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Section 3.1 : Graphing

8. Determine the \(x\)-intercepts and the \(y\)-intercepts for the following equation.

\[y = {x^2} + 6x + 58\]

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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,

\[\begin{align*}y & = {\left( 0 \right)^2} + 6\left( 0 \right) + 58\\ y & = 58\end{align*}\]

The \(y\)-intercept for this equation is then the point : \(\left( {0,58} \right)\) .

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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,

\[\begin{align*}0 & = {x^2} + 6x + 58\\ x & = \frac{{ - 6 \pm \sqrt {{6^2} - 4\left( 1 \right)\left( {58} \right)} }}{{2\left( 1 \right)}} = \frac{{ - 6 \pm \sqrt { - 196} }}{2} = \frac{{ - 6 \pm 14\,i}}{2} = - 3 \pm 7i\end{align*}\]

Because we got complex solutions to this equation we know that this equation has no x‑intercepts.