# 022 Exam 1 Sample A, Problem 7

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Find the slope of the tangent line to the graph of ${\displaystyle f(x)=x^{3}-3x^{2}-5x+7}$ at the point ${\displaystyle (3,-8)}$.

Foundations:
Recall that for a given value, ${\displaystyle f'(x)}$ is precisely the point of the tangent line through the point ${\displaystyle \left(x,f(x)\right)}$. Once we have the slope, we can then use the point-slope form for a line:
${\displaystyle y-y_{0}=m(x-x_{0}),}$
where ${\displaystyle m}$ is the known slope and ${\displaystyle \left(x_{0},y_{0}\right)}$ is a point on the line.

Solution:

Finding the slope:
Note that
${\displaystyle f'(x)\,\,=\,\,3x^{2}-6x-5,}$
so the tangent line through ${\displaystyle (3,-8)}$ has slope
${\displaystyle m\,\,=\,\,f'(3)\,\,=\,\,3(3)^{2}-6(3)-5\,\,=\,\,4.}$
Write the Equation of the Line:
Using the point-slope form listed in foundations, along with the point ${\displaystyle (3,-8)}$ and the slope ${\displaystyle m=4}$, we find
${\displaystyle y-(-8)\,\,=\,\,4(x-3),}$
or
${\displaystyle y\,\,=\,\,4x-20.}$
${\displaystyle y\,\,=\,\,4x-20.}$