10.5 Graphing quadratic equations  (Page 3/15)

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

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 $y=\mathrm{-8}$ . The y -intercept is the point $(0,\mathrm{-8})$ .
To find the x -intercept, let $y=0$ and solve for x .
Solve by factoring.

When $y=0$ , then $x=4\text{or}x=\mathrm{-2}$ . The x -intercepts are the points $\left(4,0\right)$ and $\left(\mathrm{-2},0\right)$ .

Got questions? Get instant answers now!

Got questions? Get instant answers now!

In this chapter, 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:

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.

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 $y=5x^2+x+4$ .

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.	$\begin{array}{c}\begin{array}{ccc}\hfill b^2& -\hfill & 4ac\hfill \\ \hfill 1^2& -\hfill & 4\cdot 5\cdot 4\hfill \\ 1\hfill & -\hfill & 80\hfill \end{array}\hfill \\ \hfill \mathrm{-79}\hfill \end{array}$
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 $y=4x^2-12x+9$ .

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 $(0,9)$ .
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.	$\begin{array}{c}\begin{array}{ccc}\hfill b^2& -\hfill & 4ac\hfill \\ \hfill 2^2& -\hfill & 4\cdot 4\cdot 9\hfill \\ \hfill 144& -\hfill & 144\hfill \end{array}\hfill \\ \hfill 0\hfill \end{array}$
	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 $(\frac{3}{2},0).$

Got questions? Get instant answers now!

Got questions? Get instant answers now!