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Do these results agree with our graphs? See [link] .
To find the intercepts of a parabola with equation $y=a{x}^{2}+bx+c$ :
Find the intercepts of the parabola $y={x}^{2}2x8$ .
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 $x=4\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=\mathrm{2}$ . The x intercepts are the points $\left(4,0\right)$ and $\left(\mathrm{2},0\right)$ .
Find the intercepts of the parabola $y={x}^{2}+2x8.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,\mathrm{8}\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(\mathrm{4},0\right),\left(2,0\right)$
Find the intercepts of the parabola $y={x}^{2}4x12.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,\mathrm{12}\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(6,0\right),\left(\mathrm{2},0\right)$
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 $y=5{x}^{2}+x+4$ .
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.  $\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& \hfill & 4ac\hfill \\ \hfill {1}^{2}& \hfill & 4\cdot 5\cdot 4\hfill \\ 1\hfill & \hfill & 80\hfill \end{array}\hfill \\ \hfill \mathrm{79}\phantom{\rule{0.8em}{0ex}}\hfill \end{array}$ 
Since the value of the discriminant is negative, there is no real solution to the equation.  There are no x intercepts. 
Find the intercepts of the parabola $y=3{x}^{2}+4x+4.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,4\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none}$
Find the intercepts of the parabola $y={x}^{2}4x5.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,\mathrm{5}\right);x\text{:}\phantom{\rule{0.2em}{0ex}}(5,0)\phantom{\rule{0.2em}{0ex}}(\mathrm{1},0)$
Find the intercepts of the parabola $y=4{x}^{2}12x+9$ .
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.  $\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& \hfill & 4ac\hfill \\ \hfill {2}^{2}& \hfill & 4\cdot 4\cdot 9\hfill \\ \hfill 144& \hfill & 144\hfill \end{array}\hfill \\ \hfill 0\phantom{\rule{1em}{0ex}}\hfill \end{array}$ 
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 $(\frac{3}{2},0).$ 
Find the intercepts of the parabola $y=\text{\u2212}{x}^{2}12x36.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,\mathrm{36}\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(\mathrm{6},0\right)$
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