Solving Quadratics by Graphing: Examples

Solving quadratics by graphing means finding where the parabola $y = ax^2 + bx + c$ crosses the x-axis. First get the equation in the form $ax^2 + bx + c = 0$ . Then graph the function. Each x-intercept is a solution. A quadratic can have two, one or no real solutions.

Before this lesson: You should know graphing quadratic functions, vertex form and how to find x-intercepts.

Why the x-Intercepts Are the Solutions

A quadratic equation like $x^2 - 4 = 0$ asks: which x-values make $x^2 - 4$ equal 0? On the graph of $y = x^2 - 4$ , those are the points where $y = 0$ .

Points where $y = 0$ sit on the x-axis. So the solutions are the x-intercepts. They are also called the roots or zeros of the function.

Three parabolas, each on a grid from −4 to 4. Left: y = x² − 4 crosses the x-axis at −2 and 2, so it has two solutions. Middle: y = (x − 1)² touches the x-axis only at 1, so it has one solution. Right: y = x² + 2 stays above the x-axis, so it has no real solutions. — A parabola can cross the x-axis twice, touch it once, or miss it. That is the number of real solutions.

Steps for solving by graphing

Move every term to one side so the equation reads $ax^2 + bx + c = 0$ .
Graph $y = ax^2 + bx + c$ . Find the vertex and a few points.
Find where the graph crosses or touches the x-axis.
Write each x-intercept as a solution.
Check each solution in the original equation.

Worked Examples

Example 1: Two solutions

Solve $x^2 - 2x - 3 = 0$ by graphing.

Step 1. Graph the related function $y = x^2 - 2x - 3$ . Find the axis of the vertex with $a = 1$ and $b = -2$ .

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

Step 2. Substitute $x = 1$ to find the vertex’s y-value.

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

Step 3. The vertex is $(1, -4)$ . Make a table with x-values on both sides of 1.

$x$	−2	−1	0	1	2	3	4
$y$	5	0	−3	−4	−3	0	5

Step 4. Plot the points and draw the parabola. The solutions are its x-intercepts, where $y = 0$ : $x = -1$ and $x = 3$ .

Step 5. Check $x = -1$ in the equation.

(-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0

Step 6. Check $x = 3$ .

3^2 - 2(3) - 3 = 9 - 6 - 3 = 0

Answer: $x = -1$ and $x = 3$

Example 2: One solution

Solve $x^2 - 6x + 9 = 0$ by graphing.

Step 1. Graph $y = x^2 - 6x + 9$ . Find the axis of the vertex with $a = 1$ and $b = -6$ .

x = -\frac{-6}{2(1)} = 3

Step 2. Substitute $x = 3$ .

y = 3^2 - 6(3) + 9 = 9 - 18 + 9 = 0

Step 3. The vertex $(3, 0)$ sits right on the x-axis, and the parabola opens up. It touches the x-axis only once.

Answer: The only solution is $x = 3$ .

Example 3: No real solutions

Solve $x^2 + 2x + 4 = 0$ by graphing.

Step 1. Graph $y = x^2 + 2x + 4$ . Find the axis of the vertex with $a = 1$ and $b = 2$ .

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

Step 2. Substitute $x = -1$ .

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

Step 3. The vertex $(-1, 3)$ is above the x-axis, and the parabola opens up because $a \gt 0$ . So the graph never reaches the x-axis.

Answer: There are no real solutions.

Example 4: Rearrange first

Solve $x^2 = x + 6$ by graphing.

Step 1. Get 0 on one side. Subtract $x$ from both sides (see two-step equations).

x^2 - x = 6

Step 2. Subtract 6 from both sides.

x^2 - x - 6 = 0

Step 3. Graph $y = x^2 - x - 6$ . Its vertex is at $x = \frac{1}{2}$ , so make a table around that value. The graph crosses the x-axis at $x = -2$ and $x = 3$ .

Step 4. Check $x = -2$ in the original equation. The left side is $(-2)^2 = 4$ and the right side is $-2 + 6 = 4$ .

Step 5. Check $x = 3$ . The left side is $3^2 = 9$ and the right side is $3 + 6 = 9$ .

Answer: $x = -2$ and $x = 3$

Example 5: Word problem

A model rocket’s height in meters is $h = -5t^2 + 20t$ , where $t$ is time in seconds. When does it land?

Step 1. It lands when $h = 0$ , so solve $-5t^2 + 20t = 0$ . Graph $h = -5t^2 + 20t$ . Find the vertex’s t-value.

t = -\frac{20}{2(-5)} = -\frac{20}{-10} = 2

Step 2. Substitute $t = 2$ to find the highest point.

h = -5(2)^2 + 20(2) = -20 + 40 = 20

Step 3. The vertex is $(2, 20)$ . The graph starts at $(0, 0)$ , 2 units left of the axis, so it comes back to $h = 0$ 2 units right, at $t = 4$ .

Step 4. Check $t = 4$ .

-5(4)^2 + 20(4) = -80 + 80 = 0

Step 5. The answer $t = 0$ is the launch, so it does not answer the question.

Answer: The rocket lands after 4 seconds.

Common Mistakes

Mistake	Example of the error	Fix
Not setting the equation to 0	Graphing $y = x^2$ to solve $x^2 = x + 6$	Rewrite as $x^2 - x - 6 = 0$ first, then graph.
Reading the y-intercept as a solution	$x^2 - 2x - 3 = 0$ has solution $-3$	Solutions are where the graph crosses the x-axis, not the y-axis.
Writing points instead of x-values	Solution: $(3, 0)$	The solution is the x-value: $x = 3$ .
Saying no solution means 0	$x^2 + 2x + 4 = 0$ has solution $x = 0$	No x-intercepts means no real solutions.
Trusting a rough graph	Reading $x^2 = 5$ as $x = 2$	$2^2 = 4$ , not 5. Check by substituting. The true answers are about 2.2 and −2.2.

When the Graph Doesn’t Land on a Grid Line

Some solutions are not whole numbers. For $x^2 - 5 = 0$ , the graph crosses between 2 and 3, and between −3 and −2. A graph only gives an estimate, like $x \approx 2.2$ .

For exact answers, use algebra. The next lessons show factoring, square roots, completing the square and the quadratic formula.

Graph of y = ax² + bx + c	Number of real solutions
Crosses the x-axis twice	2
Touches the x-axis at the vertex	1
Never reaches the x-axis	0

Practice Problems

Solve $x^2 - 9 = 0$ by graphing.
Solve $x^2 + 4x + 3 = 0$ by graphing.
Solve $x^2 - 4x + 4 = 0$ by graphing.
How many real solutions does $x^2 + 1 = 0$ have?
Solve $-x^2 + 2x + 8 = 0$ by graphing.
Solve $x^2 = 2x + 3$ by graphing.
A graph of $y = x^2 + x - 2$ crosses the x-axis at $(-2, 0)$ and $(1, 0)$ . Solve $x^2 + x - 2 = 0$ .
Between which two whole numbers is the positive solution of $x^2 - 7 = 0$ ?
The vertex of $y = -x^2 + 6x - 10$ is $(3, -1)$ . How many real solutions does $-x^2 + 6x - 10 = 0$ have?
A ball’s height in feet is $h = -16t^2 + 64t$ . When does it hit the ground?

Show answers

Graph $y = x^2 - 9$ . With $b = 0$ , the vertex is at $x = 0$ , and $y = 0 - 9 = -9$
The graph crosses the x-axis at $x = -3$ and $x = 3$ . Check $x = -3$ : $(-3)^2 - 9 = 9 - 9 = 0$
Check $x = 3$ : $3^2 - 9 = 9 - 9 = 0$
$x = -3$ and $x = 3$
Find the vertex’s x-value: $x = -\frac{4}{2(1)} = -2$
Find its y-value: $y = (-2)^2 + 4(-2) + 3 = 4 - 8 + 3 = -1$
The vertex $(-2, -1)$ is below the axis and the graph opens up. It crosses at $x = -3$ and $x = -1$ . Check $x = -3$ : $(-3)^2 + 4(-3) + 3 = 9 - 12 + 3 = 0$
Check $x = -1$ : $(-1)^2 + 4(-1) + 3 = 1 - 4 + 3 = 0$
$x = -3$ and $x = -1$
Find the vertex’s x-value: $x = -\frac{-4}{2(1)} = 2$
Find its y-value: $y = 2^2 - 4(2) + 4 = 4 - 8 + 4 = 0$
The vertex $(2, 0)$ is on the x-axis, so the graph touches it once
One solution: $x = 2$
With $b = 0$ , the vertex is at $x = 0$ , and $y = 0 + 1 = 1$
The vertex $(0, 1)$ is above the x-axis, and the graph opens up, so it never reaches the axis
None. There are no real solutions.
Find the vertex’s x-value: $x = -\frac{2}{2(-1)} = 1$
Find its y-value: $y = -(1)^2 + 2(1) + 8 = -1 + 2 + 8 = 9$
The vertex $(1, 9)$ is above the axis and the graph opens down. It crosses at $x = -2$ and $x = 4$ . Check $x = -2$ : $-(-2)^2 + 2(-2) + 8 = -4 - 4 + 8 = 0$
Check $x = 4$ : $-(4)^2 + 2(4) + 8 = -16 + 8 + 8 = 0$
$x = -2$ and $x = 4$
Subtract $2x$ from both sides: $x^2 - 2x = 3$
Subtract 3 from both sides: $x^2 - 2x - 3 = 0$
This is the equation from Example 1. Its graph crosses the x-axis at $x = -1$ and $x = 3$
$x = -1$ and $x = 3$
The solutions of $x^2 + x - 2 = 0$ are the x-values where the graph crosses the x-axis
$x = -2$ and $x = 1$
The graph of $y = x^2 - 7$ crosses the x-axis where $x^2 = 7$
Try whole numbers: $2^2 = 4$ , which is less than 7, and $3^2 = 9$ , which is more than 7
The positive solution is between 2 and 3.
Check the vertex: $-(3)^2 + 6(3) - 10 = -9 + 18 - 10 = -1$
The vertex $(3, -1)$ is below the x-axis, and the parabola opens down because $a \lt 0$ . So it never reaches the axis
None. There are no real solutions.
It hits the ground when $h = 0$ . Find the vertex’s t-value: $t = -\frac{64}{2(-16)} = 2$
Find the highest point: $h = -16(2)^2 + 64(2) = -64 + 128 = 64$
The graph starts at $(0, 0)$ , 2 units left of the axis, so it comes back to 0 at $t = 4$ . Check it: $-16(4)^2 + 64(4) = -256 + 256 = 0$
$t = 0$ is the throw, so it hits the ground after 4 seconds.

FAQ

How do you solve a quadratic equation by graphing?

Write it as $ax^2 + bx + c = 0$ , graph $y = ax^2 + bx + c$ , and find the x-intercepts. Each one is a solution.

What are the roots of a quadratic?

The roots are the solutions of the equation. On the graph, they are the x-intercepts, also called the zeros of the function.

Can a quadratic have no solution?

It can have no real solution. That happens when the parabola never touches the x-axis, like $y = x^2 + 2$ .

Is solving by graphing exact?

Only when the graph crosses at clear grid points. Otherwise it gives an estimate. Check answers by substituting them, or use an algebra method.

Summary

Set the equation equal to 0: $ax^2 + bx + c = 0$ .
Graph the parabola and read its x-intercepts. Those are the solutions.
Two crossings, one touch at the vertex, or no crossings: 2, 1 or 0 real solutions.
Check each solution in the original equation.
Graphs give estimates when the crossings are between grid lines.

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