# What is the equation of the line with slope 3 which is tangent to the curve f(x)=7x-x^2?

Sep 1, 2016

$y = 3 x + 4$

#### Explanation:

If $f \left(x\right) = 7 x - {x}^{2}$
then the slope (for any general $x$ value) is $f ' \left(x\right) = 7 - 2 x$

when the slope is $m = f ' \left(x\right) = 3$
then $7 - 2 x = 3$
$\textcolor{w h i t e}{\text{XX}} - 2 x = - 4$
$\textcolor{w h i t e}{\text{XX}} x = 2$

If $x = 2$ then
$\textcolor{w h i t e}{\text{XXX}} f \left(\textcolor{red}{2}\right) = 7 \cdot \left(\textcolor{red}{2}\right) - {\textcolor{red}{2}}^{2} = 14 - 4 = \textcolor{b l u e}{10}$

and the point on the curve were $f ' \left(x\right) = 3$ occurs at $\left(\textcolor{red}{2} , \textcolor{b l u e}{10}\right)$

Therefore, for the tangent, we have a slope of $\textcolor{g r e e n}{m} = 3$ and a point $\left(\textcolor{red}{2} , \textcolor{b l u e}{10}\right)$

Using the slope-point form of the equation:
$\textcolor{w h i t e}{\text{XXX}} y - \textcolor{b l u e}{10} = \textcolor{g r e e n}{3} \left(x - \textcolor{red}{2}\right)$

or
$\textcolor{w h i t e}{\text{XXX}} y = 3 x + 4$