In Exercises 81–94, begin by graphing the absolute value function...

Question

In Exercises 81–94, begin by graphing the absolute value function, f(x) = |x|. Then use transformations of this graph to graph the given function. h(x) = 2|x+3|

Accepted Answer

Welcome back. I am so glad you're here. We are asked to graph the function G. Of X where G. Of X is equal to three times the absolute value of two X plus nine. And were asked to graph it as a transformation of a five X where F. Of X is equal to the absolute value of two X. So the first step I want to take here is to get a little bit more room to draw and write some, dragging this down. And we'll start off by graphing a five X. I'm going to draw a coordinate plane, I'll draw a vertical Y axis and a horizontal X axis. And they meet together at the origin in the middle. Now if you you already know exactly what the absolute value of two X looks like. That is wonderful. If you don't you can always plot a few points. So I am going to choose three X values. I'm going to choose 10 and negative one. I'm going to plug those values in for X. In the F of X. Function. Figure out their corresponding Y values and then plot them on the coordinate plane. So plugging one in for X. That's the absolute value of two, not times X but now times one, the absolute value of two times 12 times one is two. So it's the absolute value of, to the absolute value of two is positive two. So plotting the .1, 2 on the graph, starting at the origin moving to the right one unit for our x value and then up to for our y value that's approximately there. Alright now let's plug in zero for X value have the absolute value of two times zero. So two times zero is zero. So we've got the absolute value of zero, the absolute value of zero is zero. Looking at our graph, we have a point at 00 which is right at the origin. Now for negative one we'll plug that in for X. So we've got the absolute value of two times negative one while two times negative one is negative two. So we have the absolute value of negative two. And the absolute value of negative two is positive. To looking at our graph moving from the origin to the left one unit for our X value and up to for our Y values. So if this we were to continue plotting points it would look something like this. Alright, this is our F. Of X. And both the X and Y intercepts are at 00. Now let's work on G. Of X. Well first because we're looking at it as a transformation of F of X. What's similar? Where do we find F. Of X in G. Of X. And that's here with this absolute value of two. X. Now what's different. What are the transformations that are occurring? Well this plus nine and this times three or two different transformations that we have going on here. So what do they each mean? Well this plus nine, that's inside the absolute value bars that's happening directly to the X. So it indicates a horizontal shift and what is the horizontal shift exactly. Well because we're adding nine it's going to move to the left but to figure out exactly how much it's moving to the left, the coefficient of X has to be one. So for two X plus nine to get that coefficient of X to be one, we're going to factor out A two and so then two X divided by two will be xia The coefficient of X is one and nine divided by two is nine halves or plus 4.5. So now this tells us how much it's moving to the left, it's moving to the left, 4.5 units. Alright? So if we look back at our graph instead of an X intercept at +00. Now it's going to be at negative 4.5 zero. Now let's see what else is going on with this other shift. This three being multiplied, that's on the outside, it's not happening directly to the X. It's not inside the absolute value bars. So because it's not happening directly to the exit indicates a vertical shift. And specifically because we're multiplying by a number that's greater than one, it indicates vertical stretching. So looking back here at our graph, vertical stretching would come up a little bit tighter like this, I don't know why I drew it so tall but there's our G of X. And now if we wanted to be very specific and this will help us figure out which answer choice is correct. We can find our Y intercept, our Y intercept is going to be where this hits the Y axis. And specifically our X value will be zero. So we can go back to our original G. Of X function and we can put a zero in for X. So G. Of zero. We've got three times the absolute value of two times zero plus nine. Doing what's in the absolute value bars first two times 00. So we have zero plus nine bringing this all down. Zero plus nine is nine. So we have three times the absolute value of nine and the absolute value of nine is positive nine. So we've got three times nine. Well three times nine is 27. So the value of Y when X is zero is 27. There's our Y intercept. All right, we've got this all drawn out. Now let's go see which answer choice matches what we've drawn can shrink this a little bit. Okay, got everything on screen. So now we are looking for our G. Of X to have an X intercept of negative 4.50. And our Y intercept for G. Of X. Two B. At 0 27. We look at our answer choices and this matches with answer choice C. Well done. We'll catch you on the next one

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In Exercises 81–94, begin by graphing the absolute value function, f(x) = |x|. Then use transformations of this graph to graph the given function. h(x) = 2|x+3|