Find the equation of the tangent to the curve at , where .
First, find the slope function by differentiating with respect
To find the slope when , substitute this value into the
Now find the -coordinate on the curve when by substituting
into the equation of the curve.
Therefore, the tangent has a slope of 36 and passes through the point
. Its equation can be found using the general equation of
a straight line.
Making the substitution gives