Question

answer parts (a)-(e) for the function shown below. f(x)=x³ + 2x² - x - 2 a. use the leading - coefficient test to determine the graphs end behavior. which statement describes the behavior at the ends of f(x)=x³ + 2x² - x - 2? a. the graph falls to the left and rises to the right. b. the graph rises to the left and falls to the right. c. the graph falls to the left and to the right. 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 = (type an integer or a decimal. use a comma to separate answers as needed.)

Explanation:

Step1: Analyze leading - coefficient and degree

The function $f(x)=x^{3}+2x^{2}-x - 2$ is a cubic function ($n = 3$, odd) with leading coefficient $a = 1>0$. For an odd - degree polynomial with a positive leading coefficient, as $x\to-\infty$, $y\to-\infty$ and as $x\to+\infty$, $y\to+\infty$. So the graph falls to the left and rises to the right.

Step2: Find x - intercepts

Set $f(x)=0$, so $x^{3}+2x^{2}-x - 2 = 0$. Group the terms: $(x^{3}+2x^{2})-(x + 2)=0$. Factor out common factors from each group: $x^{2}(x + 2)-(x + 2)=0$. Then $(x + 2)(x^{2}-1)=0$. Using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$ on $x^{2}-1=(x + 1)(x - 1)$, we get $(x + 2)(x + 1)(x - 1)=0$. Setting each factor equal to zero gives $x=-2,x=-1,x = 1$. Since the multiplicity of each factor is 1 (odd), the graph crosses the x - axis at each x - intercept.

Answer:

a. A. The graph falls to the left and rises to the right.
b. $x=-2,-1,1$