What is the equation of the normal line of $f(x)=sqrt(3x^2+2x)$ at $x=2$ ?

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Answer 1 · 2015-11-10T02:08:08

$y=-4/7x+36/7$

$f'(x)=frac{d}{dx}((3x^2+2x)^{1/2})$

$=1/2(3x^2+2x)^{-1/2}(6x+2)$

The tangent line should pass through the point $(2, f(2))$ , and have a gradient of $-1/{f'(2)}$ .

$f(2)=4$

$f'(2)=7/4$

Equation of tangent line:

$frac{y-f(2)}{x-2}=-1/{f'(2)}$

$frac{y-4}{x-2}=-4/7$

$y=-4/7x+36/7$

What is the equation of the normal line of f(x)=sqrt(3x^2+2x)f(x)=sqrt(3x^2+2x) at x=2 x=2 ?