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. g(x) = -|x + 4| +2

Accepted Answer

Welcome back. I am so glad you're here. We are asked to graph the function G. Of X. Which is equal to a negative absolute value of five X plus five plus three halves. And that is as a transformation of F. Of X which equals the absolute value of five X. So the first thing I'm going to do is bring this information down where there's some space to work it out. And then we are going to graph our F. Of X. And then do the transformations for our G. Of X. So I'm going to draw a rough sketch of a coordinate plane with a vertical Y axis and a horizontal X axis. They meet together at the origin in the middle. If you know exactly what the affects function, the absolute value of five X looks like great. If you don't you can always plot a few points to see what's going on. So for our X. Values I'm going to choose 10 negative one. I'm going to plug those in to our function and then figure out the corresponding Y values. Then plot those on the graph. So plugging in one for X. That will be the absolute value of five no longer times X. But now times one and five times one equals five. So it's the absolute value of five, the absolute value of five is five. Looking at the graph, starting at the origin and then going to the right one for the X value and then up five for the Y value. So approximately there. Now let's plug in zero. So that will be the absolute value of five times zero five times zero is zero. So the absolute value of zero which is zero. So we've got a point here at the origin now for negative one we take the absolute value of five times negative 15 times negative one is negative five. The absolute value of negative five is positive five. So starting at the origin moving to the left one for the X. Value and then moving up five for the corresponding Y value. So this graph looks something like this. This is our affects and it has X. And Y intercepts at 00. All right now we want to take a look at our G. Of X. And we want to ask ourselves where do we see our F. Of X in that? Where do we see the absolute value of five X. That's here and now what's happening to that? What are our transformations? And we have three of them. We've got this plus five and then we have this times and negative and then this plus three have so there are three shifts. Three transformations going on here. And recalling from previous lessons on the order of transformations, the order of operations for transformations we're going to start with what's closest to our X. The horizontal shifts. So that is what's going on here with this plus five. That is happening directly to the exits inside the absolute value functions. So this is a horizontal shift but in order to know exactly what shift is taking place, we need the coefficient of X to be one. So for five X plus five we need to have a one in front of our X. We do that by factoring out of +55 X divided by five is X. Yea look the coefficient of X is one plus five divided by five is one. So it's going to be shifting because it's being added, it's going to go to the left, It's going to shift to the Left one unit. Alright, so that's the first transformation. So I'm going to draw a temporary point starting at 00 where the F of X is and then shifting to the left one unit, two negative 10. That's temporary because we have more shifts to deal with. Okay, the next for the order of operations for transformation we don't have any stretching or shrinking going on so now we look at reflection And this is being multiplied by -1. The negative is outside of the function, it's not inside the absolute value bars. So it indicates a vertical transformation and a vertical transformation specifically when it's being multiplied by negative one, it is going to reflect over the X axis. So as we saw with our F of X that was heading up with our G of X that's going to head down. So I'm going to draw some temporary arrows going down. Alright now our last shift is this plus 3/2 we have at the end. Again, that is not happening directly to the X. So it is a vertical transformation, it's happening outside the absolute value bars and it is telling us directly which way it's going, it's going to go up because it's positive and it's going to go up three halves of a unit or 1.5 units. So from our negative 10 we are going to go up 1.5 units. Now I'm going to put this in red for our G. Of X. So up To approximately here this is at -1 1.5 and then it will head down in this direction, this is our G. Of X. Now if we want to be very specific to find our correct answer choice, we can do one more thing, we can find the Y intercept the Y intercept is going to be where we have an X value of zero. So we can plug zero in to our G. Of X. Function for X. So G of zero equals negative absolute value of five times zero plus five plus one or three halves. So five times zero is zero. So this will be just five in the absolute value bars. Still a negative in front plus three halves afterward and the absolute value of five is positive five. So we have a negative five plus three halves, we can change negative five to be a negative halves. Plus three halves. Negative 10 halves. Plus three halves will be negative seven halves. Some negative seven halves is our Why intercept all right now let's go see what matches this. Looking for our G of X function up here. See if we can fit it all on one page. Okay? So we're looking for that point where those two lines diverging lines come together is going to be at negative one negative 1.5 and it's going to be pointing down. So we look at our answer choices and this matches with answer choice. A well done, we'll catch you on the next one.

Struggling with College Algebra?

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. g(x) = -|x + 4| +2

Watch next

Common Functions practice set