Question

determine the graphs end behavior. find the x-intercepts and y-intercept. determine whether the graph has symmetry. determine the graph of the function.\\( f(x)=x^3 + 2x^2 - x - 2 \\)\
\\( \text{a. use the leading coefficient test to determine the graphs end behavior. which statement describes the behavior at the ends of } f(x)=x^3 + 2x^2 - x - 2? \\)\
\\( \bigcirc \text{a. the graph falls to the left and to the right.} \\)\
\\( \bigcirc \text{b. the graph rises to the left and falls to the right.} \\)\
\\( \bigcirc \text{c. the graph falls to the left and rises to the right.} \\)\
\\( \bigcirc \text{d. the graph rises to the left and to the right.} \\)\
\\( \text{b. what are the x-intercepts?} \\)\
\\( x = -2, -1, 1 \\) (use a comma to separate answers as needed.)\
\\( \text{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.} \\)\
\\( \bigcirc \text{a. } x = \\) (use a comma to separate answers as needed.)\
\\( \bigcirc \text{b. there are no x-intercepts at which the graph crosses the x-axis.} \\)

Explanation:

Part a

Step 1: Identify the degree and leading coefficient

The function \( f(x) = x^3 + 2x^2 - x - 2 \) is a polynomial. The degree of the polynomial is the highest power of \( x \), which is \( 3 \) (odd), and the leading coefficient is the coefficient of the highest power term, which is \( 1 \) (positive).

Step 2: Apply the leading coefficient test

For a polynomial with an odd degree:

• If the leading coefficient is positive, as \( x \to -\infty \) (left end), \( f(x) \to -\infty \) (the graph falls to the left), and as \( x \to +\infty \) (right end), \( f(x) \to +\infty \) (the graph rises to the right).

Step 1: Find x - intercepts

To find the x - intercepts, we set \( f(x)=0 \), so \( x^3 + 2x^2 - x - 2 = 0 \). We can factor the polynomial by grouping:
\[

$$\begin{align*} x^2(x + 2)-1(x + 2)&=0\\ (x^2 - 1)(x + 2)&=0\\ (x - 1)(x + 1)(x + 2)&=0 \end{align*}$$

\]
Setting each factor equal to zero gives \( x - 1 = 0\) or \( x + 1 = 0\) or \( x + 2 = 0 \), so \( x = 1\), \( x=-1\), \( x = - 2 \).

Step 2: Determine where the graph crosses the x - axis

For a polynomial function, if the multiplicity of a root (x - intercept) is odd, the graph crosses the x - axis at that root. The roots \( x=-2\), \( x = - 1\), and \( x = 1\) all have multiplicity 1 (odd), so the graph crosses the x - axis at \( x=-2\), \( x=-1\), \( x = 1 \).

Answer:

C. The graph falls to the left and rises to the right.

Part b (x - intercepts and crossing x - axis)