What is the equation of the normal line of $f(x)=x^3 + 3x^2 + 7x - 1$ at $x=-1$ ?

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Answer 1 · 2016-08-29T21:51:26

$y=x/4+23/4$

$f(x)=x ^3+3x^2+7x-1$
The gradient function is the first derivative
$f'(x)=3x^2+6x+7$

So the gradient when X=-1 is 3-6+7=4
The gradient of the normal , perpendicular, to the tangent is $-1/4$

If you are not sure about this draw a line with gradient 4 on squared paper and draw the perpendicular.

So the normal is $y=-1/4x+c$

But this line goes through the point (-1,y)
From original equation when X=-1 y=-1+3-7-1=6

So 6= $-1/4*-1+c$
$C=23/4$

What is the equation of the normal line of f(x)=x^3 + 3x^2 + 7x - 1 f(x)=x^3 + 3x^2 + 7x - 1 at x=-1 x=-1 ?