Question

answer parts (a)–(e) for the function shown below. f(x)=x³ + 2x² - x - 2
d. the graph rises to the left and to the right.
b. find the x - intercepts. state whether the graph crosses the x - axis, or touches the x - axis and turns around, at each intercept. what are the x - intercepts?
x = - 2, - 1,1 (type an integer or a decimal. use a comma to separate answers as needed.)
at which x - intercept(s) does the graph cross the x - axis? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. x = - 2, - 1,1 (type an integer or a decimal. use a comma to separate answers as needed.)
b. there are no x - intercepts at which the graph crosses the x - axis.
at which x - intercept(s) does the graph touch the x - axis and turn around? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. x = (type an integer or a decimal. use a comma to separate answers as needed.)
b. there are no x - intercepts at which the graph touches the x - axis and turns around.
c. find the y - intercept.
the y - intercept is y = (type an integer or a decimal.)

Explanation:

Step1: Find x - intercepts

Set $f(x)=0$, so $x^{3}+2x^{2}-x - 2 = 0$. Factor by grouping: $x^{2}(x + 2)-(x + 2)=0$, then $(x + 2)(x^{2}-1)=0$. Further factor $x^{2}-1$ to $(x + 1)(x - 1)$. So $(x + 2)(x + 1)(x - 1)=0$. Solving gives $x=-2,-1,1$.

Step2: Determine crossing or touching x - axis

For a polynomial function $y = f(x)$ and a root $r$ of $f(x)$, if the multiplicity of $r$ is odd, the graph crosses the x - axis at $x = r$, if even, it touches and turns around. The roots $x=-2,-1,1$ all have multiplicity 1 (odd), so the graph crosses the x - axis at $x=-2,-1,1$.

Step3: Find y - intercept

Set $x = 0$ in $f(x)$. Then $f(0)=0^{3}+2\times0^{2}-0 - 2=-2$.

Answer:

b. x - intercepts: $x=-2,-1,1$; The graph crosses the x - axis at $x=-2,-1,1$
c. y - intercept: $y=-2$