In Exercises 67-80, begin by graphing the square root function, f...

Question

In Exercises 67-80, begin by graphing the square root function, f(x) = √x. Then use transformations of this graph to graph the given function. g(x) = 2√(x+1)-1

Accepted Answer

Welcome back. I am so glad you're here. We are asked to graph H of X which is equal to three times the square root of X plus nine. And then outside of the radical minus nine. And were asked to graph this as a transformation of the square root function which is F of X equals the square root of X. So for the first step I'm just going to give us some more room to work and then we are going to work on graphing. So I'm going to draw a rough sketch of a coordinate plane will have a vertical Y axis and a horizontal X axis. They meet together at the origin in the middle. Now if you recall exactly how to graph F of X equals the square root of X. That's great. If you don't recall you can always choose a few values for X, plug them into the function, find the corresponding Y values and then plot them on the graph. We'll start by doing that. I'm going to choose positive values for X because there's a domain restriction. Anything under the radical has to equal has to be greater than or equal to zero. So I am going to choose 41 and zero for my X values. Plugging those in for X. We have the square root of four. The square to four equals two. So we have a point at four to we start at the origin on a coordinate plane. We're going to the right four units for our X value and we'll go up two units for our Y value. So approximately there. Now let's plug in one for X. We have the square root of one, the squared of one is one. So we look back starting at the origin, we're going to go to the right one unit for X. And we're going to go up one unit for Y. So approximately there now we'll plug in zero for X. The square root of zero. The squared of zero is zero. So we've got one point right at the origin. Now continuing this pattern, we'd start at the origin and we would head up toward positive infinity for both the X and Y axis. So here is our F. Of X with X and Y intercepts at 00. Now let's see what's going on with H of X where our transformations occurring. Okay, well, first of all, where is our F of X within H of X? That's right here, the square root of X. Now, what are the changes? What are the transformations? Well, first of all, we have this plus nine happening here, then we have this times three in front and then we have this minus nine at the back. So there are three transformations happening. Starting with our order of operations for transformations of functions. We're going to look at the shifts that are happening closest to our X value and that's this plus nine because that's occurring underneath the radical. It's happening directly to the X. It indicates a horizontal shift. And in particular because we are adding nine. This indicates a horizontal shift to the left nine units. I'm going to go down and where our F. Of X starts at the origin 00. This one is shifting to the left nine units. So I'm going to put a temporary dot at nine or negative 90 because it's to the left nine units. So that's at negative 90. But we'll be shifting again so that's just temporary. All right now taking a look at this three the three is not being multiplied directly to the X. It's happening outside of the radical. So that indicates it's a vertical shift. And in particular because it's being multiplied by a number greater than one. It's a vertical shrink stretching. Excuse me It is vertical stretching that is occurring because that three is greater than one. I'm not going to draw stretching on the graph just yet. Alright, and now this last shift Is this -9 at the end. Again it's not underneath the radical. It's not happening directly to our X. So it is a vertical shift. And because we are subtracting nine it indicates a vertical shift down nine units. So from our point at negative 90 we're going down nine units and we'll now have one and I'll put this in red because this is the start of our H. Of X. This is at negative nine negative nine. And now we can be specific with that vertical stretching and show what that will look like by plotting another point. So we could find where this intercepts the Y axis by plugging a zero in for X. So we can take a church of zero. This will help us find our Y intercept. So we have three times the square root of not X. But now zero plus nine and then minus nine. We do what's in the radical first. So zero plus nine is nine. So we have the square root of nine. Bring down that times three and minus nine. Well the squared of nine is three. So we have three times three minus 93 times three is nine. We have nine minus nine. And well that equals zero. So our Y intercept is at the X intercept that is at the origin at 00. So this square root H. Of X. Function is going to intercept at 00. We could plot a few more points to be more specific. But with those two points the negative nine negative nine in the 00. That should help us look at our answer choices and see which one matches. So now let's pull up there see if we've got some room. Um Kind of we'll shrink this a little bit. So again we're looking for our H. Of X. Excuse me? I didn't label that before. So we're looking at our H. Of X. And we're looking for it to start at negative nine negative nine. And it goes through the 90.0 and that matches with answer choice. A well done, we'll catch you on the next one.

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In Exercises 67-80, begin by graphing the square root function, f(x) = √x. Then use transformations of this graph to graph the given function. g(x) = 2√(x+1)-1

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