What is the equation of the line tangent to # f(x)=(x-3)^2-x^2-3# at # x=5#?

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Answer 1 · 2016-01-09T05:51:33

The equation of the tangent is-

#y=-6x+6#

Given -

#y=(x-3)^2-x^2-3#

Its slope is given by its first derivative -

#dy/dx=2(x-3)(1)-2x#

Slope at #x=5#

At #x=5; slope = 2(5-3)(1)-2(5)=4-10=-6#

At #x=5# the y-co-ordinate is -

At #x=5;y=(5-3)^2-5^2-3#

#y= 4-25-3=-24#

#(5,24)# is the point on the curve. At that point the slope is #-6#

The equation of the tangent is -

#y=mx+c#
#c+mx=y#
#c+(-6)(5)=-24#
#c-30=-24#
#c=-24+30=6#

#y=-6x+6#

Answer 2 · 2016-01-09T05:59:46

Though it appears to be a quadratic equation, it is not.

Simplify the function.

#y=(x-3)^2-x^2-3#
#y=x^2-6x+9-x^2-3#
#y=-6x+6#

In reality it is linear. Hence a tangent drawn to any point on this curve will coincide with it. The tangent at #x=5# is -

#y=-6x+6#