Question

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

Answer 1

Use the quotient rule to find `y'(x)`
The slope of the tangent line is `m = -1/(y'(-1))`
Find the y coordinate, `y_0 = y(-1)`
Use the point slope form on the equation of a line `y = m(x--1)+y_0`

Explanation:

Use the quotient rule to find `y'(x)`

`y'(x) = ((u(x))/(g(x)))' = (u'(x)g(x)-u(x)g'(x))/(g(x))^2`

let `u(x) = x^2+1 and g(x)=2x+3`, then `u'(x) =2x and g'(x)=2`

`y'(x) = ((2x)(2x+3)-2(x^2+1))/(2x+3)^2`

`y'(x) = (4x^2+6x-2x^2+2)/(2x+3)^2`

`y'(x) = (2x^2+6x+2)/(2x+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 = 1/2`

Find the y coordinate, `y_0 = y(-1)`

`y_0 = (-1^2+1)/(2(-1)+3)`

`y_0 = 0`

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

`y = 1/2(x--1)`