Algebra 1 · F.IF.C.7A

Graphing from standard form

You are handed y = ax^2 + bx + c and asked for its graph. Which few points pin the whole curve down, and how do you find them without filling in a table?

You need

The vertex and the axis of symmetryAlgebra 1

Unlocks

Roots, zeros and x-interceptsAlgebra 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

When you reach for it

You can graph any equation by making a table: pick some values of x, work out y, plot the points and join them, as in graphing from a table. For a parabola that works, but it is blind. You do not know which values of x to pick, so the table can easily miss the turning point, or show one arm and not the other.

An equation in standard form,

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

tells you more than a table does, if you know how to ask it. The sign of a says which way the curve opens. The numbers a and b together say where its axis of symmetry is. The number c is a point on the curve with no work at all. And since every parabola is symmetric, each point you find off the axis gives you a second one free.

Reach for this skill when a quadratic arrives in standard form and you want its graph, or only its key features: where it turns, how high or low it gets, and where it crosses the axes.

The procedure

Work through y = x^2 - 4x - 5 as the running example. Here a = 1, b = -4 and c = -5, each coefficient taken with its sign.

Decide which way it opens. When x is far from 0, the term ax^2 is much larger than bx + c, so it decides the sign of y. If a > 0, y is large and positive far out on both sides: the parabola opens upward and the vertex is its lowest point. If a < 0, it opens downward and the vertex is its highest point. Here a = 1 > 0, so the curve opens upward.

Find the axis of symmetry. The axis is the vertical line

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

and it is worth seeing where that comes from, because the same fact gives you a free point later. At x = 0 the height is c. Which other x gives the same height? Set ax^2 + bx + c = c, so ax^2 + bx = 0, which factors as x(ax + b) = 0. The two answers are x = 0 and \displaystyle x = -\frac{b}{a}. Every height a parabola reaches, other than its vertex’s, it reaches at exactly two points, one on each side of the axis, so these two points are mirror images and the axis runs halfway between them:

x = \frac{0 + \left(-\frac{b}{a}\right)}{2} = -\frac{b}{2a}

For the running example, \displaystyle x = -\frac{-4}{2(1)} = 2. The axis is the line x = 2.

Find the vertex. The vertex lies on the axis, so its x-coordinate is 2. Its y-coordinate is whatever the equation gives at that x:

y = 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9

The vertex is (2, -9).

Find the y-intercept and its mirror image. The y-axis is where x = 0, and there y = c, so the y-intercept is (0, c). Here that is (0, -5). It sits 2 units to the left of the axis, so its mirror image sits 2 units to the right, at (4, -5). This is the point \displaystyle x = -\frac{b}{a} from the axis argument, and you can check it: 4^2 - 4(4) - 5 = 16 - 16 - 5 = -5.

Find the x-intercepts. The x-axis is where y = 0, so solve x^2 - 4x - 5 = 0. These are the roots of the equation, found with any method you have: factoring when it works, the quadratic formula when it does not. Here

x^2 - 4x - 5 = (x - 5)(x + 1) = 0

so x = 5 or x = -1, and the x-intercepts are (5, 0) and (-1, 0). They must be mirror images too, so their average must be the axis: \displaystyle \frac{5 + (-1)}{2} = 2. It is.

A parabola can meet the x-axis twice, once or not at all. When the vertex is on the side of the x-axis the parabola opens away from, as with an upward parabola whose vertex is above the x-axis, there are no x-intercepts, and the discriminant b^2 - 4ac comes out negative to confirm it.

Plot the points and draw the curve. Mark the vertex, the axis as a dashed line, and the pairs of points. Join them with one smooth curve, symmetric about the axis, rounded at the vertex and opening the way a said.

Figure 1Five points found from the equation. The axis cuts each matching pair of points exactly in half.

Five points are usually enough. When there are fewer, because there are no x-intercepts or the y-intercept is the vertex, choose another x, work out its y, and take its mirror image as well.

Worked examples

Example 1

A parabola that opens upward

Graph y = x^2 + 2x - 3.

Solution

Here a = 1, b = 2, c = -3. Since a > 0, the parabola opens upward.

The axis is \displaystyle x = -\frac{b}{2a} = -\frac{2}{2(1)} = -1. Substituting x = -1 gives the height of the vertex:

y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4

so the vertex is (-1, -4), the lowest point on the graph.

The y-intercept is (0, -3). It is 1 unit to the right of the axis, so its mirror image is 1 unit to the left, at (-2, -3).

For the x-intercepts, solve x^2 + 2x - 3 = 0. It factors as (x + 3)(x - 1) = 0, so x = -3 or x = 1, giving (-3, 0) and (1, 0). Their average is -1, the axis, as it should be.

Plot the vertex (-1, -4), the pair (0, -3) and (-2, -3), and the pair (-3, 0) and (1, 0), and draw a smooth upward curve through all five.

Example 2

A parabola that opens downward

Graph y = -x^2 - 2x + 8.

Solution

Here a = -1, b = -2, c = 8. Since a < 0, the parabola opens downward and the vertex will be its highest point.

The axis is

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

Keep the signs with the numbers: both b and a are negative here, and it is easy to lose one of them. The vertex has height

y = -(-1)^2 - 2(-1) + 8 = -1 + 2 + 8 = 9

Note that -(-1)^2 is -1: the square is taken first, and the minus sign in front belongs to a. The vertex is (-1, 9).

The y-intercept is (0, 8), 1 unit right of the axis, so its mirror image is (-2, 8).

For the x-intercepts, solve -x^2 - 2x + 8 = 0. Multiplying both sides by -1 is allowed, since it leaves 0 as 0, and gives x^2 + 2x - 8 = 0, which factors as (x + 4)(x - 2) = 0. So the x-intercepts are (-4, 0) and (2, 0), and their average is -1.

Figure 2With a < 0 the vertex is the highest point, and the arms fall on both sides.

Example 3

An axis at a fraction

Graph y = x^2 - 3x - 4.

Solution

With a = 1 and b = -3, the axis is \displaystyle x = -\frac{-3}{2(1)} = \frac{3}{2}. An odd b with a = 1 always puts the axis halfway between two whole numbers, and nothing about the method changes. The vertex has height

y = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) - 4 = \frac{9}{4} - \frac{18}{4} - \frac{16}{4} = -\frac{25}{4}

so the vertex is \displaystyle \left(\frac{3}{2}, -\frac{25}{4}\right), which is (1.5, -6.25) for plotting.

The y-intercept (0, -4) is \displaystyle \frac{3}{2} to the left of the axis, so its mirror image is \displaystyle \frac{3}{2} to the right, at (3, -4). As a check, 3^2 - 3(3) - 4 = -4.

The equation x^2 - 3x - 4 = 0 factors as (x - 4)(x + 1) = 0, so the x-intercepts are (4, 0) and (-1, 0), whose average is \displaystyle \frac{3}{2}. The graph opens upward through these five points, with its lowest point between x = 1 and x = 2.

Example 4

A wider parabola with irrational intercepts

Graph \displaystyle y = \frac{1}{2}x^2 - 4x + 5.

Solution

Here \displaystyle a = \frac{1}{2}, b = -4, c = 5. The parabola opens upward, and since \displaystyle a = \frac{1}{2} it rises only half as fast as y = x^2 does: it is wider.

The axis is

x = -\frac{-4}{2\left(\frac{1}{2}\right)} = -\frac{-4}{1} = 4

and the vertex has height \displaystyle \frac{1}{2}(4)^2 - 4(4) + 5 = 8 - 16 + 5 = -3. The vertex is (4, -3).

The y-intercept is (0, 5), and its mirror image across x = 4 is (8, 5).

For the x-intercepts, solve \displaystyle \frac{1}{2}x^2 - 4x + 5 = 0. Multiplying both sides by 2 clears the fraction without changing the solutions: x^2 - 8x + 10 = 0. This does not factor over the integers, since b^2 - 4ac = 64 - 40 = 24 is not a perfect square, so use the quadratic formula:

x = \frac{8 \pm \sqrt{24}}{2} = \frac{8 \pm 2\sqrt{6}}{2} = 4 \pm \sqrt{6}

The exact intercepts are \displaystyle (4 + \sqrt{6}, 0) and \displaystyle (4 - \sqrt{6}, 0). To place them on a graph, use \displaystyle \sqrt{6} \approx 2.45: they are at about x \approx 6.45 and x \approx 1.55, the same distance either side of 4.

Figure 3The x-intercepts are irrational. Their decimal values are only for placing the points.

Example 5

A parabola that never meets the x-axis

Graph y = x^2 - 2x + 4.

Solution

The parabola opens upward. The axis is \displaystyle x = -\frac{-2}{2(1)} = 1, and the vertex has height 1 - 2 + 4 = 3, so the vertex is (1, 3).

The lowest point of the whole graph is 3 units above the x-axis, so no point of the graph reaches the axis: there are no x-intercepts. The discriminant agrees, since b^2 - 4ac = 4 - 16 = -12 < 0.

The y-intercept is (0, 4), and its mirror image is (2, 4). That makes only three points, all bunched near the vertex, which is not enough to show the shape. So choose another value of x away from the axis, say x = -1:

y = (-1)^2 - 2(-1) + 4 = 1 + 2 + 4 = 7

That gives (-1, 7), which is 2 units left of the axis, and its mirror image (3, 7) comes free.

Figure 4No x-intercepts, so a pair of extra points does their job.

Example 6

Reading a situation off the graph

A ball is thrown upward from the top of a building 25 meters tall. Its height above the ground, in meters, t seconds later is

h = -5t^2 + 20t + 25

Graph h against t for as long as the ball is in the air, and use the graph to find how high the ball goes and when it lands.

Solution

The variables have new names, but this is standard form with a = -5, b = 20, c = 25, and every step works as before with t in place of x and h in place of y.

Since a < 0, the graph opens downward, and its vertex is the highest point: the top of the ball’s flight. The axis is

t = -\frac{20}{2(-5)} = 2

and at t = 2 the height is -5(4) + 40 + 25 = 45. The vertex is (2, 45): the ball is highest, 45 meters up, two seconds after it is thrown.

The intercept on the h-axis is (0, 25), the moment of the throw from the roof. Its mirror image is (4, 25): at 4 seconds the ball falls back past the height of the roof.

The ball lands when h = 0. Divide -5t^2 + 20t + 25 = 0 by -5 to get t^2 - 4t - 5 = 0, which factors as (t - 5)(t + 1) = 0. So t = 5 or t = -1.

The algebra gives two answers, but only one belongs to the ball. The time t = -1 is before it was thrown, and the equation says nothing true about the ball then. The ball lands after 5 seconds, and the graph of its flight is only the piece of the parabola from t = 0 to t = 5.

Figure 5Only the part of the parabola with 0 \le t \le 5 describes the ball.

Boundaries

The coefficient a must not be 0. With a = 0 the equation is y = bx + c, a straight line, and \displaystyle -\frac{b}{2a} divides by zero.

The equation has to be in standard form first. The method reads a, b and c off y = ax^2 + bx + c. An equation such as y = (x - 1)(x + 3) or y - 2 = x^2 must be rearranged, with y alone on one side and the right side multiplied out, before those numbers mean anything. For those two forms there are also quicker routes: factored form shows the x-intercepts directly, and vertex form shows the vertex.

The y-intercept and its mirror can coincide. When b = 0, the axis is the y-axis itself, so the y-intercept is the vertex and has no separate mirror image. When c = 0, the y-intercept is the origin, which is also an x-intercept. In both cases, pick extra points.

The five points are exact; the curve between them is not. Plotting fixes the points. The curve through them is drawn from the known shape, rounded at the vertex, since near the axis y changes only by the square of a small distance, and it keeps widening without ever turning back.

A situation can use only part of the graph. The equation defines a point for every x. When x stands for a time, a length or a count, only some of those points mean anything, as with the ball, which exists on the graph only from t = 0 to t = 5.

Where this goes next

The vertex found here is the answer to every “largest” or “smallest” question a quadratic can model; finding the greatest or least value is this skill aimed at one point. The same vertex sets the range of the function, and the axis splits the graph into the part that falls and the part that rises.

Completing the square rewrites standard form as y = a(x - h)^2 + k, which puts the vertex in the equation itself, and from there graphing from vertex form is a matter of starting at (h, k) and stepping out. Seeing the graph also makes quadratic inequalities easy to read: x^2 - 4x - 5 < 0 asks where the curve in Figure 1 dips below the axis, which is between -1 and 5.

Exercises

Find the key features

For each equation, say which way the parabola opens, and find the axis of symmetry, the vertex, the y-intercept and any x-intercepts.

y = x^2 + 6x + 8
y = x^2 - 8x + 7
y = -x^2 + 6x - 5
y = x^2 - 5x + 4
y = 2x^2 + 4x - 6
y = x^2 - 4x + 1
y = -x^2 - 4x - 7
y = x^2 + 10x + 25

Graph the parabola

Graph each equation. Mark the vertex, the axis, the intercepts and the mirror image of the y-intercept, and add a pair of extra points wherever fewer than five points are found.

y = x^2 + 2x - 15
y = -x^2 + 2x + 3
y = 3x^2 - 6x + 2
y = x^2 + 4x + 6
y = -x^2 + 4
y = x^2 - 6x
\displaystyle y = -\frac{1}{2}x^2 + 2x + 6

Use the graph

A ball is kicked straight up from the ground, and its height in meters after t seconds is h = -5t^2 + 30t. How high does it go, when does it get there, and when does it land?
For which value of c does the graph of y = x^2 - 6x + c meet the x-axis at exactly one point? What is that point?
The graph of y = x^2 + bx + c crosses the x-axis at (-2, 0) and (6, 0). Find its axis, the values of b and c, and its vertex.
A parabola has axis of symmetry x = 3 and passes through (0, 5) and (1, -3). Name two more points that must be on it.

Reasoning

Show that the points (0, c) and \displaystyle \left(-\frac{b}{a}, c\right) both lie on the graph of y = ax^2 + bx + c, and explain why this puts the axis of symmetry at \displaystyle x = -\frac{b}{2a}.
Without solving anything, explain why the graph of y = 2x^2 + 3x - 7 must cross the x-axis twice. What does the same argument say whenever a and c have opposite signs?
Explain how knowing only which way a parabola opens and the y-coordinate of its vertex tells you how many x-intercepts it has.

Answers

Opens upward. Axis x = -3; vertex (-3, -1); y-intercept (0, 8); x-intercepts (-4, 0) and (-2, 0), from (x + 4)(x + 2) = 0.
Opens upward. Axis x = 4; vertex (4, -9); y-intercept (0, 7); x-intercepts (1, 0) and (7, 0), from (x - 1)(x - 7) = 0.
Opens downward. Axis x = 3, since \displaystyle -\frac{6}{2(-1)} = 3; vertex (3, 4); y-intercept (0, -5); x-intercepts (1, 0) and (5, 0), from x^2 - 6x + 5 = 0.
Opens upward. Axis \displaystyle x = \frac{5}{2}; vertex \displaystyle \left(\frac{5}{2}, -\frac{9}{4}\right), since \displaystyle \frac{25}{4} - \frac{50}{4} + \frac{16}{4} = -\frac{9}{4}; y-intercept (0, 4); x-intercepts (1, 0) and (4, 0).
Opens upward. Axis x = -1; vertex (-1, -8); y-intercept (0, -6); x-intercepts (-3, 0) and (1, 0), from 2(x + 3)(x - 1) = 0.
Opens upward. Axis x = 2; vertex (2, -3); y-intercept (0, 1); x-intercepts at \displaystyle x = 2 \pm \sqrt{3}, about 0.27 and 3.73, from the quadratic formula.
Opens downward. Axis x = -2, since \displaystyle -\frac{-4}{2(-1)} = -2; vertex (-2, -3); y-intercept (0, -7); no x-intercepts, because the highest point is below the axis (b^2 - 4ac = 16 - 28 = -12).
Opens upward. Axis x = -5; vertex (-5, 0); y-intercept (0, 25); one x-intercept, (-5, 0), which is the vertex, since x^2 + 10x + 25 = (x + 5)^2.
Opens upward. Vertex (-1, -16) on the axis x = -1; y-intercept (0, -15) and its mirror (-2, -15); x-intercepts (-5, 0) and (3, 0).
Opens downward. Vertex (1, 4) on the axis x = 1; y-intercept (0, 3) and its mirror (2, 3); x-intercepts (-1, 0) and (3, 0).
Opens upward, narrow. Vertex (1, -1) on the axis x = 1; y-intercept (0, 2) and its mirror (2, 2); x-intercepts at \displaystyle x = \frac{3 \pm \sqrt{3}}{3}, about 0.42 and 1.58.
Opens upward. Vertex (-2, 2) on the axis x = -2; y-intercept (0, 6) and its mirror (-4, 6); no x-intercepts. Extra points: (-1, 3) and its mirror (-3, 3).
Opens downward. Axis x = 0, the y-axis, so the vertex is the y-intercept, (0, 4), and it has no mirror; x-intercepts (-2, 0) and (2, 0). Extra points: (1, 3) and (-1, 3).
Opens upward. Vertex (3, -9) on the axis x = 3; the y-intercept is the origin (0, 0), which is also an x-intercept, and its mirror (6, 0) is the other x-intercept. That is only three points, so add a pair: x = 1 gives 1 - 6 = -5, so (1, -5) and its mirror (5, -5).
Opens downward, wide. Axis x = 2, since \displaystyle -\frac{2}{2(-\frac{1}{2})} = 2; vertex (2, 8); y-intercept (0, 6) and its mirror (4, 6); x-intercepts (-2, 0) and (6, 0), from x^2 - 4x - 12 = 0 after multiplying by -2.
It reaches 45 meters at t = 3 seconds and lands at t = 6 seconds. The axis is \displaystyle t = -\frac{30}{2(-5)} = 3, and h = -45 + 90 = 45 there; -5t(t - 6) = 0 gives t = 0 (the kick) and t = 6.
c = 9, and the point is (3, 0). The graph meets the axis once exactly when the vertex is on it; the vertex is at x = 3 with height 9 - 18 + c = c - 9, which is 0 when c = 9.
Axis x = 2, halfway between -2 and 6; b = -4 and c = -12, since (x + 2)(x - 6) = x^2 - 4x - 12; vertex (2, -16).
(6, 5) and (5, -3). The point (0, 5) is 3 left of the axis, so its mirror is 3 right; (1, -3) is 2 left, so its mirror is 2 right.
a(0)^2 + b(0) + c = c, and \displaystyle a\left(\frac{b^2}{a^2}\right) + b\left(-\frac{b}{a}\right) + c = \frac{b^2}{a} - \frac{b^2}{a} + c = c. Both points have height c, and a parabola reaches a height other than its vertex’s at two mirror-image points, so the axis is halfway between 0 and \displaystyle -\frac{b}{a}, at \displaystyle x = -\frac{b}{2a}. (If b = 0 the two points are one point, the vertex, on the axis x = 0.)
Here a = 2 > 0, so the graph opens upward, and the y-intercept (0, -7) is below the x-axis. The vertex is the lowest point, so it is at or below -7 too. From there the arms rise without limit on both sides, so each crosses the axis once. The same holds whenever a and c have opposite signs: the y-intercept lies on the side of the axis that the parabola opens away from, so the curve must cross once on each side.
If it opens upward, the vertex is the lowest point: a vertex above the axis means no x-intercepts, on the axis means one, below it means two, since the arms rise from there on both sides. If it opens downward, the vertex is the highest point and the three cases swap: below the axis means none, on it one, above it two.