What is the equation of the line normal to $f (x) = \sqrt{\frac{x}{x + 2}}$ $f(x)=sqrt(x/(x+2)$ at $x = - 3$ $x=-3$ ?

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Answer 1 · 2016-01-15T11:21:11

The Normal Line: $y = - \sqrt{3} x - 2 \sqrt{3}$ $y=-sqrt(3)x-2 sqrt(3)$

Given $f (x) = \sqrt{\frac{x}{x + 2}}$ $f(x) = sqrt (x/(x+2))$

$f (- 3) = \sqrt{3}$ $f(-3)=sqrt(3)$

the point $(- 3, \sqrt{3})$ $(-3, sqrt(3))$

$f' (-3)$ = slope for tangent line

$-1/(f' (-3))$ = slope for the Normal Line= $-sqrt(3)$

What is the equation of the line normal to f(x)=sqrt(x/(x+2) f(x)=√xx+2 f(x)=sqrt(x/(x+2) at x=-3x=−3 x=-3?