Question

How do you find the critical points to sketch the graph `y=x(x+2)^2`?

Answer 1

Please see below.

Explanation:

A critical number for a function `f` is a number in the domain of `f` at which either the derivative of `f` is `0` or the derivative fails to exist.

`f(x) = x(x+2)^2` has domain `(-oo,oo)`

To find `f'(x)` either expand the polynomial or use the product rule.

`f'(x) = (x+2)^2 + x 2(x+2)`

`= (x+2)[(x+2)+2x)`

`= (x+2)(3x+2)`

`f'(x) = (x+2)(3x+2)`

`f'(x)` is defined for all `x` and `f'(x) = 0` at

`x=-2` and at `x=-2/3`, both of which are in the domain of `f`.

The critical numbers are `-2` and `-2/3`.

If you have been taught that critical point is the same as critical number , then you are done.

You will answer: the critical points are `-2` and `-2/3`.

If you have been taught that a critical point is a point on a graph , then you still need to find the corresponding `y` values using
`y = f(x) = x(x+2)^2`

`x=-2` `rArr` `y = 0`
`x=-2/3` `rArr` `y = -32/27`

So you will answer: the critical points are `(-2,0)` and `(-2/3,-32/27)`