How do you find the equation of a line normal to the function y=(x^2-1)/(2x+3)y=x212x+3 at x=-1?

1 Answer
Mar 23, 2017

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)