## Precalculus (6th Edition) Blitzer

To find the x-intercepts of the function, equate the function $f\left( x \right)={{x}^{2}}+2x-8$ to zero. \begin{align} & {{x}^{2}}+2x-8=0 \\ & {{x}^{2}}+4x-2x-8=0 \\ & x\left( x+4 \right)-2\left( x+4 \right)=0 \\ & \left( x+4 \right)\left( x-2 \right)=0 \end{align} Therefore, $x=-4$ and $x=2$ are the x-intercepts. To find the y-intercept of the function, find the value of $f\left( 0 \right)$. \begin{align} & f\left( 0 \right)={{0}^{2}}+2\left( 0 \right)-8 \\ & =-8 \end{align} Thus, the y-intercept is -8. Substitute x with $-x$ to check the symmetry of the function: \begin{align} & f\left( -x \right)={{\left( -x \right)}^{2}}+2\left( -x \right)-8 \\ & ={{x}^{2}}-2x-8 \end{align} Since $f\left( x \right)\ne f\left( -x \right)$, the graph is not symmetric with respect to the y-axis and since $f\left( -x \right)\ne -f\left( x \right)$, the graph is not symmetric through the origin.