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What is the equation of the line normal to # f(x)=(x+5)(x+2)+2x^2-8x+2# at # x=0#?

Answer 1

It is #y=x+12#.

The strategy is to find first the slope of the tangent line than we can easily find the orthogonal.

The slope of the tangent is given by the derivative evaluate in #x=0# Before doing the derivative I do the product between the parenthesis and reorder the terms

#f(x)=(x+5)(x+2)+2x^2-8x+2=x^2+2x+5x+10+2x^2-8x+2# #=3x^2-x+12#.

Now for the derivative

#f'(x)=d/dxf(x)=d/dx 3x^2-x+12=6x-1# then when #x=0# we have #f'(0)=-1#.

This is the the slope of the tangent line passing from #x=0#.

The slope of the normal line is the opposite inverse of this:

#m=-1/-1=1#. We need now to find the intercept and to do this we impose the passage for the point #(0, f(0))#. #f(0)=3*0^2-0+12=12#, so the normal line passes from #(0,12)#.

The equation is #y=mx+q# with #m=1# it is #y=x+q# and not I substitute the point #(0, 12)# #12=q#

Then the final equation of the normal line is

#y=x+12#.

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Answer 2

Emily Ammerman

The equation of the line normal to f(x)=(x+5)(x+2)+2x^2-8x+2 at x=0 is y = -5x + 2.

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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.