Problem 69 - Algebra and Trigonometry

Question

Graph the rational function and find all vertical asymptotes, \(x\) - and \(y\) -intercepts, and local extrema, correct to the nearest decimal. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. \(y=\frac{2 x^{2}-5 x}{2 x+3}\)

Step-by-Step Solution

Answer

The vertical asymptote is at \( x = -\frac{3}{2} \); x-intercepts are \( x = 0 \) and \( \frac{5}{2} \), y-intercept at \( (0, 0) \). End behavior resembles \( y = x - \frac{19}{12} \).

Step 1: Find Vertical Asymptotes

A vertical asymptote occurs where the denominator of a rational function is zero, assuming the numerator is not zero at the same point. For the function \( y = \frac{2x^2 - 5x}{2x + 3} \), set the denominator equal to zero: \( 2x + 3 = 0 \). Solving for \( x \), we get \( x = -\frac{3}{2} \). Therefore, there is a vertical asymptote at \( x = -\frac{3}{2} \).

Step 2: Find x-intercepts

The x-intercepts of a function occur where the numerator is zero, and the function is defined. For \( y = \frac{2x^2 - 5x}{2x + 3} \), set the numerator equal to zero: \( 2x^2 - 5x = 0 \). Factoring gives \( x(2x - 5) = 0 \). Therefore, \( x = 0 \) and \( x = \frac{5}{2} \) are the x-intercepts.

Step 3: Find y-intercept

The y-intercept occurs where \( x = 0 \). Substituting \( x = 0 \) into the function \( y = \frac{2x^2 - 5x}{2x + 3} \), we have \( y = \frac{0}{3} = 0 \). The y-intercept is at \( (0, 0) \).

Step 4: Determine Local Extrema

Local extrema occur where the first derivative \( y' \) is zero or undefined. We first find the derivative of \( y = \frac{2x^2 - 5x}{2x + 3} \) using the quotient rule: \( y' = \frac{(4x - 5)(2x + 3) - (2x^2 - 5x)(2)}{(2x + 3)^2} \). Simplifying gives \( y' = \frac{8x^2 + 12x - 10x - 15 - 4x^2 + 10x}{(2x + 3)^2} = \frac{4x^2 + 7x - 15}{(2x + 3)^2} \). Setting \( y' = 0 \), solve \( 4x^2 + 7x - 15 = 0 \) using the quadratic formula: \( x = \frac{-7 \pm \sqrt{7^2 - 4\cdot4\cdot(-15)}}{2\cdot4} \). Calculating gives \( x \approx -2.42 \) and \( x \approx 1.55 \). Check these x-values in the second derivative or by plotting to confirm the nature of these extrema.

Step 5: Long Division for End Behavior

Perform polynomial long division on \( \frac{2x^2 - 5x}{2x + 3} \) to find a polynomial that models the end behavior. Dividing, we find the quotient as \( x - \frac{19}{12} \) with remainder \( \frac{-\frac{15}{2}}{2x + 3} \). The polynomial part, \( x - \frac{19}{12} \), represents the end behavior.

Step 6: Graph the Functions

Plot the rational function \( y = \frac{2x^2 - 5x}{2x + 3} \), the polynomial \( y = x - \frac{19}{12} \), and their asymptotes (at \( x = -\frac{3}{2} \)) in a sufficiently large viewing window. The x-intercepts are at \( x = 0 \) and \( x = \frac{5}{2} \), and the y-intercept is at \( (0, 0) \). The local extrema and the behavior near the vertical asymptote can also be visualized. Verify that both the rational function and the polynomial have the same end behavior as \( x \to \infty \) and \( x \to -\infty \).

Key Concepts

Vertical Asymptotes

Vertical asymptotes are the vertical lines where a rational function is undefined because the denominator equals zero. Understanding vertical asymptotes helps to visualize where the graph of the function will tend to infinity. To find them in the function \(y = \frac{2x^2 - 5x}{2x + 3}\), we need to solve for \(x\) where the denominator \(2x + 3 = 0\). This happens when \(x = -\frac{3}{2}\).

At this point, the function does not exist, and as you approach the asymptote from either side, the values of the function increase or decrease without bound, resulting in a sharp change or break in the graph. Keep in mind, vertical asymptotes are not part of the graph itself, but lines the graph approaches.

• Vertical asymptote: \( x = -\frac{3}{2}\).

x-intercepts

The \(x\)-intercepts of a rational function occur where the graph crosses the x-axis, which happens when the numerator is zero. These points tell us where the outputs of the function are zero. For the function \(y = \frac{2x^2 - 5x}{2x + 3}\), setting the numerator \(2x^2 - 5x = 0\) gives us the solutions by factoring: \(x(2x - 5) = 0\).

This tells us the \(x\)-intercepts are at \(x = 0\) and \(x = \frac{5}{2}\). It's crucial to ensure that the function remains defined at these points by checking that the denominator isn't also zero, which would otherwise result in a removable discontinuity and not an \(x\)-intercept.

• \(x\)-intercepts: \(x = 0\) and \(x = \frac{5}{2}\).

y-intercepts

The \(y\)-intercept of a rational function is the point where the graph crosses the y-axis. This occurs when \(x = 0\) in the function equation. For the function \(y = \frac{2x^2 - 5x}{2x + 3}\), substituting \(x = 0\) yields \(y = \frac{0}{3} = 0\).

This gives a \(y\)-intercept at the origin, \((0, 0)\). Unlike \(x\)-intercepts, there is only one \(y\)-intercept for a given function because there's only one \(y\)-axis to cross. Knowing the \(y\)-intercept can help in sketching the graph and understanding its behavior around the origin.

• \(y\)-intercept: \((0, 0)\).

Local Extrema

Local extrema are the points where the graph reaches a local minimum or maximum; these points occur where the first derivative is zero or undefined. Finding local extrema helps to understand the turning points of the graph and how it changes direction.

For the function \(y = \frac{2x^2 - 5x}{2x + 3}\), the derivative, calculated using the quotient rule, is \(y' = \frac{4x^2 + 7x - 15}{(2x + 3)^2}\). Setting \(y' = 0\) allows us to find where these extrema occur: solving \(4x^2 + 7x - 15 = 0\) results in \(x \approx -2.42\) and \(x \approx 1.55\). Finally, verifying these points using the second derivative test can confirm the nature of these extrema.

When graphing, these points are either peaks or valleys, illustrating key features of the graph's shape.

End Behavior

End behavior describes what happens to the values of a rational function as \(x\) approaches infinity or negative infinity. For \(y = \frac{2x^2 - 5x}{2x + 3}\), dividing the numerator by the denominator through polynomial long division gives families that share similar long-term behavior.

In this case, dividing results in \(y \approx x - \frac{19}{12}\), indicating that as \(x\) goes towards infinity or negative infinity, the behavior of the rational function resembles that of a straight line \(y = x - \frac{19}{12}\).

• End behavior mimics the polynomial: \(y = x - \frac{19}{12}\).

Observe that as \(x\) moves towards large positive or negative values, the fractional component \(\frac{-\frac{15}{2}}{2x + 3}\) diminishes, leaving the polynomial part to govern the graph. Understanding end behavior is critical for sketching the graph accurately over a large domain.