What is the equation of the normal line of $f(x)=5x^3-2x^2-3x-1$ at $x=-8$ ?

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Answer 1 · 2017-07-30T06:27:55

Equation of normal is $x+989y+2635693=0$

At $x=-8$ , $f(x)=5(-8)^3-2(-8)^2-3(-8)-1$

$=-2560-128+24-1=-2665$

Hencewe are seeking normal at $(-8,-2665)$

Slope of tangent is given by $f'(-8)$ and as $f'(x)=15x^2-4x-3$

slope of tangent is $15(-8)^2-4(-8)-3=989$

and hence slope of normal is $-1/989$ and

equation of normal is $y+2665=-1/989(x+8)$

or $x+989y+2635693=0$

What is the equation of the normal line of f(x)=5x^3-2x^2-3x-1f(x)=5x^3-2x^2-3x-1 at x=-8x=-8?