Algebra - Graphing

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Section 3.1 : Graphing

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9. Determine the $x$ -intercepts and the $y$ -intercepts for the following equation.

y = {(x + 3)}^{2} - 8

$y = {\left( {x + 3} \right)^2} - 8$

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Start Solution

Recall that in order to find the $y$ -intercept all we need to do is plug $x = 0$ into the equation and solve for $y$ . Doing that for this equation gives,

\begin{matrix} y & = {(0 + 3)}^{2} - 8 y & = 1 \end{matrix}

$\begin{align*}y & = {\left( {0 + 3} \right)^2} - 8\\ y & = 1\end{align*}$

The $y$ -intercept for this equation is then the point : $\left( {0,1} \right)$ .

Show Step 2

Finding the $x$ -intercept is similar to the $y$ -intercept. All we do is plug in $y = 0$ and solve for $x$ . Doing that for this equation gives,

\begin{matrix} 0 & = {(x + 3)}^{2} - 8 {(x + 3)}^{2} & = 8 x + 3 & = \pm \sqrt{8} x & = - 3 \pm \sqrt{8} \end{matrix}

$\begin{align*}0 & = {\left( {x + 3} \right)^2} - 8\\ {\left( {x + 3} \right)^2} & = 8\\ x + 3 & = \pm \sqrt 8 \\ x & = - 3 \pm \sqrt 8 \end{align*}$

The $x$ -intercepts for this equation are then the two points : $\left( { - 3 - \sqrt 8 ,0} \right)$ and $\left( { - 3 + \sqrt 8 ,0} \right)$ .

Don’t worry about the “messy” answers here. This kind of intercept will show up occasionally so we need to be able to deal with them when they do.