For #f(x) =(x-3)/(x-4)#, what is the equation of the line tangent to #x =3#?

1 Answer
Mar 10, 2018

#y=-x+3#

Explanation:

In order to find the equation of the tangent line at #x=3#, we first need to find the gradient of this line at #x=3#.

To do this we find the derivative of #f(x)#. In this case we need to use the quotient rule:

The quotient rule states that for functions #f(x),g(x)#:

#(f/g)^'=(f^'g-fg^')/g^2#

Let:

#f(x)=x-3# and #g(x)=x-4#

#:.#

#((x-3)/(x-4))^'=(1(x-4)-(x-3)1)/(x-4)^2=-1/(x-4)^2#

We now plug #x=3# into this derivative.

#-1/((3)-4)^2=-1#

So gradient #m=-1#

Using the original function to calculate a #y# value:

#f(3)=(x-3)/(x-4)=((3)-3)/((3)-4)=0#

Using point slope form of a line:

#(y_2-y_1)=m(x_2-x_1)#

#y-0=-1(x-3)#

#y=-x+3#

This is the equation of the tangent line at #x=3#

Graph:

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