Question

Consider the function below. f(x)= x^3 + 2x^2 - x - 2 plot the x and y intercepts of the function

Answer 1

Plot all four of these points on the graph (-2,0) (-1,0) (0,-2) (1,0)

Explanation:

The other person is correct

Answer 2

In the Figure below is shown the graph of this function. We have the following function:

`f(x)=x^3+2x^2-x-2`

The
`y-intercept` occurs when
`x=0`, so:

`f(0)=(0)^3+2(0)^2-(0)-2=-2`

Therefore, the
`y-intercept` is the given by the point:

`\boxed{(0,-2)}`

From the figure we have three
`x-intercepts`:

`\boxed{P_(1)(-2,0)} \\ \boxed{P_(2)(-1,0)} \\ \boxed{P_(3)(1,0)}`

So, the
`x-intercepts` occur when
`y=0`. Thus, proving this:

`f(x)=x^3+2x^2-x-2 \\ \\ For \ P_(1):\\ If \ x=-2, \ y=(-2)^3+2(-2)^2-(-2)-2=0 \\ \\ For \ P_(2):\\ If \ x=-1, \ y=(-1)^3+2(-1)^2-(-1)-2=0 \\ \\ For \ P_(3):\\ If \ x=1, \ y=(1)^3+2(1)^2-(1)-2=0`