Answer
The equation of the tangent line to $f(x)$ at $(0,0)$ is $y=5x$.
The graphs are below. The second intersection point is $(6,30)$.
Work Step by Step
$$y=f(x)=x^3-6x^2+5x$$
a) Find $f'(x)$: $$f'(x)=3x^2-12x+5$$
The slope of the tangent line to $f(x)$ at $(0,0)$ is $f'(0)=3\times0^2-12\times0+5=5$.
The equation of the tangent line to $f(x)$ at $(0,0)$ is $$y-0=5(x-0)$$
$$y=5x$$
b) The graphs of the functions are enclosed below.
We see that the tangent intersects another point of the curve. Tracing the coordinates of the intersected point, we guess that the point is $(6,30)$.
c) To find the intersection point, as stated, we solve the equation for the curve and the tangent simultaneously: $$x^3-6x^2+5x=5x$$ $$x^3-6x^2=0$$ $$x^2(x-6)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=6$$
- $x=0$ is already the point of tangency.
- At $x=6$: $f(6)=6^3-6\times6^2+5\times6=6^3-6^3+30=30$
Therefore, the second intersection point is indeed $(6,30)$.
