Questions: Let p(x)=-4(x+6)(4x-3)^2. Part 1 - y-intercept Find the y-intercept of the graph of y=p(x). (x, y)=(0,-216) Part 2 - End Behavior Describe the end behavior of p using proper notation: As x -> infinity, p(x) -> -infinity As x -> -infinity, p(x) -> infinity Part 3 - Zeros Determine the zeros of p and their multiplicities. x= (smaller value) with multiplicity x= (larger value) with multiplicity

Solution Steps

To find the $y$-intercept, we need to evaluate the polynomial $p(x)$ at $x = 0$.

The end behavior of the polynomial can be determined by examining the leading term when the polynomial is expanded. Since the polynomial is of degree 3 (from the term $-4(x+6)(4x-3)^2$), we can infer the end behavior based on the leading coefficient and the degree.

To find the zeros of the polynomial, we need to solve for $x$ when $p(x) = 0$. The zeros are the values of $x$ that make each factor zero, and their multiplicities are determined by the exponents of the factors.

To find the $y$-intercept of the graph of \( y = p(x) \), we need to evaluate \( p(x) \) at \( x = 0 \).

Given: \[ p(x) = -4(x+6)(4x-3)^2 \]

Substitute \( x = 0 \): \[ p(0) = -4(0+6)(4 \cdot 0 - 3)^2 \] \[ p(0) = -4(6)(-3)^2 \] \[ p(0) = -4(6)(9) \] \[ p(0) = -4 \cdot 54 \] \[ p(0) = -216 \]

Thus, the $y$-intercept is: \[ (x, y) = (0, -216) \]

To describe the end behavior of \( p(x) \), we need to analyze the leading term of the polynomial when expanded. The polynomial \( p(x) \) is given by: \[ p(x) = -4(x+6)(4x-3)^2 \]

First, expand \( (4x-3)^2 \): \[ (4x-3)^2 = 16x^2 - 24x + 9 \]

Now, multiply by \( (x+6) \): \[ p(x) = -4(x+6)(16x^2 - 24x + 9) \] \[ p(x) = -4 \left( 16x^3 - 24x^2 + 9x + 96x^2 - 144x + 54 \right) \] \[ p(x) = -4 \left( 16x^3 + 72x^2 - 135x + 54 \right) \] \[ p(x) = -64x^3 - 288x^2 + 540x - 216 \]

The leading term is \( -64x^3 \). Since the coefficient of the leading term is negative and the degree is odd, the end behavior is: \[ \begin{array}{l} \text{As } x \rightarrow \infty, p(x) \rightarrow -\infty \\ \text{As } x \rightarrow -\infty, p(x) \rightarrow \infty \end{array} \]

To find the zeros of \( p(x) \), we set \( p(x) = 0 \): \[ -4(x+6)(4x-3)^2 = 0 \]

This equation is satisfied when either factor is zero:

1. \( x + 6 = 0 \) \[ x = -6 \]

2. \( (4x-3)^2 = 0 \) \[ 4x - 3 = 0 \] \[ 4x = 3 \] \[ x = \frac{3}{4} \]

The multiplicities are determined by the exponents of the factors:

• \( x = -6 \) has a multiplicity of 1.
• \( x = \frac{3}{4} \) has a multiplicity of 2.

Final Answer