# How do you find the equation of a line normal to the function y=(x^2-1)/(2x+3) at x=-1?

Mar 23, 2017

Use the quotient rule to find $y ' \left(x\right)$
The slope of the tangent line is $m = - \frac{1}{y ' \left(- 1\right)}$
Find the y coordinate, ${y}_{0} = y \left(- 1\right)$
Use the point slope form on the equation of a line $y = m \left(x - - 1\right) + {y}_{0}$

#### Explanation:

Use the quotient rule to find $y ' \left(x\right)$

$y ' \left(x\right) = \left(\frac{u \left(x\right)}{g \left(x\right)}\right) ' = \frac{u ' \left(x\right) g \left(x\right) - u \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

let $u \left(x\right) = {x}^{2} + 1 \mathmr{and} g \left(x\right) = 2 x + 3$, then $u ' \left(x\right) = 2 x \mathmr{and} g ' \left(x\right) = 2$

$y ' \left(x\right) = \frac{\left(2 x\right) \left(2 x + 3\right) - 2 \left({x}^{2} + 1\right)}{2 x + 3} ^ 2$

$y ' \left(x\right) = \frac{4 {x}^{2} + 6 x - 2 {x}^{2} + 2}{2 x + 3} ^ 2$

$y ' \left(x\right) = \frac{2 {x}^{2} + 6 x + 2}{2 x + 3} ^ 2$

The slope of the tangent line is:

m = -1/y'(-1)

m = -(2(-1)+3)^2/((2(-1)^2+6(-1)+2)#

$m = \frac{1}{2}$

Find the y coordinate, ${y}_{0} = y \left(- 1\right)$

${y}_{0} = \frac{- {1}^{2} + 1}{2 \left(- 1\right) + 3}$

${y}_{0} = 0$

Use the point slope form on the equation of a line $y = m \left(x - - 1\right) + k$

$y = \frac{1}{2} \left(x - - 1\right)$