Graphing

In Exercises 69–76,...

Question

Graphing

In Exercises 69–76, graph each function not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.14–1.17 and applying an appropriate transformation.

y = (1/2x) − 1

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says, graph the function Y is equal to 1 divided by the quantity of 3 X in quantity plus 2, using transformations on the graph of Y is equal to 1 divided by X. Below the problem we're given a graph of our function of Y is equal to 1 divided by X, and below that, we're we have another empty graph on which to plot our other function. So, for this problem, we need to graph the function Y is equal to 1 divided by the quantity of 3 X in quantity plus 2, using transformations of the graph of Y is equal to 1 divided by X. So, if we look at our function, Y is equal to 1 divided by the quantity of 3 X in quantity plus 2, and compare that to Y is equal to 1 divided by X, we see that the factor of 3 in front of the X variable in the denominator. is going to compress our x axis by a factor of 3, and the plus 2 that's being added to the fraction is going to shift our graph of Y is equal to 1 divided by X up by 2 units. That means we're going to take the point of X. Y on the graph of Y is equal to 1 divided by X, and it's going to be transformed to the point of X divided by 3 Y + 2 on the graph of Y is equal to 1 divided by the quantity of 3 X in quantity plus 2. Now, if we look at our graph of Y is equal to 1 divided by X, we see that we have a vertical asymptote as well as a horizontal asymptote. If we look at a vertical astote, that is located at X equal to 0. And so we'll need to transform this vertical asymptote to find its new location. So, since we're dealing with an X coordinate here, we'll divide 0 by 3, and when I do so, I get X is equal to 0 again. So, our vertical asymptote is going to be located at the same X value on both graphs, but we have a horizontal asymptote at Y equal to 0. Now, since I'm dealing with a Y coordate here, I'm gonna need to add 22 to this to get its location on our new graph. So, our new horizontal astote is going to be at Y equal to 2. And now all we need to do is locate some key points and transform their coordinates from the graph of Y is equal to 1 divided by X to the graph of Y is equal to 1 divided by the quantity of 3 X in quantity plus 2. The first point we're going to look at is 1.1. And if we apply our transformation to this point, on our new graph, we're going to get 1 divided by 33. And another point that we can look at is -1-1. And when we transform this point, This is going to go to. -1 divided by 31. And so we're going to use these transform points along with our transformations for our aseptotes to now draw our new graph. So we'll start off by drawing our lines for our asymptotes. We'll have a dashed vertical line at X equal to 0 for vertical asymptotes. And we're going to have a horizon line at Y equal to 2 for our horizontal asymptote. And now we're going to plot our two points. And so we'll have 1 divided by 33 as 1 point. And then we'll have -1 divided by 3.1. As our second point, So now we just need to draw a smooth curve, approaching each of our asymptotes from each of these points. And so drawing that We'll see that we get a graph that is similar to our original graph of Y equal to 1 divided by X, but the slope is a lot steeper around the origin. And so the graph that we have drawn on the screen is going to be the answer to this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Graphing

In Exercises 69–76, graph each function not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.14–1.17 and applying an appropriate transformation.

y = (1/2x) − 1

Key Concepts

Standard Functions

Function Transformations

Linear Functions

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