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?