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

Recall that for a given value, $f'(x)$ is precisely the point of the tangent line through the point $\left(x,f(x)\right)$. Once we have the slope, we can then use the pointslope form for a line:

 $yy_{0}=m(xx_{0}),$

where $m$ is the known slope and $\left(x_{0},y_{0}\right)$ is a point on the line.

Solution:
Finding the slope:

Note that

 $f'(x)\,\,=\,\,3x^{2}6x5,$

so the tangent line through $(3,8)$ has slope

 $m\,\,=\,\,f'(3)\,\,=\,\,3(3)^{2}6(3)5\,\,=\,\,4.$

Write the Equation of the Line:

Using the pointslope form listed in foundations, along with the point $(3,8)$ and the slope $m=4$, we find

 $y(8)\,\,=\,\,4(x3),$

or

 $y\,\,=\,\,4x20.$

Final Answer:

 $y\,\,=\,\,4x20.$

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