Answer
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.
Work Step by Step
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.
