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]}