Algebra 1

The shape of a parabola

Square the input and plot what comes out. Why does the picture bend into the same U every time, and what decides which way the U points?

You need

What makes an expression quadraticAlgebra 1 Graphing a line from a table of valuesMath 8

Unlocks

The vertex and the axis of symmetryAlgebra 1 The trapezoidal rule and Simpson's ruleCalculus 1

Where you are 4 in Quadratic expressions and parabolas

Algebra9

Quadratics12

Quadratic expressions and parabolas4

Graphing from standard form

The shape of a parabola

The vertex and the axis of symmetry What makes an expression quadratic

Why this exists

The graph of a linear equation is a straight line, and two points are enough to draw it. Try the same with y = x^2 and it fails at once. Plot (0, 0) and (2, 4), join them with a ruler, and the line says the point above x = 1 is at height 2. It is at height 1.

So the graph of a quadratic is not a line. It is always the same shape, though: a U, or an upside-down U, the same on both sides and turning exactly once. This lesson explains why squaring produces that shape, and how to tell from the equation alone which way the U points and how narrow it is.

Definition

The graph of an equation

y = ax^2 + bx + c \qquad (a \neq 0)

is called a parabola: the set of all points (x, y) that make the equation true, just as the graph of y = 2x + 1 is.

To see one, make a table and plot it. For y = x^2, take whole numbers from -3 to 3:

\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y = x^2 & 9 & 4 & 1 & 0 & 1 & 4 & 9 \end{array}

Join the seven points with a smooth curve, not straight pieces: the inputs in between have outputs too.

Figure 1Every height above 0 is reached twice, once on each side, at the same distance from the y-axis.

The words that go with the picture:

The parabola opens upward when it is a U, and opens downward when it is an upside-down U.
The point where it turns around is the vertex. On an upward parabola it is the lowest point; on a downward one, the highest.
The vertical line the parabola folds onto, so that one half lands exactly on the other, is its axis of symmetry, or fold line. For y = x^2 it is the y-axis.

Intuition

The two sides match because a number and its negative have the same square. Look along any row of the table: x = 2 and x = -2 both give 4, and x = 3 and x = -3 both give 9. Every point on the right has a partner on the left at the same height and the same distance from the y-axis, which is what the dashed segments in Figure 1 show. A picture made of such pairs is its own mirror image.

The curve bends because the outputs grow faster and faster. Step the input along by 1 each time, from 0 to 1 to 2 to 3. The outputs go 0, 1, 4, 9, so the rises are 1, 3, 5. On a straight line every step across gives the same rise. Here every rise is 2 more than the one before, so the curve gets steeper at every step.

Figure 2Each step across is the same, but each rise is 2 more than the last. That is why the graph bends.

It turns at the bottom because a square is never negative. x^2 is 0 at x = 0 and positive everywhere else, so the graph of y = x^2 touches the x-axis once and lies above it everywhere else. The lowest point is (0, 0), and from there the curve climbs in both directions.

Everything else is a copy of this one. Changing a stretches the U or squashes it, and a negative a flips it over. Adding bx + c slides it without changing its shape, which vertex form makes exact. Every parabola is the U of y = x^2, stretched, perhaps flipped, and moved.

Results

Symmetry. Every parabola has exactly one vertical fold line. Every height except the vertex’s is reached by a pair of points, one on each side of it and the same distance from it. The vertex sits on the line, alone at its height.

Which way it opens. For y = ax^2 + bx + c:

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

Only the sign of a decides this. The signs of b and c never do.

How narrow it is. Of two parabolas y = ax^2, the one with the larger |a| is the narrower: it climbs faster and reaches every height closer to its fold line. y = ax^2 and y = -ax^2 have the same width and are mirror images across the x-axis.

Figure 3A negative coefficient on x^2 turns every output over, so the U points down.

Why it is true

Equal heights come in pairs about one line. When there is no x term this is immediate: y = ax^2 + c gives the same output at x and at -x, because (-x)^2 = x^2, so the fold line is the y-axis. With an x term the pairs are no longer centered on 0, but they still share one center: the axis of symmetry proves that every pair of equal heights has the same midpoint, and finds that midpoint from a and b.

The sign of a decides the direction. For x \neq 0, factor out x^2:

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

As x grows large in either direction, \displaystyle \frac{b}{x} and \displaystyle \frac{c}{x^2} shrink toward 0, so the bracket ends up with the same sign as a, while x^2 is positive and huge. When a > 0 the outputs grow without limit at both ends, which is a U; when a < 0 they fall without limit at both ends, which is an upside-down U.

It turns exactly once. Write f(x) = ax^2 + bx + c and step the input by 1, from n to n + 1. The rise is

f(n + 1) - f(n) = a(2n + 1) + b

and each rise is 2a more than the one before. When a > 0 the rises only ever increase, so the graph can switch from falling to rising once and never back: one turn, at the bottom. When a < 0 the one turn is at the top. (Steps of any size give the same.)

Width. At any input x \neq 0, the graph of y = ax^2 is |a| times as far from the x-axis as the graph of y = x^2. A larger |a| reaches each height sooner, which is what narrower means.

Worked examples

Example 1

Plotting from a table

Make a table of values for y = x^2 - 4 with x from -3 to 3, plot it, and describe the graph.

Solution

Substitute each input and subtract 4 from its square:

\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y = x^2 - 4 & 5 & 0 & -3 & -4 & -3 & 0 & 5 \end{array}

Every output is 4 less than the matching output of y = x^2, so the whole U of Figure 1 has moved down 4 without changing shape.

Figure 4y = x^2 - 4 is the U of y = x^2 moved down 4.

The coefficient of x^2 is 1, which is positive, so the parabola opens upward. The outputs pair up, -3 with 3, -2 with 2, -1 with 1, so the fold line is the y-axis and the vertex is the lowest point, (0, -4). The graph crosses the x-axis where y = 0, at (-2, 0) and (2, 0).

Example 2

Up or down

Decide whether each parabola opens upward or downward.

(a) y = -2x^2 + 5x - 1 \qquad (b) y = 3 - x + 4x^2 \qquad (c) y = (1 - x)(x + 3)

Solution

The direction depends only on a, the coefficient of x^2, so find it in each.

(a) a = -2, which is negative, so the parabola opens downward. The +5x does not change that.

(b) The terms are in a different order, but the coefficient of x^2 is still what matters: a = 4, positive, so it opens upward.

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

Now a = -1, so the parabola opens downward. Without expanding: for large x, 1 - x is negative and x + 3 positive, so the output is large and negative.

Example 3

Narrow and wide

Compare the graphs of y = 3x^2, y = x^2 and \displaystyle y = \frac{1}{2}x^2. Where does each reach the height 12?

Solution

At every input the three outputs are in the ratio \displaystyle 3 : 1 : \frac{1}{2}. At x = 2, for instance, they are 12, 4 and 2. So y = 3x^2 climbs fastest and \displaystyle y = \frac{1}{2}x^2 slowest.

To find where each reaches 12, set it equal to 12 and solve:

3x^2 = 12 \;\Rightarrow\; x^2 = 4 \;\Rightarrow\; x = \pm 2

x^2 = 12 \;\Rightarrow\; x = \pm 2\sqrt{3} \approx \pm 3.46

\tfrac{1}{2}x^2 = 12 \;\Rightarrow\; x^2 = 24 \;\Rightarrow\; x = \pm 2\sqrt{6} \approx \pm 4.90

At height 12 the first U is 4 units wide, the second about 6.9, the third about 9.8. All three open upward and turn at the origin; only the width differs. The graph of y = -3x^2 would have the width of y = 3x^2, flipped.

Figure 5The larger the coefficient of x^2, the faster the curve climbs and the narrower the U.

Example 4

Finishing a table with symmetry

A table for y = x^2 - 6x + 5 begins

\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 5 & 0 & -3 & -4 & -3 \end{array}

Find the fold line and the vertex, then find y at x = 5 and x = 6 without substituting.

Solution

The inputs 2 and 4 give the same output, -3, so they are a mirrored pair, and the fold line passes halfway between them: x = 3.

The vertex sits on the fold line, at (3, -4). It is the lowest point, because a = 1 > 0.

Now use the mirror. x = 5 is 2 to the right of the fold line, and its partner is 2 to the left, x = 1, where y = 0. So y = 0 at x = 5. Likewise x = 6 mirrors x = 0, so y = 5.

Check. 5^2 - 6(5) + 5 = 25 - 30 + 5 = 0 and 6^2 - 6(6) + 5 = 36 - 36 + 5 = 5.

Example 5

The path of a ball

A ball is thrown straight up. Its height after t seconds is h = -5t^2 + 20t meters. Does the graph of h open upward or downward? Find the height at t = 1 and at t = 3, and use them to find when the ball is highest and how high it gets.

Solution

The coefficient of t^2 is -5, so the graph opens downward: the height rises, turns once at a highest point, and falls, as a thrown ball does.

h(1) = -5 + 20 = 15 \qquad h(3) = -45 + 60 = 15

The heights are equal, so t = 1 and t = 3 are a mirrored pair and the fold line is halfway, at t = 2. The turning point, the top of the flight, is on that line:

h(2) = -20 + 40 = 20

The ball is highest at t = 2 seconds, 20 meters up.

Figure 6Equal heights come in pairs, and the top of the arch sits halfway between each pair.

The same mirror says more. The ball leaves the ground at t = 0, which is 2 seconds before the top, so it lands 2 seconds after it, at t = 4. Indeed h(4) = -80 + 80 = 0.

Example 6

When the coefficient is unknown

For which values of k does the graph of y = (k - 2)x^2 + 3x - 1 open downward? What happens when k = 2?

Solution

Here the coefficient of x^2 is k - 2, and the parabola opens downward exactly when that is negative:

k - 2 < 0 \qquad\Longrightarrow\qquad k < 2

When k > 2 the coefficient is positive and it opens upward. When k = 2 the x^2 term vanishes and the equation is y = 3x - 1, a straight line. It is not a parabola at all, so it opens neither way; this is why the definition insists on a \neq 0.

Boundaries

a must not be 0. Without the x^2 term there is nothing to bend the graph, and it is a straight line. Every result above leans on a \neq 0.

The fold line is always vertical. The graph of y = ax^2 + bx + c is mirrored left to right, never top to bottom. A sideways parabola such as x = y^2 exists, but it is not the graph of any equation of this form, since one input would have two outputs. It belongs to the parabola as a conic.

A table shows only the points you chose. Whole-number inputs may miss the vertex. For y = x^2 - x the table gives 0 at both x = 0 and x = 1 and nothing lower; the true lowest point is between them, at \displaystyle \left(\frac{1}{2}, -\frac{1}{4}\right).

Not every U is a parabola. A chain hanging between two hooks makes a symmetric U that turns once, but its rises do not grow evenly, so it is a different curve.

Where this goes next

The axis of symmetry finds the fold line straight from the equation, and the vertex finds the turning point on it. With those two landmarks, graphing from standard form replaces a long table with a handful of well-chosen points.

Where the parabola meets the x-axis, ax^2 + bx + c = 0, which is why roots, zeros and x-intercepts are one idea seen three ways. The arch of the thrown ball is the start of projectile problems.

Further on, the same curve is a slice of a cone, with a reflective property that shapes satellite dishes. In calculus, Simpson’s rule estimates areas by fitting a parabola over each pair of strips, because a parabola bends where a straight line cannot.

Exercises

Tables and plotting

Make a table for y = x^2 + 1 at x = -2, -1, 0, 1, 2, and name the lowest point of its graph.
Make a table for y = -x^2 + 3 at x = -2, -1, 0, 1, 2, and name the highest point of its graph.
Make a table for y = 2x^2 - 1 at x = -2, -1, 0, 1, 2, and name the vertex.
For y = x^2 - 2x, find y at x = -1 and at x = 3. What does the result tell you about the fold line?

Up or down

Decide whether each parabola opens upward or downward.

y = 4x^2 - x + 7
y = -x^2 + 10x
y = 5 + 2x - 3x^2
y = (x - 4)(x + 1)
y = (2 - x)(x + 5)
y = x(6 - x) + 2x^2

Narrow and wide

Put the graphs of y = 2x^2, y = -5x^2 and \displaystyle y = \frac{1}{3}x^2 in order from narrowest to widest.
How are the graphs of y = 4x^2 and y = -4x^2 related? Explain.
How high is \displaystyle y = \frac{1}{3}x^2 at x = 3, and how high is y = 3x^2 there? At which inputs does \displaystyle y = \frac{1}{3}x^2 reach the height 27?

Using symmetry

A parabola passes through (1, 6) and (7, 6). What is its fold line? It also passes through (3, -2); name one more point it must pass through.
A parabola passes through (-1, 8), (0, 3), (1, 0), (2, -1) and (3, 0). Find its fold line and its outputs at x = 4 and x = 5.
A ball’s height after t seconds is h = -5t^2 + 30t meters. Find h at t = 2 and at t = 4, and use them to find when the ball is highest, how high it gets, and when it lands.
For y = x^2 + 6x + 5, show that x = -1 and x = -5 give the same output. Name the fold line and the vertex.
For which values of m does y = (m + 3)x^2 - x + 4 open upward? What is the graph when m = -3?

Reasoning

Explain why y = x^2 never has a negative output, and what that says about where its graph lies.
Use the outputs of y = x^2 at x = 0, 1 and 2 to explain why its graph cannot be a straight line.
Explain why a parabola can never pass through three different points that all have the same height.
On the window from x = 0 to x = 10, the graph of y = 0.01x^2 - 5x looks like a falling straight line. Explain why it must eventually turn and rise, and find where it turns.
Explain why the lowest point of an upward parabola must lie on its fold line.

Answers

Outputs 5, 2, 1, 2, 5; the lowest point is (0, 1). Each output is 1 more than x^2.
Outputs -1, 2, 3, 2, -1; the highest point is (0, 3). The coefficient of x^2 is -1, so the graph opens downward.
Outputs 7, 1, -1, 1, 7; the vertex is (0, -1), the lowest point, since a = 2 > 0.
Both give y = 3: (-1)^2 - 2(-1) = 3 and 3^2 - 2(3) = 3. Equal outputs are a mirrored pair, so the fold line is halfway between, at x = 1.
Upward, since a = 4 > 0.
Downward, since a = -1 < 0.
Downward. The x^2 term is -3x^2, so a = -3.
Upward. Expanding gives x^2 - 3x - 4, so a = 1.
Downward. Expanding gives -x^2 - 3x + 10, so a = -1.
Upward. Expanding gives 6x - x^2 + 2x^2 = x^2 + 6x, so a = 1.
y = -5x^2, then y = 2x^2, then \displaystyle y = \frac{1}{3}x^2. Width depends on the size of the coefficient, \displaystyle 5 > 2 > \frac{1}{3}; the minus sign only flips the graph.
They are mirror images of each other across the x-axis, with the same width. At every input the outputs are equal in size and opposite in sign, so the first opens upward and the second downward.
\displaystyle \frac{1}{3}(3)^2 = 3 and 3(3)^2 = 27. \displaystyle y = \frac{1}{3}x^2 reaches 27 at x = \pm 9, from x^2 = 81: three times as far out as y = 3x^2, which reaches it at x = \pm 3.
The fold line is x = 4, halfway between 1 and 7. The point (5, -2): 3 is 1 left of the line, so its partner is 1 right of it.
The fold line is x = 2, since x = 1 and x = 3 both give 0. Then y = 3 at x = 4 (mirroring x = 0) and y = 8 at x = 5 (mirroring x = -1).
h(2) = 40 and h(4) = 40, so the fold line is t = 3. The ball is highest at t = 3 seconds, h(3) = 45 meters. It leaves the ground at t = 0, 3 seconds before the top, so it lands at t = 6.
Both give 0: 1 - 6 + 5 = 0 and 25 - 30 + 5 = 0. The fold line is x = -3, halfway between, and the vertex is (-3, -4), since 9 - 18 + 5 = -4.
Upward when m > -3, since then m + 3 > 0. When m = -3 the graph is the straight line y = -x + 4.
A square is a number times itself: positive times positive and negative times negative are both positive, and 0 \cdot 0 = 0. So the graph never goes below the x-axis; it touches it only at the origin.
The points are (0, 0), (1, 1) and (2, 4). From the first to the second the rise over one step is 1, from the second to the third it is 3. A straight line rises by the same amount over every equal step, so these three points are not on one line.
Points at equal heights come in mirrored pairs about the one fold line, and each input has exactly one mirror image, so at most two points share any height. Equivalently, ax^2 + bx + c = k is a quadratic equation, with at most two solutions.
The coefficient of x^2 is 0.01 > 0, so the parabola opens upward however small that is: \displaystyle 0.01x^2 - 5x = x^2\left(0.01 - \frac{5}{x}\right), and the bracket is positive once x > 500. The outputs at x = 0 and x = 500 are both 0, so the fold line, and the turn, is at x = 250.
If the lowest point were off the fold line, its mirror image would be a second point at the same lowest height. But an upward parabola turns only once, so it has one lowest point, and that point must be its own mirror image, which only a point on the line is.