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What is the equation of the line normal to #f(x)=x^2-x # at #x=-2#?

Answer 1

#y=1/5x+32/5#

The normal line will intersect the curve at #(-2,f(-2))#. Since #f(-2)=(-2)^2-(-2)=6#, we know the normal line passes through the point #(2,6)#.

To find the slope of the normal line, first find the slope of the tangent line at that point by finding the value of the derivative at #x=-2#. Then, since the normal line is perpendicular to the tangent line, take the opposite reciprocal of the slope of the tangent line.

Through the power rule, we see that #f'(x)=2x-1#. We then see that the slope of the tangent line at #x=-2# is #f'(-2)=2(-2)-1=-5#.

The slope of the normal line is then #-1/(-5)=1/5#.

Using the point #(-2,6)# and slope of #1/5#, we can write the equation of the line from #y=mx+b#:

#6=1/5(-2)+b" "=>" "b=32/5#

The normal line is:

#y=1/5x+32/5#

Graphed are #f(x)# and the normal line:

graph{(y-x^2+x)(-y+1/5x+32/5)=0 [-15.61, 12.86, -1.46, 12.77]}

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Answer 2

Cameron Alexander

The equation of the line normal to f(x)=x^2-x at x=-2 is y = -5x - 6.

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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.