Problem 56 - Single Variable Calculus

Question

In Exercises \(55-58\), plot the graph of the function. $$ f(x)=\frac{x^{2}+x}{3 x^{2}+x-1} $$

Step-by-Step Solution

Answer

The graph of the function \(f(x) = \frac{x^2 + x}{3x^2 + x - 1}\) has x-intercepts at x = 0 and x = -1, and a y-intercept at (0,0). It has vertical asymptotes at \(x = \frac{1 + \sqrt{13}}{6}\) and \(x = \frac{1 - \sqrt{13}}{6}\), and a horizontal asymptote at \(y = \frac{1}{3}\). The curve is bounded between the vertical asymptotes and converges to the horizontal asymptote as x goes to positive or negative infinity.

Step 1: Find the Intercepts

To find the x-intercepts of the function, we set f(x) = 0, and solve for x: $$ 0 = \frac{x^2 + x}{3x^2 + x - 1} $$ Since a fraction is equal to zero when its numerator is zero, we have: $$ x^2 + x = 0 $$ Factoring out an x, we get: $$ x(x + 1) = 0 $$ So, the x-intercepts are x = 0 and x = -1. To find the y-intercept, we set x = 0 and solve for f(0): $$ f(0) = \frac{0^2 + 0}{3(0^2) + 0 - 1} = \frac{0}{1} = 0 $$ So, the y-intercept is also (0,0), which is the same point as one of our x-intercepts.

Step 2: Find the Asymptotes

To find the vertical asymptotes, we find the values of x where the denominator equals 0: $$ 3x^2 + x - 1 = 0 $$ This is a quadratic equation which doesn't factor easily, so we use the quadratic formula to find the roots: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{(-1)^2 - 4(3)(-1)}}{2(3)} $$ Calculating the values, we get: $$ x = \frac{1 \pm \sqrt{13}}{6} $$ So the vertical asymptotes are at \(x = \frac{1 + \sqrt{13}}{6}\) and \(x = \frac{1 - \sqrt{13}}{6}\). For horizontal or oblique asymptotes, since the degree of the numerator is equal to the degree of the denominator, we can find the horizontal asymptote by taking the ratio of the leading coefficients: Horizontal asymptote: \(y = \frac{1}{3}\)

Step 3: Analyze the Function's Behavior

Now we analyze the behavior of the function around the key points found in steps 1 and 2. 1. Between the vertical asymptotes: As x approaches either of the vertical asymptotes, the denominator gets close to 0, causing the function to approach infinity or negative infinity, depending on the sign of the denominator. 2. To the left or right of the vertical asymptotes: As x goes to positive or negative infinity, the higher-degree terms dominate, causing the function to approach the horizontal asymptote at \(y = \frac{1}{3}\).

Step 4: Plot the Graph

Based on the analysis and the key points found, we plot the graph of the function. We include the x-intercepts, y-intercept, vertical asymptotes, and horizontal asymptote. Finally, sketch the function, showing how the graph behaves in different regions: 1. Between the vertical asymptotes: The curve will be bounded by the two asymptotes, crossing the x-axis at x=-1 and x=0 (which is also where it crosses the y-axis). 2. To the left or right of the vertical asymptotes: As x goes to positive or negative infinity, the graph will converge to the horizontal asymptote at \(y = \frac{1}{3}\). With all of the information included, we can now plot the complete graph of the function \(f(x) = \frac{x^2 + x}{3x^2 + x - 1}\).

Key Concepts

x-intercepts

For a rational function like \( f(x) = \frac{x^2 + x}{3x^2 + x - 1} \), x-intercepts are found by setting the numerator equal to zero. This happens because a fraction is zero when its numerator is zero, as long as the denominator is not zero at the same point. Thus, we solve the equation \( x^2 + x = 0 \).

To solve this, factor the numerator:

• \( x(x + 1) = 0 \)

Setting each factor equal to zero gives the x-intercepts:

• \( x = 0 \)
• \( x = -1 \)

These are the points where the graph crosses the x-axis, indicating where the function output is zero.

vertical asymptotes

Vertical asymptotes of a rational function occur where the denominator equals zero and the numerator does not also equal zero. For the function \( f(x) = \frac{x^2 + x}{3x^2 + x - 1} \), set the denominator to zero:

• \( 3x^2 + x - 1 = 0 \)

This quadratic equation doesn't factor easily, so we use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

where \( a = 3 \), \( b = 1 \), and \( c = -1 \). Calculate using the formula:

• \( x = \frac{-1 \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \)
• \( x = \frac{1 \pm \sqrt{13}}{6} \)

Thus, the vertical asymptotes are at \( x = \frac{1 + \sqrt{13}}{6} \) and \( x = \frac{1 - \sqrt{13}}{6} \). These lines represent where the function's value becomes very large or very small, extending towards infinity or negative infinity.

horizontal asymptotes

Horizontal asymptotes indicate the behavior of a function as \( x \) goes to positive or negative infinity. For the function \( f(x) = \frac{x^2 + x}{3x^2 + x - 1} \), the degrees of the numerator and the denominator are both 2.

When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. Here, the leading coefficient of the numerator is 1 (from \( x^2 \)) and from the denominator is 3 (from \( 3x^2 \)):

• Horizontal asymptote: \( y = \frac{1}{3} \)

This line tells us where the function's value tends to stabilize as \( x \) becomes very large or very small, approaching this constant value from both sides.

quadratic formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). This method is especially helpful when factoring is difficult or impossible. For our function's denominator, \( 3x^2 + x - 1 = 0 \), we apply the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Plugging in the coefficients \( a = 3 \), \( b = 1 \), and \( c = -1 \), we get:

• Discriminant: \( b^2 - 4ac = 1^2 - 4 \cdot 3 \cdot (-1) \)
• Solution: \( x = \frac{-1 \pm \sqrt{13}}{6} \)

This discriminant helps determine whether the roots are real and distinct, real and repeated, or complex. Here, since the discriminant is positive (13), there are two distinct real roots, showing where the vertical asymptotes are located in the graph of our function.