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

Answer 1

#x-6y=-52#

Given #f(x)=(5-x)^2# #color(white)("XXX")rarr f(x)=25-10x+x^2# Its general slope is given by the derivative: #color(white)("XXX")f'(x) = 2x-10#

At #x=2# this slope will be #color(white)("XXX")2(2)-10 = -6# and a line normal to it (i.e. perpendicular to it) will have a slope #color(white)("XXX")=-(1/(-6))=1/6#

At #x=2, f(2)= (5-2)^2=9# giving a point #(2,9)#

Using the slope-point form for the normal line #color(white)("XXX")y-9=1/6(x-2)#

We can simplify this as #color(white)("XXX")6y-54=x-2# or (in standard form) #color(white)("XXX")x-6y=-52# graph{(y-(5-x)^2)(x-6y+52)=0 [-7.83, 14.67, -0.585, 10.665]}

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Answer 2

Paisley Johnson

The equation of the line normal to f(x)=(5-x)^2 at x=2 is y = -3x + 19.

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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.