Let be the quadratic function

Find values for and such that the graph of has a tangent line given by at the point .

First, we compute the derivative

So, if the line is tangent to at the point we must have

Next, the point must be on the graph of , i.e., we must have . Thus,

So, the values are and .

why is f'(2)=2 ?

Because the tangent at point (2, 4) is 2 (i.e. the slope of line 2x is 2).

I think there might be an error in this answer, why do you have f'(2) = 2?

is asking for the the tangent to be equal to y = 2x. f'(x) = 2x + a, so at (2,4) a should = 0.

and therefore b = 0 as well…

No, the answer is correct. It is asking for the tangent line to be equal to at the point , not for the derivative to be . (The derivative gives the slope of the tangent line, it is not a formula for the tangent line.) So, since is tangent to the curve at and this line has slope 2, we know the derivative must be equal to 2 at that point. Hence, . Does that make sense?