Answer
The y-coordinate of the vertex of $f\left( x \right)=4{{x}^{2}}-16x+300$ is $f\left( 2 \right)$. False.
Work Step by Step
Consider the given function $f\left( x \right)=4{{x}^{2}}-16x+300$ and put $x=0,f\left( 0 \right)$ to get the y-coordinate.
$\begin{align}
& f\left( 0 \right)=4{{\left( 0 \right)}^{2}}-16\left( 0 \right)+300 \\
& =300 \\
&
\end{align}$
Now, Solve for $f\left( 2 \right)$ ,
$\begin{align}
& f\left( 2 \right)=4{{\left( 2 \right)}^{2}}-16\times 2+300 \\
& =16-32+300 \\
& =284
\end{align}$
So, the y-coordinate is not equal to $f\left( 2 \right)$.
