## Precalculus (6th Edition) Blitzer To find the x-intercepts of the function equate the function $f\left( x \right)={{x}^{3}}-4{{x}^{2}}-x+4$ to zero: \begin{align} & {{x}^{3}}-4{{x}^{2}}-x+4=0 \\ & {{x}^{2}}\left( x-4 \right)-1\left( x-4 \right)=0 \\ & \left( {{x}^{2}}-1 \right)\left( x-4 \right)=0 \\ & \left( x+1 \right)\left( x-1 \right)\left( x-4 \right)=0 \end{align} $x=-1,1,4$ 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)=\left( 0+1 \right)\left( 0-1 \right)\left( 0-4 \right) \\ & f\left( 0 \right)=4 \\ \end{align} Thus, the y-intercept is 4. Substitute x with $-x$ to check the symmetry of the function: $f\left( x \right)={{x}^{3}}-4{{x}^{2}}-x+4$: \begin{align} & f\left( -x \right)={{\left( -x \right)}^{3}}-4{{\left( -x \right)}^{2}}-\left( -x \right)+4 \\ & =-{{x}^{3}}-4{{x}^{2}}+x+4 \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.