10.5 Graphing quadratic equations  (Page 3/15)

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

Find the intercepts of a parabola.

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

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

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 $\left(-2,0\right)$ .

Find the intercepts of the parabola $y={x}^{2}+2x-8.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-8\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4,0\right),\left(2,0\right)$

Find the intercepts of the parabola $y={x}^{2}-4x-12.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-12\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(6,0\right),\left(-2,0\right)$

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:

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

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.

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 $\left(0,4\right)$ . 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.

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 $y={x}^{2}-4x-5.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-5\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(5,0\right)\phantom{\rule{0.2em}{0ex}}\left(-1,0\right)$

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 $\left(\frac{3}{2},0\right).$

Find the intercepts of the parabola $y=\text{−}{x}^{2}-12x-36.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-36\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-6,0\right)$