Problem 49 - Calculus Early Transcendentals: Pearson New International Edition

Question

Graph \(f(x)=(3 x-4) /\left(x^{2}+x-6\right)\) on the domain \([-6,6]\). (a) Determine the \(x\)-and \(y\)-intercepts. (b) Determine the range of \(f\) for the given domain. (c) Determine the vertical asymptotes of the graph. (d) Determine the horizontal asymptote for the graph when the domain is enlarged to the natural domain.

Step-by-Step Solution

Answer

x-intercept: \(x = \frac{4}{3}\), y-intercept: \(y=\frac{2}{3}\). Range: \([-\infty, \infty]\), Vertical asymptotes: \(x=-3, 2\), Horizontal asymptote: \(y=0\).

Step 1: Determine x-intercepts

The x-intercepts occur where the numerator is zero. Solve the equation for the numerator: \[3x - 4 = 0\]Setting this equal to zero gives:\[3x = 4\]\[x = \frac{4}{3}\]This means the x-intercept is at \(x = \frac{4}{3}\).

Step 2: Determine y-intercepts

The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the function \(f(x)\):\[f(0) = \frac{3(0) - 4}{0^2 + 0 - 6} = \frac{-4}{-6} = \frac{2}{3}\]Thus, the y-intercept is at \(y = \frac{2}{3}\).

Step 3: Identify Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is not zero. Solve the denominator:\[x^2 + x - 6 = 0\]Factorizing gives:\[(x - 2)(x + 3) = 0\]So, the roots are \(x = 2\) and \(x = -3\). These are the lines of the vertical asymptotes.

Step 4: Identify Horizontal Asymptote

For horizontal asymptotes, compare the degrees of the numerator and denominator. The degrees are 1 for the numerator and 2 for the denominator. Since the degree of the denominator is greater, the horizontal asymptote is \(y = 0\).

Key Concepts

X-intercepts

In the graphing of rational functions, **x-intercepts** are the points where the graph touches or crosses the x-axis. To find these points, set the numerator of the function equal to zero because these are the values of x that make the overall function equal to zero. For the function we are working with, \(f(x)=(3x-4)/(x^2+x-6)\), the numerator is \(3x-4\).

• Setting this equal to zero: \(3x - 4 = 0\)
• Solving for \(x\): \(3x = 4\), thus \(x = \frac{4}{3}\)

This means the graph will have an x-intercept at the point \(x = \frac{4}{3}\). At this point, the value of the function \(f(x)\) is zero.

Y-intercepts

The **y-intercepts** of a rational function are found by determining where the graph crosses the y-axis. This occurs when \(x = 0\). To find the y-intercept, substitute \(x = 0\) into the function and solve.For the function \(f(x)\), this looks like:

• Substituting: \(f(0) = \frac{3(0) - 4}{0^2 + 0 - 6} = \frac{-4}{-6}\)
• Simplifying gives \(\frac{2}{3}\)

Thus, the y-intercept is at the point \(y = \frac{2}{3}\). This tells us that when \(x\) is zero, the graph touches or crosses the y-axis at \(y = \frac{2}{3}\).

Vertical Asymptotes

**Vertical asymptotes** represent the x-values that make the denominator of the function zero (where the function is undefined as it approaches infinity or negative infinity). To find vertical asymptotes:Factor the denominator and set it equal to zero:

• The denominator is \(x^2 + x - 6\)
• Factorized, it becomes \((x - 2)(x + 3) = 0\)

Solving, we find the roots:

• \(x = 2\)
• \(x = -3\)

These are the points where the graph will have vertical lines, meaning the function approaches infinity or negative infinity, but never actually touches these lines.

Horizontal Asymptote

**Horizontal asymptotes** help us determine the behavior of a rational function as \(x\) approaches infinity. To find horizontal asymptotes, compare the degrees of the numerator and the denominator.

• The numerator \(3x-4\) has a degree of 1.
• The denominator \(x^2+x-6\) has a degree of 2.

If the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is \(y = 0\). This implies that as \(x\) increases or decreases towards infinity, \(f(x)\) approaches zero. Thus, for our function, the horizontal asymptote is \(y = 0\). This line indicates the leveling off of the graph towards the x-axis without ever crossing it.