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

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Answer 1 · 2016-03-26T03:02:59

#y=7x-8#

Find the point the tangent line will pass through by finding

#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#

Differentiating term by term, we see that the function's derivative 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 can also be written as

#y=7x-8#

Graphed are the function and its tangent line:

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