Question

1. which of these polynomial functions is graphed below?

a. y= 2x³ -3x² b. f(x)=x³ - 2x² c. y= -2x³ +3x² d. y = x³ - 3

1. . what is the factored form of x³ + 2x² - 3x = 0?

a. x (x + 1)(x + 3)
b. (x + 1)(x - 3)
c. x (x - 1)(x + 3)
d. x (x - 1)(x - 3)

1. factor the polynomial, then determine the real zeros of the function: f(x) = x⁴ - 81

a. 3i, -3i, 3, -3
b. 0
c. 3, -3
d. 9

1. . state the number of x-intercepts for the polynomial function: h(x) = x(x + 5)(x - 4)²

a. 4 b. 3 c. 6 d. 7

1. what is the domain and range for the following graph?

a. domain (-∞, ∞); range (-∞, ∞)
b. domain (-2.5, ∞); range (-22.438, ∞)
c. domain (-∞, ∞); range (-22.438, ∞)
d. domain (-2.5, ∞); range (-∞, ∞)

Explanation:

Question 5

Step1: Analyze end behavior

The graph rises to the right, so the leading coefficient is positive. The degree is 3 (cubic).

Step2: Analyze roots/zeros

The graph has a root at \( x = 0 \) (with multiplicity 2, since it touches and turns) and another root. Let's check the options:

• Option A: \( y = 2x^3 - 3x^2 = x^2(2x - 3) \). Roots at \( x = 0 \) (multiplicity 2) and \( x = \frac{3}{2} \). The graph's shape (steepness, turning points) matches when we consider the leading coefficient 2 (steeper than option B).
• Option B: \( f(x) = x^3 - 2x^2 = x^2(x - 2) \). Root at \( x = 2 \), but the graph's root is around \( x = 1 \) or less, so B is out.
• Option C: Leading coefficient negative, graph would fall to the right, so C is out.
• Option D: \( y = x^3 - 3 \). Root at \( \sqrt[3]{3} \approx 1.44 \), but no double root at 0, so D is out.

Step1: Factor out GCF

Factor \( x \) from \( x^3 + 2x^2 - 3x \): \( x(x^2 + 2x - 3) \)

Step2: Factor quadratic

Factor \( x^2 + 2x - 3 \). We need two numbers that multiply to -3 and add to 2: 3 and -1. So \( x^2 + 2x - 3 = (x + 3)(x - 1) \)

Step3: Combine factors

So \( x^3 + 2x^2 - 3x = x(x - 1)(x + 3) \)

Step1: Factor as difference of squares

\( x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9) \)

Step2: Factor \( x^2 - 9 \)

\( x^2 - 9 = (x - 3)(x + 3) \)

Step3: Analyze real zeros

\( x^2 + 9 = 0 \) has no real solutions (roots are \( \pm 3i \)). So real zeros are from \( (x - 3)(x + 3) \), which are \( x = 3 \) and \( x = -3 \)

Answer:

A. \( y = 2x^3 - 3x^2 \)

Question 6