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

Question

Christopher Allers · Accepted Answer

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# 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)#

Isabella Ackerman · Answer

undefined To find the equation of a line normal to a function at a given point, follow these steps:

1. Find the derivative of the function.
2. Evaluate the derivative at the given point to find the slope of the tangent line.
3. Determine the negative reciprocal of the slope to find the slope of the line normal to the function.
4. Use the point-slope form of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope of the line, to find the equation of the line normal to the function.

For the given function y = (x^2 - 1)/(2x + 3) and x = -1:

1. Find the derivative of the function:
   y' = (2x(2x + 3) - (x^2 - 1)(2))/(2x + 3)^2

2. Evaluate the derivative at x = -1:
   y'(-1) = (2(-1)(2(-1) + 3) - ((-1)^2 - 1)(2))/(2(-1) + 3)^2

3. Simplify the expression to find the slope of the tangent line:
   y'(-1) = -8/25

4. Determine the negative reciprocal of the slope:
   Slope of the line normal = -1/(y'(-1)) = -25/8

5. Use the point-slope form with the given point (-1, y(-1)):
   y - y₁ = m(x - x₁)
   y - y(-1) = (-25/8)(x - (-1))

6. Simplify the equation:
   y + (x + 1)(25/8) = 0

Therefore, the equation of the line normal to the function y = (x^2 - 1)/(2x + 3) at x = -1 is y + (x + 1)(25/8) = 0.