Question

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

Answer 1

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:

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

Answer 2

Answer:

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

Step-by-step explanation:

The other person is correct