# What is the equation of the line normal to # f(x)=sqrt(1/(x+2) # at # x=-1#?

Its equation is

graph{(ysqrt(x+2)-1)(2x-y+3)=0x^2 [-10, 10, -5, 5]}

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The equation is

We start by finding the y-coordinate at the given x-point.

We can start by simplifying the function.

The normal line is perpendicular to the tangent, that's to say its slope is the negative reciprocal of that of the tangent.

Hopefully this helps!

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The equation of the line normal to f(x) = sqrt(1/(x+2)) at x = -1 is y = -2(x+1) + 1.

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