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

2 Answers

4 votes

Answer:

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

Explanation:

The other person is correct

User Leviathanbadger
by
7.4k points
2 votes

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

Consider the function below. f(x)= x^3 + 2x^2 - x - 2 plot the x and y intercepts-example-1
User Torjescu Sergiu
by
8.1k points

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