# What is the equation of the line normal to f(x)=2x^2-x - 1 at x=5?

Dec 25, 2015

$y = - \frac{1}{19} x + 44 \frac{5}{19}$

or

#y = -1/19 x + 841/19

#### Explanation:

Remember $m = f ' \left(x\right)$ is slope of tangent line

Slope of normal line is ${m}_{2} = \frac{1}{m}$ as it is perpendicular to the tangent line

$f \left(x\right) = 2 {x}^{2} - x - 1$ at $x = 5$

Step 1: Find derivative, to determine the slope of tangent line

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

$f ' \left(x\right) = 4 x - 1$

Step 2: Find the slope of the tangent line

$m = f ' \left(5\right) = 4 \left(5\right) - 1 = 19$

Slope of normal line is ${m}_{2} = \frac{- 1}{19}$

Step 3 Determine the $y$ coordinate, of $f \left(x\right)$ , when $x = 5$

$f \left(5\right) = 2 {\left(5\right)}^{2} - 5 - 1$

$f \left(5\right) = 50 - 5 - 1$

$f \left(5\right) = 44$

Ordered pair $\left(5 , 44\right)$

Step 4: Write the equation of the line using point slope formula

${m}_{2} = - \frac{1}{19} , \text{ " } \left(5 , 44\right)$

$y - 44 = - \frac{1}{19} \left(x - 5\right)$

$y - 44 = - \frac{1}{19} x + \frac{5}{19}$

$y = - \frac{1}{19} x + \frac{5}{19} + 44$

$y = - \frac{1}{19} x + 44 \frac{5}{19}$

or
$y = - \frac{1}{19} x + \frac{841}{19}$