Question

determine the graph’s 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\\)\
a. use the leading coefficient test to determine the graph’s end behavior. which statement describes the behavior at the ends of \\(f(x)=x^{3}+2x^{2}-x - 2\\)?\
\\(\bigcirc\\) a. the graph falls to the left and to the right.\
\\(\bigcirc\\) b. the graph rises to the left and falls to the right.\
\\(\bigcirc\\) c. the graph falls to the left and rises to the right.\
\\(\bigcirc\\) d. the graph rises to the left and to the right.\
b. what are the x-intercepts?\
\\(x = \square\\) (use a comma to separate answers as needed.)

Explanation:

Part a

To determine the end - behavior of a polynomial function \(f(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\) using the leading - coefficient test:

• If the degree \(n\) of the polynomial is odd:
• If the leading coefficient \(a_n>0\), the graph falls to the left and rises to the right.
• If the leading coefficient \(a_n < 0\), the graph rises to the left and falls to the right.
• If the degree \(n\) of the polynomial is even:
• If the leading coefficient \(a_n>0\), the graph rises to the left and rises to the right.
• If the leading coefficient \(a_n < 0\), the graph falls to the left and falls to the right.

For the function \(f(x)=x^3 + 2x^2-x - 2\), the degree \(n = 3\) (which is odd) and the leading coefficient \(a_n=1>0\). So, the graph falls to the left and rises to the right.

Step 1: Factor the polynomial

We are given the function \(f(x)=x^3+2x^2 - x - 2\). We can factor by grouping.
Group the terms as follows: \((x^3 + 2x^2)+(-x - 2)\)
Factor out the greatest common factor from each group:
From the first group \(x^3+2x^2\), we can factor out \(x^2\) to get \(x^2(x + 2)\).
From the second group \(-x - 2\), we can factor out \(- 1\) to get \(-1(x + 2)\).
So, \(f(x)=x^2(x + 2)-1(x + 2)\)
Now, factor out the common binomial factor \((x + 2)\): \(f(x)=(x + 2)(x^2-1)\)

Step 2: Factor the difference of squares

We know that \(a^2 - b^2=(a + b)(a - b)\). For \(x^2-1\), where \(a = x\) and \(b = 1\), we have \(x^2-1=(x + 1)(x - 1)\)
So, the factored form of the function is \(f(x)=(x + 2)(x + 1)(x - 1)\)

Step 3: Find the x - intercepts

To find the x - intercepts, we set \(f(x)=0\).
So, \((x + 2)(x + 1)(x - 1)=0\)
Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\) (or both).
Set each factor equal to zero:

• \(x+2 = 0\) gives \(x=-2\)
• \(x + 1=0\) gives \(x=-1\)
• \(x - 1=0\) gives \(x = 1\)

Answer:

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

Part b