Problem 49 - College Algebra with Corequisite Support

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. $$ w(x)=\frac{(x-1)(x+3)(x-5)}{(x+2)^{2}(x-4)} $$

Step-by-Step Solution

Answer

Horizontal intercepts: \( x = 1, -3, 5 \). Vertical intercept: \( \left(0, \frac{15}{16}\right) \). Vertical asymptotes: \( x = -2, 4 \). Horizontal asymptote: \( y = 1 \).

Step 1: Identifying the Horizontal Intercepts

To find the horizontal intercepts, set the numerator equal to zero because the values of \( x \) that make the numerator zero are the horizontal intercepts.\[(x-1)(x+3)(x-5) = 0\]Solve for \( x \):- \( x = 1 \)- \( x = -3 \)- \( x = 5 \)Thus, the horizontal intercepts are at \( x = 1 \), \( x = -3 \), and \( x = 5 \).

Step 2: Finding the Vertical Intercept

To find the vertical intercept, set \( x = 0 \) in the function and solve for \( w(x) \).\[w(0) = \frac{(0-1)(0+3)(0-5)}{(0+2)^{2}(0-4)} = \frac{(-1)(3)(-5)}{4(4)} = \frac{15}{16}\]Thus, the vertical intercept is \( \left(0, \frac{15}{16}\right) \).

Step 3: Determining Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is not zero.\[(x+2)^{2}(x-4) = 0\]Solve for \( x \):- \( x = -2 \)- \( x = 4 \)Thus, vertical asymptotes occur at \( x = -2 \) and \( x = 4 \).

Step 4: Finding the Horizontal or Slant Asymptote

The degrees of the numerator and denominator determine the horizontal asymptote.The numerator \((x-1)(x+3)(x-5)\) is degree 3, and the denominator \((x+2)^2 (x-4)\) is also degree 3. When the degrees are equal, the horizontal asymptote is \( y = \frac{a}{b} \), where \( a \) and \( b \) are the leading coefficients.For this function:\[y = \frac{1}{1} = 1\]Thus, there is a horizontal asymptote at \( y = 1 \).

Key Concepts

Horizontal Intercepts

Horizontal intercepts are the points where the graph of a function crosses the x-axis. This happens when the function's output, or y-value, equals zero.
To find these intercepts of a rational function like \( w(x) = \frac{(x-1)(x+3)(x-5)}{(x+2)^2(x-4)} \), you need to look at the numerator of the function.
Specifically, set the numerator \((x-1)(x+3)(x-5)\) equal to zero and solve for \( x \):

• Set \((x-1) = 0\) to get \(x = 1\)
• Set \((x+3) = 0\) to get \(x = -3\)
• Set \((x-5) = 0\) to get \(x = 5\)

These values - \( x = 1 \), \( x = -3 \), and \( x = 5 \) - are the horizontal intercepts of the function. Each of these points lies on the x-axis where the output is zero.

Vertical Intercepts

Vertical intercept, also known as the y-intercept, is the point where the graph of a function crosses the y-axis. This occurs when \( x \) is equal to zero. To find the vertical intercept of the rational function \( w(x) \), substitute \( x = 0 \) into the equation.Substitute it as follows:\[ w(0) = \frac{(0-1)(0+3)(0-5)}{(0+2)^2(0-4)} = \frac{(-1)(3)(-5)}{4(4)} = \frac{15}{16} \]Thus, the vertical intercept is at \( \left( 0, \frac{15}{16} \right) \).
This means that when \( x \) is zero, the function's output is \( \frac{15}{16} \).
A graph of this function would cross the y-axis at this point.

Vertical Asymptotes

Vertical asymptotes are lines that the graph of a function approaches but never touches or crosses. They occur at values of \( x \) that make the denominator zero but not the numerator.
For the given function \( w(x) = \frac{(x-1)(x+3)(x-5)}{(x+2)^2(x-4)} \), set the denominator \((x+2)^2(x-4)\) equal to zero to find potential asymptotes:

• \((x+2)^2 = 0\), solve to get \(x = -2\)
• \((x-4) = 0\) gives \(x = 4\)

These values, \( x = -2 \) and \( x = 4 \), are where vertical asymptotes occur.
This means the function will get closer and closer to these lines but never actually cross them.

Horizontal Asymptote

Horizontal asymptotes discuss the behavior of a function as \( x \) approaches infinity or negative infinity.
To find the horizontal asymptote of \( w(x) = \frac{(x-1)(x+3)(x-5)}{(x+2)^2(x-4)} \), compare the degrees of the numerator (which is degree 3) and the denominator (also degree 3).
When both degrees are equal, the horizontal asymptote is the ratio of the leading coefficients of those polynomials. For this function, both the numerator and the denominator have leading coefficients equal to 1, so the horizontal asymptote is\[ y = \frac{1}{1} = 1 \]This tells us that as \( x \) becomes very large or very small, the function's output will get closer and closer to \( y = 1 \).
The graph of the function will approach this line but not necessarily reach it until \( x \) is extremely large or small.