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

How do you obtain the graph of \(y=f(x+2)\) from the graph of \(y=f(x) ?\)

Short Answer

Expert verified
Answer: The parameter "+2" inside the argument of the function y=f(x+2) causes a horizontal shift of the graph of y=f(x) 2 units to the left.

Step by step solution

01

Identify the transformation

In this case, the transformation involves the addition of 2 inside the function argument. This corresponds to a horizontal shift. Adding a positive number inside the parentheses will cause a shift to the left, while subtracting a positive number will cause a shift to the right.
02

Determine the direction of the shift

Because we have a positive number (+2) inside the argument of the function f(x), the graph of y=f(x+2) will be shifted to the left by 2 units compared to the graph of y=f(x).
03

Create the new graph

Starting from the graph of y=f(x), move every point on the graph 2 units to the left to obtain the graph of y=f(x+2). This means that for every point (x, f(x)) on the original graph, the corresponding point on the new graph will be (x-2, f(x+2)).
04

Verify the transformation

Once you have made the shift, double-check your work by ensuring that every point on the new graph is 2 units to the left of the corresponding point on the original graph. If this is the case, then you have successfully obtained the graph of y=f(x+2) from the graph of y=f(x).

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

Surface area of a sphere The surface area of a sphere of radius \(r\) is \(S=4 \pi r^{2} .\) Solve for \(r\) in terms of \(S\) and graph the radius function for \(S \geq 0\).

Evaluating inverse trigonometric functions Without using a calculator, evaluate or simplify the following expressions. $$\csc ^{-1}(\sec 2)$$

Assume that \(b>0\) and \(b \neq 1 .\) Show that \(\log _{1 / b} x=-\log _{b} x\)

Suppose the probability of a server winning any given point in a tennis match is a constant \(p,\) with \(0 \leq p \leq 1\).Then the probability of the server winning a game when serving from deuce is $$f(p)=\frac{p^{2}}{1-2 p(1-p)}$$,a. Evaluate \(f(0.75)\) and interpret the result. b. Evaluate \(f(0.25)\) and interpret the result. (Source: The College Mathematics Journal 38, 1, Jan 2007).

Field goal attempt Near the end of the 1950 Rose Bowl football game between the University of California and Ohio State University, Ohio State was preparing to attempt a field goal from a distance of 23 yd from the end line at point \(A\) on the edge of the kicking region (see figure). But before the kick, Ohio State committed a penalty and the ball was backed up 5 yd to point \(B\) on the edge of the kicking region. After the game, the Ohio State coach claimed that his team deliberately committed a penalty to improve the kicking angle. Given that a successful kick must go between the uprights of the goal posts \(G_{1}\) and \(G_{2},\) is \(\angle G_{1} B G_{2}\) greater than \(\angle G_{1} A G_{2} ?\) (In \(1950,\) the uprights were \(23 \mathrm{ft} 4\) in apart, equidistant from the origin on the end line. The boundaries of the kicking region are \(53 \mathrm{ft} 4\) in apart and are equidistant from the \(y\) -axis. (Source: The College Mathematics Journal 27, 4, Sep 1996).

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