#### Answer

At the point $(2,4)$, $f(x)$ is parallel with the given line.

#### Work Step by Step

$$f(x)=3x^2-4x$$
1) Find $f'(x)$: $$f'(x)=(3x^2-4x)'=6x-4$$
2) The slope of the tangent line to $f(x)$ at $x=a$ is $f'(a)=6a-4$.
We know that parallel lines have the same slope. Therefore, all the tangent lines to $f(x)$ parallel to $y=8x+5$ must have $f'(a)=8$.
In other words, in order to find the required points, we now have to find $a$ such that $f'(a)=6a-4=8$: $$6a-4=8$$
$$6a=12$$
$$a=2$$
For $a=2$: $$f(a)=3\times2^2-4\times2=3\times4-8=12-8=4$$
Therefore, at the point $(2,4)$, $f(x)$ is parallel with the given line.