What is the equation of the line normal to $f(x)=-4x^2 -5x +1$ at $x=-3$ ?

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Answer 1 · 2016-02-20T23:15:45

$y=-x/19-383/19$

Let $y=-4x^2-5x+1$

$y'=-8x-5$

This is the gradient of the tangent $m$ .

At $x=-3$ this becomes:

$m=(-8xx-3)-5=19$

If the gradient of the normal is $m'$ then:

$m'm=-1$

$:.m'=-1/19$

To get the value of $y$ at $x=3rArr$

$y=-4xx(-3)^2-(5xx-3)+1$

$y=-20$

The equation of the normal line is of the form:

$y=m'x+c$

$:.-20=(-1/19xx-3)+c$

$:.c=-20-3/19=-380/19-3/19=-383/19$

So the equation is:

$y=-x/19-383/19$

This is shown here:

graph{(-4x^2-5x+1-y)(-x/19-383/19-y)=0 [-80, 80, -40, 40]}

What is the equation of the line normal to f(x)=-4x^2 -5x +1 f(x)=-4x^2 -5x +1 at x=-3x=-3?