## Precalculus (6th Edition) Blitzer

The statement is False. The graph of $f\left( x \right)=2{{\left( x-5 \right)}^{2}}-1$ has one y-intercept and no x-intercept exists.
The general form of a parabolic function is defined as $f\left( x \right)={{\left( x-h \right)}^{2}}+k$. Now, to calculate the y-intercept, put $x=0$ in $f\left( x \right)$: \begin{align} & f\left( 0 \right)=-2{{\left( 0+4 \right)}^{2}}-8 \\ & =-2\left( 16 \right)-8 \\ & =-32-8 \\ & =-40 \end{align} Thus, the y-intercept is $\left( 0,-40 \right)$. To calculate the x-intercept, put $f\left( x \right)=0$: $-2{{\left( x+4 \right)}^{2}}-8=0$ Now, adding 8 to both sides of the above equation: $-2{{\left( x+4 \right)}^{2}}=8$ Dividing the above equation by –2, ${{\left( x+4 \right)}^{2}}=-4$ Taking the square root on both sides: \begin{align} & \sqrt{{{\left( x+4 \right)}^{2}}}=\sqrt{-4} \\ & x+4=\pm 2i \end{align} Thus, $x=4\pm 2i$ Therefore, there are no real x intercepts of the function. Hence, the provided statement is false. The graph of $f\left( x \right)=2{{\left( x-5 \right)}^{2}}-1$ has one y-intercept and no x-intercept exists.