# What is the equation of the normal line of f(x)= -2x^2-10/x  at x=2 ?

Mar 22, 2016

$y = \frac{2}{11} x - \frac{147}{11}$

#### Explanation:

Things we need to do:
• Part 1 - Solve for the point at $x = 2$
• Part 2 - Getting the derivative of $f \left(x\right)$
• Part 3 - Getting slope of the tangent line at $x = 2$
• Part 4 - Getting slope of the normal line at $x = 2$
• Part 5 - Getting the equation of the normal line at $x = 2$

Part 1 - Solve for the point at $x = 2$

$\left[1\right] \text{ } - 2 {x}^{2} - \frac{10}{x}$

Simply plug substitute $2$ into $x$.

$\left[2\right] \text{ } = - 2 {\left(2\right)}^{2} - \frac{10}{2}$

$\left[3\right] \text{ } = - 8 - 5$

$\left[4\right] \text{ } = - 13$

So the point at $x = 2$ is: $\textcolor{b l u e}{\left(2 , - 13\right)}$

Part 2 - Getting the derivative of $f \left(x\right)$

$\left[1\right] \text{ } f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} - \frac{10}{x}\right)$

$\left[2\right] \text{ } f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} - 10 {x}^{-} 1\right)$

By difference rule, $\frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} - 10 {x}^{-} 1\right) = \frac{d}{\mathrm{dx}} \left(- 2 {x}^{2}\right) - \frac{d}{\mathrm{dx}} \left(10 {x}^{-} 1\right)$

$\left[3\right] \text{ } f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(- 2 {x}^{2}\right) - \frac{d}{\mathrm{dx}} \left(10 {x}^{-} 1\right)$

By power rule, $\frac{d}{\mathrm{dx}} \left(- 2 {x}^{2}\right) = - 4 x$
By power rule, $\frac{d}{\mathrm{dx}} \left(10 {x}^{-} 1\right) = - 10 {x}^{-} 2 = - \frac{10}{x} ^ 2$

$\left[4\right] \text{ } \textcolor{b l u e}{f ' \left(x\right) = - 4 x + \frac{10}{x} ^ 2}$

Part 3 - Getting slope of the tangent line at $x = 2$

$\left[1\right] \text{ } f ' \left(x\right) = - 4 x + \frac{10}{x} ^ 2$

$\left[2\right] \text{ } f ' \left(2\right) = - 4 \left(2\right) + \frac{10}{2} ^ 2$

$\left[3\right] \text{ } f ' \left(2\right) = - 8 + \frac{5}{2}$

$\left[4\right] \text{ } \textcolor{b l u e}{f ' \left(2\right) = {m}_{T L} = - \frac{11}{2}}$

Part 4 - Getting slope of the normal line at $x = 2$

Since the normal line is the line perpendicular to the tangent line, we can easily find the slope of the normal line by getting the negative reciprocal of the slope of the tangent line.

$\left[1\right] \text{ } {m}_{N L} = - \frac{1}{M} _ \left(T L\right)$

$\left[2\right] \text{ } {m}_{N L} = - \frac{1}{- \frac{11}{2}}$

$\left[3\right] \text{ } \textcolor{b l u e}{{m}_{N L} = \frac{2}{11}}$

Part 5 - Getting the equation of the normal line at $x = 2$

We will use the point-slope form to find the equation of the normal line, where $m$ is the slope and $\left({x}_{0} , {y}_{0}\right)$ is a point on the line.

$\left[1\right] \text{ } y - {y}_{0} = m \left(x - {x}_{0}\right)$

Plug in the slope of the normal line.

$\left[2\right] \text{ } y - {y}_{0} = \frac{2}{11} \left(x - {x}_{0}\right)$

Plug in the point $\left(2 , - 13\right)$.

$\left[3\right] \text{ } y - \left(- 13\right) = \frac{2}{11} \left(x - 2\right)$

$\left[4\right] \text{ } y = \frac{2}{11} x - \frac{4}{11} - 13$

$\left[5\right] \text{ } y = \frac{2}{11} x - \frac{4}{11} - \frac{143}{11}$

$\left[6\right] \text{ } \textcolor{b l u e}{y = \frac{2}{11} x - \frac{147}{11}}$