Problem 49 - Calculus Early Transcendentals: Pearson New International Edition

Question

To use technology effectively, you need to discover its capabilities, its strengths, and its weaknesses. We urge you to practice graphing functions of various types using your own computer package or calculator. 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

(a) x-intercept: \((\frac{4}{3},0)\), y-intercept: \((0,\frac{2}{3})\). (b) Range is all real except at vertical asymptotes. (c) Vertical asymptotes: \(x=2\) and \(x=-3\). (d) Horizontal asymptote: \(y=0\).

Step 1: Find the x-intercepts

To find the x-intercepts, set \( f(x) = 0 \). This implies solving \( 3x - 4 = 0 \). Thus, \( x = \frac{4}{3} \). The x-intercept is \( (\frac{4}{3}, 0) \).

Step 2: Find the y-intercepts

To find the y-intercept, substitute \( x = 0 \) into the function. So, \( f(0) = \frac{3(0) - 4}{0^2 + 0 - 6} = \frac{-4}{-6} = \frac{2}{3} \). The y-intercept is \( (0, \frac{2}{3}) \).

Step 3: Identify vertical asymptotes

Vertical asymptotes occur where the function is undefined, or where the denominator is zero and the numerator is not zero. Solve \( x^2 + x - 6 = 0 \), which factors into \( (x-2)(x+3) = 0 \). Thus, \( x = 2 \) and \( x = -3 \) are the vertical asymptotes.

Step 4: Determine the horizontal asymptote

To find the horizontal asymptote, consider the ratio of the leading terms of the numerator and denominator. Since the degree of the denominator is higher, the horizontal asymptote is \( y = 0 \).

Step 5: Analyze the range of the function on the given domain

By graphing the function over \([-6, 6]\) excluding the asymptotes, observe the behavior and numerical outputs using the graphing tool. From the graph, you can infer the range considering the real values of \( f(x) \) over the given domain. Exclude any discontinuities at the vertical asymptotes.

Key Concepts

X-Intercepts

Finding the x-intercepts of a graph is like discovering where the graph touches or crosses the x-axis. These points represent where the function has a value of zero. To find the x-intercepts of the function \(f(x)=(3x-4)/(x^2+x-6)\), you should set the function equal to zero and solve for \(x\).

For this exercise, we set \(3x - 4 = 0\). Solving this quickly gives us \(x = \frac{4}{3}\). Therefore, the x-intercept is at \( (\frac{4}{3}, 0) \).

Remember, x-intercepts only occur where the numerator of the rational function is zero. It's crucial to ensure that the denominator at these points is not zero, as that would make the function undefined.

Y-Intercepts

Y-intercepts occur where the graph crosses the y-axis, and they can be found by evaluating the function at \(x=0\). This gives the vertical position of the graph when it meets the y-axis.

To find the y-intercept of \(f(x)=(3x-4)/(x^2+x-6)\), substitute \(x = 0\) into the equation and solve:

• \( f(0) = \frac{3(0) - 4}{0^2 + 0 - 6} \)
• This simplifies to \( f(0) = \frac{-4}{-6} = \frac{2}{3} \).

The y-intercept is thus the point \((0, \frac{2}{3})\). Always check this calculation by substituting to verify if the function simplifies correctly where needed.

Vertical Asymptotes

Vertical asymptotes in a graph indicate points where the function "shoots" towards infinity or negative infinity, essentially showing where the function is undefined due to the denominator being zero. Identifying these can guide us in understanding the discontinuity of the function.

For \(f(x)=(3x-4)/(x^2+x-6)\), vertical asymptotes occur where the denominator equals zero. Solve \(x^2 + x - 6 = 0\) for these points:

• Factor the quadratic to get \((x-2)(x+3)=0\).
• Thus, \(x = 2\) and \(x = -3\) are the roots.

These values are where the vertical asymptotes exist: at \(x=2\) and \(x=-3\). It's always good to interpret these visually by reviewing a graph, where you'll see the curve never quite reaching the lines \(x=2\) and \(x=-3\).

Horizontal Asymptotes

Horizontal asymptotes reveal the behavior of the graph as \(x\) approaches very large or very small values. For rational functions, these can be determined by analyzing the degrees of the polynomial terms in the numerator and denominator.

In \(f(x)=(3x-4)/(x^2+x-6)\), the degree of the numerator is 1 and the degree of the denominator is 2. Since the degree of the denominator is greater, the horizontal asymptote is at \(y=0\). This means as \(x\) becomes extremely large or small, the value of \(f(x)\) approaches zero.

Understanding horizontal asymptotes helps to predict the function's end behavior and ensure your interpretations of the graph align with mathematical expectations. The literal line \(y=0\) is a boundary the graph approaches but doesn't cross for large values of \(x\).