What is the slope of the line normal to the tangent line of f(x) = e^(x^2-1)+2x-2  at  x= 1 ?

May 16, 2017

$y = - \frac{1}{82.34} x + 22.09$

Explanation:

Given

$y = {e}^{{x}^{2} - 1} + 2 x - 2$

At $x = 2$

$y = {2.71828}^{{2}^{2} - 1} + 2 \left(2\right) - 2$
$y = {2.71828}^{3} + 4 - 2$
$y = 20.085 + 2 = 22.085$

At Point $\left(2 , 22.085\right)$ There is a tangent to the curve.
Look at the graph

The slope of the tangent is equal to the slope of the given curve.

The slope of the given curve is-

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {e}^{{x}^{2} - 1} + 2$

At $x = 2$ the slope of the curve is

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(2\right) \left({2.71828}^{3}\right) + 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 \times 20.085 + 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 82.34$

The slope of the tangent ${m}_{1} = 82.34$

The slope of the normal is ${m}_{2} = - 1 \times \frac{1}{82.34} = \frac{- 1}{82.34}$

The equation of the Normal is -

$y = {m}_{2} x + c$
$22.085 = - \frac{1}{82.34} \left(2\right) + c$
$22.085 + \frac{1}{82.34} = c$
$22.09 = c$

$y = - \frac{1}{82.34} x + 22.09$