/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Sketch the graph of each rationa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sketch the graph of each rational function. Specify the intercepts and the asymptotes. (a) \(y=3 x /[(x-1)(x+3)]\) (b) \(y=3 x^{2} /[(x-1)(x+3)]\)

Short Answer

Expert verified
For (a): Intercepts at (0,0), asymptotes: vertical at x=1, x=-3, horizontal y=0. For (b): Intercepts at (0,0), asymptotes: vertical at x=1, x=-3, horizontal y=3.

Step by step solution

01

Identify Intercepts for Function (a)

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the function \(y = \frac{3x}{(x-1)(x+3)}\):\[y = \frac{3(0)}{(0-1)(0+3)} = 0\]The y-intercept is \((0, 0)\).To find the x-intercepts, set \(y = 0\):\[\frac{3x}{(x-1)(x+3)} = 0\]This occurs when the numerator is zero, so \(3x = 0\). Therefore, the x-intercept is \(x = 0\).
02

Determine Vertical Asymptotes for Function (a)

Vertical asymptotes occur where the denominator is zero and the function is undefined. Set the denominator to zero:\[(x-1)(x+3) = 0\]The solutions are \(x = 1\) and \(x = -3\). Thus, there are vertical asymptotes at \(x = 1\) and \(x = -3\).
03

Identify Horizontal Asymptote for Function (a)

To find the horizontal asymptote, compare the degrees of the numerator and denominator. The numerator degree is 1, and the denominator degree is 2, so the horizontal asymptote is at \(y = 0\), because the degree of the numerator is less than the degree of the denominator.
04

Identify Intercepts for Function (b)

For the y-intercept, set \(x = 0\):\[y = \frac{3(0)^2}{(0-1)(0+3)} = 0\]The y-intercept is \((0, 0)\).For the x-intercepts, set \(y = 0\):\[\frac{3x^2}{(x-1)(x+3)} = 0\]The x-intercepts occur when the numerator is zero, so \(3x^2 = 0\). Therefore, \(x = 0\) is the only x-intercept.
05

Determine Vertical Asymptotes for Function (b)

Vertical asymptotes occur where the denominator is zero:\[(x-1)(x+3) = 0\]So, \(x = 1\) and \(x = -3\) are vertical asymptotes for this function as well.
06

Identify Horizontal Asymptote for Function (b)

For the horizontal asymptote, compare the degrees: the numerator is degree 2 and the denominator is degree 2. Thus, the horizontal asymptote is \(y = \frac{3}{1} = 3\), the ratio of the leading coefficients.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intercepts
Understanding intercepts is crucial when graphing rational functions. They help in determining where the graph crosses the axes.

For the y-intercept, set the value of x to zero in the equation of the function. This gives the point where the graph intersects the y-axis. For example, in function (a) \(y = \frac{3x}{(x-1)(x+3)}\), substituting \(x = 0\) leads to \(y = 0\). Thus, the y-intercept is at point \((0, 0)\). Similarly, for function (b) \(y = \frac{3x^2}{(x-1)(x+3)}\), substituting \(x = 0\) results in the same y-intercept at \((0, 0)\).

For the x-intercept, set y to zero and solve for x. This calculates where the graph crosses the x-axis. In the case of function (a), the solution of \(3x = 0\) leads to the x-intercept \(x = 0\). For function (b), solving \(3x^2 = 0\) also points to \(x = 0\) as the x-intercept.
  • Y-intercept: set \(x = 0\)
  • X-intercept: set \(y = 0\)
Vertical Asymptotes
Vertical asymptotes indicate the values of x where the function shoots off to infinity or negative infinity. They occur where the denominator of the rational function becomes zero because division by zero is undefined.

For instance, in functions (a) and (b), the denominator is \((x-1)(x+3)\). Setting it to zero, we have \(x - 1 = 0\) and \(x + 3 = 0\). So, solving each gives the vertical asymptotes as \(x = 1\) and \(x = -3\).

These vertical lines, \(x = 1\) and \(x = -3\), represent positions on the graph where the function heads towards infinity, thus splitting the graph into different sections.
  • Occurs when the denominator is zero
  • Find by setting denominator to zero
  • Example: \((x-1)(x+3) = 0\)
Horizontal Asymptotes
Horizontal asymptotes indicate the behavior of a function as x approaches infinity or negative infinity. To find them, compare the degrees of the numerator and the denominator.

In rational function (a), \(y = \frac{3x}{(x-1)(x+3)}\), the degree of the numerator is 1, while for the denominator it is 2. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).

For function (b), \(y = \frac{3x^2}{(x-1)(x+3)}\), the degrees of both numerator and denominator are equal, which establishes the horizontal asymptote by taking the ratio of the leading coefficients: \(y = \frac{3}{1} = 3\). This means as x grows very large, the function's value approaches 3.
  • Compare degrees of numerator and denominator
  • If numerator < denominator, asymptote is \(y = 0\)
  • If numerator = denominator, use leading coefficients
Graphing Rational Functions
Graphing rational functions combines all insights gained from intercepts and asymptotes. It visually represents the behavior of the function across its domain.

The intercepts tell us where the function crosses the axes. For function (a), both x and y intercepts are at \((0,0)\). In function (b), while it shares the same intercept, the horizontal asymptote causes the graph to behave differently as x approaches very large or very small numbers.

The vertical asymptotes, found by setting the denominator to zero, split the graph into segments. They define points where the function is undefined, often creating a dramatic shift in the graph's direction.

The horizontal asymptote shows the long-term direction of the graph. For function (a), it approaches \(y=0\), while for function (b), it nears \(y=3\) as x becomes very large.
  • Start by plotting intercepts
  • Draw vertical asymptotes as dashed lines to indicate undefined regions
  • Consider horizontal asymptote to see end-behavior
Graphing rational functions guides you to visualize complex relationships in math.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

By completing the square, show that the coordinates of the vertex of the parabola \(y=a x^{2}+b x+c\) are \((-b / 2 a,-D / 4 a),\) where \(D=b^{2}-4 a c\).

(a) An open-top box is to be constructed from a 6 -by- 8 -in. rectangular sheet of tin by cutting out equal squares at each corner and then folding up the resulting flaps. Let \(x\) denote the length of the side of each cutout square. Show that the volume \(V(x)\) is $$V(x)=x(6-2 x)(8-2 x)$$ (b) What is the domain of the volume function in part (a)? [The answer is not \((-\infty, \infty) .]\) (c) Use a graphing utility to graph the volume function, taking into account your answer in part (b). (d) By zooming in on the turning point, estimate to the nearest one-hundredth the maximum volume.

The following table and scatter plot show global coal consumption for the years \(1990-1995\). $$\begin{array}{cc} \hline \text { Year } x & \text { Coal consumption } y \\ \hline x=0 \leftrightarrow 1990 & \text { (billion tons) } \\ \hline 0 & 3.368 \\ 1 & 3.285 \\ 2 & 3.258 \\ 3 & 3.243 \\ 4 & 3.261 \\ 5 & 3.311 \\ \hline \end{array}$$ (GRAPH CAN'T COPY) (a) Use a graphing utility to find a quadratic model for the data. Then use the model to make estimates for global coal consumption in 1989 and 1996 (b) Use the following information to show that, in terms of percentage error, the 1996 estimate is better than the 1989 estimate, but in both cases the percentage error is less than \(2 \% .\) The actual figures for coal consumption in 1989 and 1996 are 3.408 and 3.428 billion tons, respectively. (c) Use the model to project worldwide coal consumption in \(1998 .\) Then show that the percentage error is more than \(9 \%,\) given that the actual 1998 consumption was 3.329 billion tons.

Find quadratic functions satisfying the given conditions. The vertex is \((3,-1),\) and one \(x\) -intercept is 1.

First, use a graphing utility to estimate to the nearest one-tenth the maximum value of the function. Then use algebra to determine the exact value, and check that your answer is consistent with the graphical estimate. (a) \(f(x)=\sqrt{-x^{2}+4 x+12}\) (b) \(g(x)=\sqrt[3]{-x^{2}+4 x+12}\) (c) \(h(x)=-x^{4}+4 x^{2}+12\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.