What is the equation of the line normal to $f(x)=-2x^2 +3x - 4$ at $x=-2$ ?

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Answer 1 · 2016-05-27T07:12:39

$y=-1/11x-18 2/11$

Given -

$y=-2x^2+3x-4$

The first derivative gives the slope at any given point

$dy/dx=-4x+3$

At $x=-2$ its slope is

$dy/dx=-4(-2)+3=8+3=11$

It can be taken as the slope of the Tangent, drawn to that point on the curve.

Y- co-ordinate of the point -

$y=-2(-2)^2+3(-2)-4$
$y=-8-6-4=-18$

The point on the curve, the normal passing through is $(-2, -18)$

If the two lines are perpendicular then -

$m_2 xxm_2=-1$
$11 xx m_2=-1$
$m_2=(-1)/11$

The equation of the normal is -

$mx+c = y$
$-1/11(-2)+c=-18$
$2/11+c=-18$
$c=-18-2/11=18 2/11$

Equation is -

$y=-1/11x-18 2/11$

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What is the equation of the line normal to f(x)=-2x^2 +3x - 4 f(x)=-2x^2 +3x - 4 at x=-2x=-2?