Answer
False.
The vertex of the parabola described by $f(x)=2(x-5)^2-1$ is at $(5,-1)$.
Work Step by Step
The given function is a quadratic function:
$f(x)=2(x-5)^2-1$.
The standard form is
$f(x)=a(x-h)^2+k$
Consider both equations
$a=2,h=5$ and $k=-1$.
The vertex is $(h,k)=(5,-1)$.
Hence, the given statement is false.
A true statement is:
The vertex of the parabola described by $f(x)=2(x-5)^2-1$ is at $(5,-1)$.
