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### Discussion

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

uh, I went ahead and started by choosing my five values of X that I want to plug into my equation. And so essentially, I just chose values that we're going to be to the left of to the right of and at X is equal to three because I can see from looking at my my function that we're going to have a horizontal shift of three units to the right here so we can go ahead and start evaluating our function. Um, at F zero at the beginning of our table. And when we plug this in, we're going to get F zero is equal to the absolute value of zero minus three plus two. So this is equal to the absolute value of negative three plus two, which will simplify to five. And we can go ahead and do the same thing for f of one. So in this case, uh, it will be equal to one minus three, putting in or one for X there close to. So this is going to become the absolute value of negative two plus two, which will simplify to four. And we can continue down our left column the same way and solve for our remaining values in our right column. And so when we do that, um, our remaining why values are going to be too for and five. And our last step in this problem is going to be graphing our function so we can go ahead and start plotting her points. So we have 05 14 32 54 And finally 65 So we can from here, go ahead and connect our points in a straight line, and now we have our absolute value.

University of California, Berkeley
##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp