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

Question

Charles Anctil · Accepted Answer

Using the first derivative #(dy)/(dx)# we can find the gradient of the tangent, and then use point-gradient formula #y-y_1=m(x-x_1)# First, finding #(y_1,x_1)# for where #x=-1#
#y=1/(2-(-1))=1/3#
So #(x_1,y_1)# is #(-1,1/3)# Now, deriving the function, 
#y=(2-x)^-1->#changing to index form
#(dy)/(dx)=-(2-x)^-2#
Subbing in #x=-1# to #(dy)/(dx)# to find the gradient of the tangent,
#m=-(2-(-1))^-2=-1/9# Subbing #(x_1,y_1)# and #m# into the point-gradient formula,  Eq of tangent: #y-1/3=-1/9(x+1)#
#9y-3=-x-1# #:.x+9y-2=0# is the equation of the tangent to the curve at #x=-1#

Adam Adkins · Answer

undefined To find the equation of a line tangent to the function y=1/(2-x) at x=-1, we need to find the slope of the tangent line at that point and then use the point-slope form of a line to write the equation.

To find the slope, we can take the derivative of the function y=1/(2-x) with respect to x.

The derivative of y=1/(2-x) is dy/dx = 1/(2-x)^2.

Substituting x=-1 into the derivative, we get dy/dx = 1/(2-(-1))^2 = 1/9.

So, the slope of the tangent line at x=-1 is 1/9.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute x1=-1, y1=1/(2-(-1)) = 1/3, and m=1/9 into the equation.

Therefore, the equation of the line tangent to the function y=1/(2-x) at x=-1 is y - 1/3 = (1/9)(x - (-1)).