frac{5}{4}.
\displaystyle \left(\frac{5}{4}, \frac{25}{8}\right), highest. \displaystyle h = -\frac{5}{2(-2)} = \frac{5}{4}, and \displaystyle -2 \cdot \frac{25}{16} + \frac{25}{4} = -\frac{25}{8} + \frac{50}{8} = \frac{25}{8}.
\displaystyle \left(\frac{1}{2}, -4\right), lowest. \displaystyle h = -\frac{-4}{8} = \frac{1}{2}, and 1 - 2 - 3 = -4.
(5, -9). Halfway between 2 and 8 is 5, and (5 - 2)(5 - 8) = 3(-3) = -9.
(2, 9). Halfway between -1 and 5 is 2, and -(2 + 1)(2 - 5) = -(3)(-3) = 9.
(-2, -12). The roots are 0 and -4, halfway is -2, and 3(-2)(-2 + 4) = -12.
\displaystyle x = -\frac{3}{2}. The intercepts are -4 and 1, and \displaystyle \frac{-4 + 1}{2} = -\frac{3}{2}. The factor -3 stretches and flips the graph but does not move its intercepts.
\displaystyle x = -\frac{9}{2}. The intercepts are 0 and -9.
x = 5. Vertex form with h = 5.
x = -7, since (x + 7)^2 = (x - (-7))^2, so h = -7.
(7, 8). (-1, 8) is 4 units left of x = 3, so its mirror image is 4 units right, at the same height.
The other intercept is (7, 0), and c = -21. The axis is \displaystyle x = -\frac{-4}{2} = 2, and -3 is 5 units left of it, so the partner is 5 units right. Then (-3)^2 - 4(-3) + c = 0 gives 21 + c = 0. Check: x^2 - 4x - 21 = (x + 3)(x - 7).
m = -2, since \displaystyle -\frac{8}{2m} = 2 gives -8 = 4m.
y = x^2 + 2x + 3. The point (0, 3) gives c = 3. The axis gives \displaystyle -\frac{b}{2a} = -1, so b = 2a. The point (1, 6) gives a + 2a + 3 = 6, so a = 1 and b = 2.
Highest at t = 3; 27 meters up again at t = 5; back at 2 meters at t = 6. The axis is \displaystyle t = -\frac{30}{2(-5)} = 3. The time 1 is 2 seconds before it, so its partner is 2 seconds after; the start, t = 0, is 3 seconds before, so its partner is 3 seconds after.
The least value is 1, at x = 7. \displaystyle h = -\frac{-14}{2} = 7, and 49 - 98 + 50 = 1.
The greatest value is 16, at x = 2. In standard form this is -x^2 + 4x + 12, so \displaystyle h = -\frac{4}{2(-1)} = 2, and 12 + 8 - 4 = 16.
c = 14. h = -3, and the height there is 9 - 18 + c = c - 9, so c - 9 = 5.
The greatest height is 22 meters, at t = 2 seconds. \displaystyle h = -\frac{20}{2(-5)} = 2, and -5(4) + 40 + 2 = 22.
The largest product is 225, from 15 and 15. With the numbers x and 30 - x, the product is x(30 - x) = -x^2 + 30x, which opens downward with vertex at x = 15, and 15 \cdot 15 = 225.
A point a distance d to one side of the axis has its partner a distance d to the other side. For the vertex d = 0, so its “partner” is itself. Put another way, every other height the graph reaches is reached once on the way down and once on the way up, but the height at the turn is reached only once.
Let d = p - h, the distance from the axis to p. Then p = h + d and 2h - p = h - d, and the axis gives f(h - d) = f(h + d) = q, so (2h - p, q) is on the parabola. When p = h the two points are the same point: a point on the axis is its own mirror image, which is why the vertex has no partner.
The mirror image of (4, 0) in the axis is on the parabola and at the same height, 0, so it is on the x-axis too. The only such point is (4, 0) itself, so the reflection leaves (4, 0) where it is, and a point a reflection leaves in place lies on the mirror. The axis is x = 4.
x = -1. The axis comes from the shape of the curve, f(-1 + d) = f(-1 - d) = d^2 + 4, not from any crossing with the x-axis, and \displaystyle -\frac{2}{2(1)} = -1.
The vertex is (3, 1): h = 3 and 9 - 18 + 10 = 1. The parabola opens upward, so 1 is its least value, and every point of the graph has y \ge 1. It never reaches y = 0.
The y-axis is the line x = 0, so the vertex is on it exactly when \displaystyle h = -\frac{b}{2a} = 0. Since a \neq 0, that fraction is zero exactly when b = 0.
Substitute \displaystyle x = -\frac{b}{2a}. The first term becomes \displaystyle a \cdot \frac{b^2}{4a^2} = \frac{b^2}{4a}. The middle term becomes \displaystyle -\frac{b^2}{2a}, which is \displaystyle -\frac{2b^2}{4a}. So the height is \displaystyle \frac{b^2}{4a} - \frac{2b^2}{4a} + c = c - \frac{b^2}{4a}.
The x-coordinate is \displaystyle h = -\frac{b}{2a} = 2, which uses only a and b, not c. The height is 4 - 8 + c = c - 4, so as c grows the vertex moves straight up along the line x = 2, one unit for each unit of c.
The roots are \displaystyle -\frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a} and \displaystyle -\frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}. Adding them, the square roots cancel, so their average is \displaystyle -\frac{b}{2a}, the axis.
Both are x = 3. For the first, \displaystyle -\frac{-6}{2} = 3; for the second, \displaystyle -\frac{12}{2(-2)} = 3.