How do you find the equation of the tangent and normal line to the curve #y=x^2# at x=-3?

1 Answer
Aug 26, 2016

Equation of the tangent

#y=-6x-9#

Equation of the Normal

#y=1/6x+19/2#

Explanation:

Given -

#y=x^2#

It is a quadratic function. The curve is a upward facing parabola.

Its first derivative gives the slope at any given point on the curve.

#dy/dx=2x#

Slope exactly at #x=-3#

#m_1=2(-3)=-6#

At #x=-3 ; y=(-3)^2=9#

The tangent and Normal are passing through the point #(-3, 9)#

Slope of the tangent is #m_2=-6#

Steps to get Equation of the tangent -

#mx+c=y#

#(-6)(-3)+c=9#

#18+c=9#

#c=9-18=-9#

Equation of the tangent is -

#y=-6x-9#

Steps to get Equation of the Normal -

Normal's slope #m_3=-(1/(m_2))=-(1/(-6))=1/6#

#mx+c=y#
#1/6(-3)+c=9#
#-1/2+c=9#
#c=9+1/2=(18+1)/2=19/2#

Equation of the Normal -

#y=1/6x+19/2#

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