Problem 50 - College Algebra

Question

For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph. $$ z(x)=\frac{(x+2)^{2}(x-5)}{(x-3)(x+1)(x+4)} $$

Step-by-Step Solution

Answer

Intercepts: \((-2, 0)\), \( (5, 0)\), and intercept \((0, \frac{5}{3})\). Vertical asymptotes at \(x = 3, -1, -4\), horizontal asymptote at \(y = 1\).

Step 1: Identifying Horizontal Intercepts

To find the horizontal intercepts, set the numerator equal to zero and solve for \(x\). This gives \((x+2)^2(x-5) = 0\). So the horizontal intercepts are \(x = -2\) (with multiplicity 2) and \(x = 5\).

Step 2: Identifying Vertical Intercept

To find the vertical intercept, evaluate \(z(x)\) at \(x = 0\). Therefore, \(z(0) = \frac{(0+2)^2(0-5)}{(0-3)(0+1)(0+4)} = \frac{4(-5)}{-12} = \frac{20}{12} = \frac{5}{3}\). So the vertical intercept is \(\left(0, \frac{5}{3}\right)\).

Step 3: Finding Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is not simultaneously zero. Therefore, solve \((x-3)(x+1)(x+4) = 0\) to determine \(x = 3\), \(x = -1\), and \(x = -4\) as the locations of the vertical asymptotes.

Step 4: Finding Horizontal Asymptote

To identify the horizontal asymptote, compare the degrees of the numerator and denominator. The numerator degree is 3 and the denominator degree is 3. Since the degrees are equal, the horizontal asymptote is \(y = \frac{1}{1} = 1\).

Key Concepts

Horizontal Intercepts

Horizontal intercepts, also known as x-intercepts, occur where a function crosses the x-axis. For rational functions like \(z(x) = \frac{(x+2)^2(x-5)}{(x-3)(x+1)(x+4)}\), these intercepts are found by setting the numerator equal to zero. The numerator, \((x+2)^2(x-5)\), provides the x-values where the function will touch or cross the x-axis.
Each factor in the expression gives a root:

• \(x = -2\) with multiplicity 2, which implies the graph touches the x-axis and turns around at this point.
• \(x = 5\), meaning the graph intersects the x-axis here.

Understanding horizontal intercepts is crucial because they demonstrate where the graph of the rational function will be at y=0. If a factor has an even multiplicity, the graph only touches the x-axis without crossing it.

Vertical Asymptotes

Vertical asymptotes occur in the graph of a rational function when the value of the denominator approaches zero, causing the function to tend towards infinity. These asymptotes are not actual lines on the graph, but rather they act as boundaries the graph approaches but never touches or crosses. For our function, \(z(x) = \frac{(x+2)^2(x-5)}{(x-3)(x+1)(x+4)}\), the asymptotes are found by setting the denominator equal to zero while ensuring the numerator is not simultaneously zero:

• \(x = 3\): Eliminates \((x-3)\), creating one asymptote.
• \(x = -1\): From \((x+1)\) in the denominator.
• \(x = -4\): Asymptote arising from \((x+4)\)

Vertical asymptotes give insight into the behavior of the function near specific x-values. Observe how the graph steeply increases or decreases on either side of these lines.

Vertical Intercept

The vertical intercept of a function occurs when it crosses the y-axis. To find it for a rational function, substitute \(x = 0\) into the expression. For our function, substitution leads to: \[ z(0) = \frac{(0+2)^2(0-5)}{(0-3)(0+1)(0+4)} = \frac{4(-5)}{-12} = \frac{20}{12} = \frac{5}{3} \]
Thus, the vertical intercept is at \( \left(0, \frac{5}{3} \right) \). Understanding the vertical intercept is essential for sketching the initial position of the graph relative to the y-axis and helps in visualizing where the graph starts before any other values of x are considered.

Horizontal Asymptote

A horizontal asymptote indicates the value that a function's outputs approach as the inputs become extremely large or small. In rational functions, this is determined by comparing the degrees (highest powers) of the numerator and the denominator. For \(z(x) = \frac{(x+2)^2(x-5)}{(x-3)(x+1)(x+4)}\), both the numerator and the denominator have a degree of 3.
When these degrees are equal, the horizontal asymptote is found by taking the leading coefficients of the numerator and the denominator, which gives us: \[ y = \frac{1}{1} = 1 \]
Thus, the horizontal asymptote is at \(y = 1\). Knowing the horizontal asymptote helps gauge the long-term trend of a graph; it is especially important when predicting how the function behaves as x approaches positive or negative infinity.