Ulearngo

Finding the Intercepts of a Parabola

Finding the Intercepts of a Parabola

When we graphed linear equations, we often used the x- and y-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y-intercept the value of \(x\) is zero. So, to find the y-intercept, we substitute \(x=0\) into the equation.

Let’s find the y-intercepts of the two parabolas shown in the figure below.

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2. Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the statement “x equals 0”. Next to that is the equation of the graph with 0 plugged in for x which gives y equals 0 squared plus4 times 0 plus 3. This simplifies to y equals 3. Below the equation is the statement “y-intercept (0, 3)”. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2. Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the statement “x equals 0”. Next to that is the equation of the graph with 0 plugged in for x which gives y equals negative quantity 0 squared plus 4 times 0 plus 3. This simplifies to y equals 3. Below the equation is the statement “y-intercept (0, 3)”.

At an x-intercept, the value of \(y\) is zero. To find an x-intercept, we substitute \(y=0\) into the equation. In other words, we will need to solve the equation \(0=a{x}^{2}+bx+c\) for \(x\).

\(\begin{array}{}\\ \\ y=a{x}^{2}+bx+c\hfill \\ 0=a{x}^{2}+bx+c\hfill \end{array}\)

But solving quadratic equations like this is exactly what we have done earlier in this tutorial.

We can now find the x-intercepts of the two parabolas shown in the figure above.

First, we will find the x-intercepts of a parabola with equation \(y={x}^{2}+4x+3\).

.
Let \(y=0\). .
Factor. .
Use the zero product property. .
Solve. .
The x intercepts are \(\left(\text{−}1,0\right)\) and \(\left(\text{−}3,0\right).\)

Now, we will find the x-intercepts of the parabola with equation \(y=-{x}^{2}+4x+3\).

.
Let \(y=0\). .
This quadratic does not factor, so we use the Quadratic Formula. .
\(a=-1\), \(b=4\), \(c=3\) .
Simplify. .
.
.
.
The x intercepts are \(\left(2+\sqrt{7},0\right)\) and \(\left(2-\sqrt{7},0\right)\).

We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph.

\(\begin{array}{cccc}\left(2+\sqrt{7},0\right)\approx \left(4.6,0\right)\hfill & & & \left(2-\sqrt{7},0\right)\approx \left(-0.6,0\right)\hfill \end{array}\)

Do these results agree with our graphs? See the figure below.

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (-2, -1). Three points are plotted on the curve at (-3, 0), (-1, 0), and (0, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2. Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the statement “y-intercept (0, 3)”. Below that is the statement “x-intercepts (-1, 0) and (-3, 0)”. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (2, 7). Three points are plotted on the curve at (-0.6, 0), (4.6, 0), and (0, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2. Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the statement “y-intercept (0, 3)”. Below that is the statement “x-intercepts (2 plus square root of 7, 0) is approximately equal to (4.6, 0) and (2 minus square root of 7, 0) is approximately equal to (-0.6, 0).”

Find the intercepts of a parabola.

To find the intercepts of a parabola with equation \(y=a{x}^{2}+bx+c\):

\(\begin{array}{cccc}\hfill {\text{y}}\mathbf{\text{-intercept}}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{\text{x}}\mathbf{\text{-intercepts}}\hfill \\ \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}x=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{Let}\phantom{\rule{0.2em}{0ex}}y=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}x.\hfill \end{array}\)

Example

Find the intercepts of the parabola \(y={x}^{2}-2x-8\).

Solution

.
To find the y-intercept, let \(x=0\) and solve for y. .
When \(x=0\), then \(y=-8\).
The y-intercept is the point \(\left(0,-8\right)\).
.
To find the x-intercept, let \(y=0\) and solve for x. .
Solve by factoring. .
.

When \(y=0\), then \(x=4\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=-2\). The x-intercepts are the points \(\left(4,0\right)\) and \(\left(-2,0\right)\).

In this tutorial, we have been solving quadratic equations of the form \(a{x}^{2}+bx+c=0\). We solved for \(x\) and the results were the solutions to the equation.

We are now looking at quadratic equations in two variables of the form \(y=a{x}^{2}+bx+c\). The graphs of these equations are parabolas. The x-intercepts of the parabolas occur where \(y=0\).

For example:

\(\begin{array}{cccc}\hfill \mathbf{\text{Quadratic equation}}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\mathbf{\text{Quadratic equation in two variables}}\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}y={x}^{2}-2x-15\hfill \\ \hfill \begin{array}{ccc}\hfill {x}^{2}-2x-15& =\hfill & 0\hfill \\ \hfill \left(x-5\right)\left(x+3\right)& =\hfill & 0\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{let}\phantom{\rule{0.2em}{0ex}}y=0\phantom{\rule{1em}{0ex}}\begin{array}{c}0={x}^{2}-2x-15\hfill \\ 0=\left(x-5\right)\left(x+3\right)\hfill \end{array}\hfill \\ \hfill \begin{array}{cccccccc}\hfill x-5& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 5\hfill & & & \hfill x& =\hfill & -3\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\begin{array}{cccccccc}\hfill x-5& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 5\hfill & & & \hfill x& =\hfill & -3\hfill \end{array}\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}\left(5,0\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(-3,0\right)\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}x\text{-intercepts}\hfill \end{array}\)

The solutions of the quadratic equation are the \(x\) values of the x-intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the equations give the x-intercepts of the graphs, the number of x-intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic equation of the form \(a{x}^{2}+bx+c=0\). Now, we can use the discriminant to tell us how many x-intercepts there are on the graph.

This figure shows three graphs side by side. The leftmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. Below the graph is the inequality b squared minus 4 a c greater than 0. Below that is the statement “Two solutions”. Below that is the statement “ Two x-intercepts”. The middle graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is on the x-axis. Below the graph is the equation b squared minus 4 a c equals 0. Below that is the statement “One solution”. Below that is the statement “ One x-intercept”. The rightmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper left quadrant. Below the graph is the inequality b squared minus 4 a c less than 0. Below that is the statement “No real solutions”. Below that is the statement “ No x-intercept”.

Before you start solving the quadratic equation to find the values of the x-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Example

Find the intercepts of the parabola \(y=5{x}^{2}+x+4\).

Solution

.
To find the y-intercept, let \(x=0\) and solve for y. .
.
When \(x=0\), then \(y=4\).
The y-intercept is the point \(\left(0,4\right)\).
.
To find the x-intercept, let \(y=0\) and solve for x. .
Find the value of the discriminant to predict the number of solutions and so x-intercepts. \(\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& -\hfill & 4ac\hfill \\ \hfill {1}^{2}& -\hfill & 4\cdot 5\cdot 4\hfill \\ 1\hfill & -\hfill & 80\hfill \end{array}\hfill \\ \hfill -79\phantom{\rule{0.8em}{0ex}}\hfill \end{array}\)
Since the value of the discriminant is negative, there is no real solution to the equation. There are no x-intercepts.

Example

Find the intercepts of the parabola \(y=4{x}^{2}-12x+9\).

Solution

.
To find the y-intercept, let \(x=0\) and solve for y. .
.
When \(x=0\), then \(y=9\).
The y-intercept is the point \(\left(0,9\right)\).
.
To find the x-intercept, let \(y=0\) and solve for x. .
Find the value of the discriminant to predict the number of solutions and so x-intercepts. \(\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& -\hfill & 4ac\hfill \\ \hfill {2}^{2}& -\hfill & 4\cdot 4\cdot 9\hfill \\ \hfill 144& -\hfill & 144\hfill \end{array}\hfill \\ \hfill 0\phantom{\rule{1em}{0ex}}\hfill \end{array}\)
Since the value of the discriminant is 0, there is no real solution to the equation. So there is one x-intercept.
Solve the equation by factoring the perfect square trinomial. .
Use the Zero Product Property. .
Solve for x. .
.
When \(y=0\), then \(\frac{3}{2}=x.\)
The x-intercept is the point \(\left(\frac{3}{2},0\right).\)

This lesson is part of:

Introducing Quadratic Equations

View Full Tutorial

Track Your Learning Progress

Sign in to unlock unlimited practice exams, tutorial practice quizzes, personalized weak area practice, AI study assistance with Lexi, and detailed performance analytics.