I am having trouble solving this problem:
Find numbers $a$, $b$, and $c$ so that the graph of $f(x) = ax^2 + bx + c$ has
$x$-intercepts at $(0,0)$ and $(8,0)$ and a tangent with slope $16$ where $x = 2$.
I've tried to find the values by using elimination, finding equations using the $x$ intercepts, but I do not know which direction to take.
Best Answer
Firstly, finding the slope is easy. The derivative of the function at a point $x$ gives the slope at that point. The derivative of the function is $$f'(x)=2ax+b.$$
When $x=2$ the slope is $16$, therefore $2a(2)+b=16$, or $4a+b=16.$ We also know that there is an intercept at $(0,0)$, therefore $$0=a(0)^2+b(0)+c,$$ thus $c=0$. Another point is, $$0=a(8)^2+b(8)+c$$ $$\therefore 64a+8b=0.$$ Solving the two equations $$4a+b=16$$ $$64a+8b=0.$$
We have $a=-4$ and $b=32$, thus $$f(x)=-4x^2+32x.$$