Home > Calculus > Tangent Line to a Curve

What is the equation of the line tangent to # f(x)=-x^2+4x-5/x+1# at # x=1 #?

Answer 1

#y=7x-8#

Locate the point through which the tangent line will pass by using

#f(1)=-(1^2)+4(1)-5/1+1=-1#

We know the tangent line will pass through the point #(1,-1)#.

Now, all we need to know is the slope of the tangent line, which we can find by evaluating the derivative of the function at #x=1#.

To find the derivative of #f(x)#, first rewrite the function:

#f(x)=-x^2+4x-5x^-1+1#

When we differentiate term by term, we observe that the derivative of the function equals

#f'(x)=-2x+4+5x^-2=-2x+4+5/x^2#

Hence the slope of the tangent line at #x=1# is

#f'(1)=-2(1)+4+5/1^2=7#

We can relate the slope of the tangent line, #7#, with a point we know it to pass through, #(1,-1)#, as a line in point-slope form:

#y-(-1)=7(x-1)#

which is also able to be written as

#y=7x-8#

The function and its tangent line are graphed:

graph{[-2, 6, -11, 10]} = -x^2+4x-5x^-1+1-y)(y-7x+8)=0

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Elijah Abram

The equation of the line tangent to f(x)=-x^2+4x-5/x+1 at x=1 is y = -2x + 1.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.