Problem 38 - Algebra for College Students

Question

For Problems \(35-42\), (a) find the \(y\) intercepts, (b) find the \(x\) intercepts, and (c) find the intervals of \(x\) where \(f(x)>0\) and those where \(f(x)<0\). Do not sketch the graphs. $$ f(x)=(x-4)^{2}(x+3)^{3} $$

Step-by-Step Solution

Answer

y-intercept: (0, 432); x-intercepts: (4, 0), (-3, 0); f(x) > 0 for x ∈ (-∞, -3) and (4, ∞); f(x) < 0 for x ∈ (-3, 4).

Step 1: Find the y-intercept

To find the y-intercept of the function, evaluate the function at \(x = 0\). This means substituting \(x = 0\) into \(f(x) = (x-4)^{2}(x+3)^{3}\): \[ f(0) = (0 - 4)^{2}(0 + 3)^{3} = (16)(27) = 432. \] Hence, the y-intercept is \((0, 432)\).

Step 2: Find the x-intercepts

To find the x-intercepts, set the function equal to zero and solve for \(x\): \((x-4)^{2}(x+3)^{3} = 0\). This occurs when either \((x-4)^{2} = 0\) or \((x+3)^{3} = 0\). Solving these gives: \[ x-4 = 0 \Rightarrow x = 4 \] \[ x+3 = 0 \Rightarrow x = -3 \] Thus, the x-intercepts are \((4, 0)\) and \((-3, 0)\).

Step 3: Determine intervals where \(f(x) > 0\) and \(f(x) < 0\)

To determine where the function is positive or negative, analyze the sign of \(f(x)\) between and beyond the x-intercepts, at \(x = -3\) and \(x = 4\). **Intervals:** \((-\infty, -3)\), \((-3, 4)\), \((4, \infty)\). - For \(x < -3\), both \(x-4\) and \(x+3\) are negative, so \(f(x) > 0\). - For \(-3 < x < 4\), \(x-4\) is negative and \(x+3\) is positive, so \(f(x) < 0\). - For \(x > 4\), both \(x-4\) and \(x+3\) are positive, so \(f(x) > 0\). Thus, \(f(x) > 0\) on \((-\infty, -3)\) and \((4, \infty)\), and \(f(x) < 0\) on \((-3, 4)\).

Key Concepts

Understanding Y-intercepts

The y-intercept of a polynomial function is where the graph crosses the y-axis. To find the y-intercept, we substitute zero for every x in the function, since the y-axis is at x = 0. For our function, \( f(x) = (x-4)^2(x+3)^3 \), we calculate \( f(0) \). This requires evaluating the expression by substituting \( x = 0 \) into the function, resulting in:\[f(0) = (0 - 4)^2 (0 + 3)^3 = 16 \times 27 = 432.\]
The calculation shows the y-intercept to be (0, 432). This point tells us where the curve touches or crosses the y-axis on a graph.

Finding X-intercepts

X-intercepts occur where the graph of the function crosses the x-axis. This happens when the output of the function, \( f(x) \), equals zero. For our polynomial, set the expression to zero:\[(x-4)^2(x+3)^3 = 0\]
Solve for x by setting each factor equal to zero individually:

• \( (x-4)^2 = 0 \) gives \( x = 4 \)
• \( (x+3)^3 = 0 \) gives \( x = -3 \)

Thus, the x-intercepts are \((4, 0)\) and \((-3, 0)\). Each intersept indicates the function meets the x-axis at these coordinates, which are pivotal when analyzing the graph's shape and behavior.

Analyzing Function Intervals

Analyzing intervals involves understanding different parts of the number line and how the function behaves in each segment. Consider the x-intercepts at \( x = -3 \) and \( x = 4 \), which break the x-axis into distinct intervals:

• \(( -\infty, -3 )\)
• \( (-3, 4) \)
• \( (4, \infty) \)

In essence:
- Between \( ( -\infty, -3 ) \), substitute any point in this range into \( (x-4) \) and \( (x+3) \). Both factors are negative here, meaning their product is positive.- Within \( (-3, 4) \), the terms \( (x-4) \) becomes negative while \( (x+3) \) remains positive, yielding a negative outcome.- Beyond \( (4, \infty) \), both parts \( (x-4) \) and \( (x+3) \) are positive, resulting in positive values.
These behaviors guide us in identifying important properties of the function, necessary for sketching graphs or understanding real-world applications.

Positivity and Negativity of Functions

To identify positivity and negativity of the function \( f(x) = (x-4)^2(x+3)^3 \), observe where the function results in positive or negative values. Such examination depends on the sign of the function over determined intervals:
**Positive intervals** occur where the product of the factors yields a positive value. For example:

• **\(( -\infty, -3 )\):** Both factors are negative, resulting in a positive product.
• **\(( 4, \infty )\):** Both factors are positive, resulting in a positive product.

**Negative intervals** are characterized by a negative product of the factors. Between:

• **\(( -3, 4 )\):** The mixture of one negative and one positive factor outcomes in a negative product.

These intervals reflect where the graph is positioned relative to the x-axis (above or below), offering insight into the reality of solutions or roots that might exist for applied scenarios.