I attached the question number. I found the answer to A and B but I'm stuck on C.
For B I got y=x+7
but the tangent line for equation is 3x^2 -2x -4
Thank you.
Oh, dear. $\displaystyle 3x^2- 2x- 4$ is a parabola, NOT any line at all! What you mean is that the derivative of $\displaystyle x^3- x^2- 4x+ 4$ so gives the slope of the tangent line.
Given that, at x= a, the derivative is $\displaystyle 3a^2- 2a- 4$ and the value is $\displaystyle a^3- a^2- 4a+ 4$. That means that the equation of the tangent line, at x= a, is $\displaystyle y= (3a^2- 2a- 4)(x- a)+ a^3- a^2- 4a+ 4$. Saying that line goes through (0, -8) mean that x= 0, y= -8 makes that a true equation: solve $\displaystyle (3a^2- 2a- 4)(0- a)+ a^3- a^2- 4a+ 4= -2a^3+ a^2+ 4= -8$.
for c,
remember that $\displaystyle (a,b)$ and $\displaystyle (0,-8)$ lie on the same line,i.e the tangent line. So the slope of the line is$\displaystyle \frac{8+b}{a}$.
Again you know that the slope of the $\displaystyle 3x^2-2x-4$(which you got by differentiating ) and thus for $\displaystyle (a,b)$ the slope is $\displaystyle 3a^2-2a-4$.
equate the slopes:
$\displaystyle \frac{8+b}{a}=3a^2-2a-4 \implies 8+b=3a^3-2a^2-4$ and $\displaystyle b=a^3-a^2-4a-4$ from the original equation:
so $\displaystyle 8+a^3-a^2-4a-4=3a^3-2a^2-4a \implies 2a^3-a^2-12=0$ hence $\displaystyle a=2$ and $\displaystyle b=2^3-2^2-4(2)+4=0$