What is the equation of the line normal to # f(x)=(x-1)(x+2) # at # x=0#?

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Answer 1 · 2018-01-14T22:39:44

#y=x-2#

#f(x) = (x-1)(x+2)#

Firstly, just expand the brackets so:

#f(x) = x^2+x-2#

Differentiate:

#f'(x)= 2x+1#

Find #f'(0)#

#f'(0) = 2(0)+1=1#

So we now the gradient of the line is is #1#.

We need a point on the line and we know it should pass through the point: #(0,f(0))#. So find this point:

#f(0) = (0)^2+(0)-2=-2#

So #m=1# and #(a,b)=(0,-2)#.

Substitute into the equation of a line:

#y-b=m(x-a)#

#y-(-2)=(1)(x-0)#

#->y=x-2#

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