Question

Find the intercepts and relative extrema for the graph of the function.
1. f(x)=1/4(x+2)(x-1)^2

Answer 1

for x-intercept solve:
1/4(x+2)(x-1)^2=0
for y- intercept solve:
1/4(0+2)(0-1)^2=f(0)
for extrema find derivative f'(x) and solve f'(x)=0

Answer 2

for x-inter I got -2 ,1
for y-inter I got 1/2
but im confused how to find the extrema

Answer 3

1º find the derivative

Answer 4

use the rule of derivative of product: (uv)'=u'v+uv'
where u=1/4(x+2) and v=(x-1)^2

Answer 5

((1/4)(x+2)(x-1)^2)=(1/4)(x+2)(x-1)^2+(1/4)(x+2)(x-1)^2

Answer 6

((1/4)(x+2)(x-1)^2)=1/2 (-1+x)^2 (2+x) ??

Answer 7

u'=1/4
v'=2(x-1)=2x-2
So:
(uv)'=u'v+uv'=1/4(x-1)^2+1/4(x+2)2(x-1)=1/4(x-1)(x-1+2x+4)=1/4(x-1)(3x+3)
now set it to 0:
1/4(x-1)(3x+3)=0
so at x=1 and x=-1 are the extrema.
To find the actual value of the extrema plug this x values nto original equation and find f(x)