<< Chapter < Page Chapter >> Page >

Do these results agree with our graphs? See [link] .

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 + b x + c :

y -intercept x -intercepts Let x = 0 and solve for y . Let y = 0 and solve for x .

Find the intercepts of the parabola y = x 2 2 x 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 ( 0 , −8 ) .
.
To find the x -intercept, let y = 0 and solve for x . .
Solve by factoring. .
.

When y = 0 , then x = 4 or x = −2 . The x -intercepts are the points ( 4 , 0 ) and ( −2 , 0 ) .

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Find the intercepts of the parabola y = x 2 + 2 x 8 .

y : ( 0 , −8 ) ; x : ( −4 , 0 ) , ( 2 , 0 )

Got questions? Get instant answers now!

Find the intercepts of the parabola y = x 2 4 x 12 .

y : ( 0 , −12 ) ; x : ( 6 , 0 ) , ( −2 , 0 )

Got questions? Get instant answers now!

In this chapter, we have been solving quadratic equations of the form a x 2 + b x + 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 + b x + c . The graphs of these equations are parabolas. The x -intercepts of the parabolas occur where y = 0 .

For example:

Quadratic equation Quadratic equation in two variables y = x 2 2 x 15 x 2 2 x 15 = 0 ( x 5 ) ( x + 3 ) = 0 let y = 0 0 = x 2 2 x 15 0 = ( x 5 ) ( x + 3 ) x 5 = 0 x + 3 = 0 x = 5 x = −3 x 5 = 0 x + 3 = 0 x = 5 x = −3 ( 5 , 0 ) and ( −3 , 0 ) x -intercepts

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 + b x + 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.

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 ( 0 , 4 ) .
.
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. b 2 4 a c 1 2 4 5 4 1 80 −79
Since the value of the discriminant is negative, there is no real solution to the equation. There are no x -intercepts.
Got questions? Get instant answers now!
Got questions? Get instant answers now!

Find the intercepts of the parabola y = 3 x 2 + 4 x + 4 .

y : ( 0 , 4 ) ; x : none

Got questions? Get instant answers now!

Find the intercepts of the parabola y = x 2 4 x 5 .

y : ( 0 , −5 ) ; x : ( 5 , 0 ) ( −1 , 0 )

Got questions? Get instant answers now!

Find the intercepts of the parabola y = 4 x 2 12 x + 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 ( 0 , 9 ) .
.
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. b 2 4 a c 2 2 4 4 9 144 144 0
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 3 2 = x .
The x -intercept is the point ( 3 2 , 0 ) .
Got questions? Get instant answers now!
Got questions? Get instant answers now!

Find the intercepts of the parabola y = x 2 12 x 36 .

y : ( 0 , −36 ) ; x : ( −6 , 0 )

Got questions? Get instant answers now!
Practice Key Terms 6

Get the best Elementary algebra course in your pocket!





Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Elementary algebra' conversation and receive update notifications?

Ask