Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Plot the graph of the following functions: (i) \(y=\frac{\cos x}{\cos 2 x}\) (ii) \(y=\frac{x^{2}+2 x-3}{x^{2}+2 x-8}\)

Short Answer

Expert verified
Answer: The vertical asymptotes for the function \(y=\frac{\cos x}{\cos 2 x}\) are at \(x = \frac{(2n + 1) \pi}{4}\) for integer values of \(n\). The vertical asymptotes for the function \(y=\frac{x^{2}+2 x-3}{x^{2}+2 x-8}\) are at \(x = -4\) and \(x = 2\).

Step by step solution

01

Identify the domain and range.

The domain of a function is the set of all possible inputs (values of \(x\)) for which the function is defined. For this function, the domain will be all values of \(x\) except the points where the denominator is zero (i.e. \(\cos 2 x = 0\)). We can find these points by setting \(\cos 2x = 0\). The general solution is \(x = \frac{(2n + 1) \pi}{4}\), with \(n\) being an integer. Thus, the domain of the function will be \(x \in \mathbb{R}\) except for \(x = \frac{(2n + 1) \pi}{4}\). The range of the function is the set of all possible output values (\(y\)), which will be all real numbers, since cosine functions have a range between -1 and 1.
02

Identify asymptotes and key points.

Vertical asymptotes occur at the points where the function is not defined. In this case, vertical asymptotes will be at \(x = \frac{(2n + 1) \pi}{4}\) for integer values of \(n\). There are no horizontal asymptotes, as the range of the function is all real numbers.
03

Plot the graph.

With the information we have, let's plot the graph of the function. First, plot the vertical asymptotes at \(x = \frac{(2n + 1) \pi}{4}\) using dashed lines. Then, plot the function using the cosine function as a guide, keeping in mind that the maximum and minimum values of the cosine function are 1 and -1, respectively. Since the function has no horizontal asymptotes, the graph will continue to oscillate around the asymptotes. (ii) \(y=\frac{x^{2}+2 x-3}{x^{2}+2 x-8}\)
04

Identify the domain and range.

The domain of the function is the set of all x-values for which the function is defined. In this case, it will be all values of \(x\) except for the points where the denominator is zero. To find these points, factor the denominator: \(x^{2}+2 x-8 = (x+4)(x-2)\) Thus, the domain will be \(x \in \mathbb{R}\) except for \(x = -4, 2\). Since the function is a ratio of polynomials, the range is also all real numbers.
05

Identify asymptotes and key points.

Vertical asymptotes occur at the points where the function is not defined. In this case, vertical asymptotes will be at \(x = -4\) and \(x = 2\). To find a horizontal asymptote, we look at the degree of the polynomials in the numerator and denominator. Since they are the same, there is a horizontal asymptote at \(y = \frac{\text{coefficient of leading term in the numerator}}{\text{coefficient of leading term in the denominator}} = \frac{1}{1} = 1\).
06

Plot the graph.

With the information we've gathered, let's plot the graph of the function. First, plot the vertical asymptotes at \(x = -4\) and \(x = 2\) using dashed lines. Then, plot the horizontal asymptote at \(y = 1\) using a dashed line as well. Finally, plot the graph of the function using these asymptotes as guides.

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 Vaia!

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

For what value of a does the straight line \(\mathrm{y}=\mathrm{a}\) bisects the area of the figure bounded by the lines \(y=0, y=2+x-x^{2} ?\)

Find the area of the region bounded by the graphs of \(y=\frac{2 x}{\sqrt{x^{2}+9}}, y=0, x=0\), and \(x=4\).

A rectangle with edges parallel to the coordinate axeshas one vertex at the origin and the diagonally opposite vertex on the curve \(y=k x^{m}\) at the point where \(\mathrm{x}=\mathrm{b}(\mathrm{b}>0, \mathrm{k}>0\), and \(\mathrm{m} \geq 0)\). Show that the fraction of the area of the rectangle that lies between the curveand the \(x\)-axis depends on \(m\) but not on \(\mathrm{k}\) or \(\mathrm{b}\).

(a) If \(f(y)=-y^{2}+y+2\), sketch the region bounded by the curve \(x=f(y)\), the \(y\)-axis, and the lines \(\mathrm{y}=0\) and \(\mathrm{y}=1\). Find its area. (b) Find the area bounded by the curve \(x=-y^{2}+\) \(\mathrm{y}+2\) and the \(\mathrm{y}\)-axis. (c) The equation \(\mathrm{x}+\mathrm{y}^{2}=4\) can be solved for \(\mathrm{x}\) as a function of \(\mathrm{y}\), or for \(\mathrm{y}\) as plus or minus a function of \(x\). Sketch the region in the first quadrant bounded by the curve \(x+y^{2}=4\) and find its area first by integrating a function of \(\mathrm{y}\) and then by integrating a function of \(\mathrm{x}\).

Express with the aid of an integral the area of a figure bounded by : (i) The coordinate axes, the straight line \(\mathrm{x}=3\) and the parabola \(\mathrm{y}=\mathrm{x}^{2}+1\). (ii) The \(x\)-axis, the straight lines \(x=a, x=b\) and the curve \(y=e^{x}+2(b>a)\).

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.

Sign-up for free