For each equation, (a) give a table with at least three ordered p...

Question

For each equation, (a) give a table with at least three ordered pairs that are solutions, and (b) graph the equation. See Examples 7 and 8. y=|x+4|

Accepted Answer

Hello everybody. I hope we're doing all right. Today, today we're gonna be looking at this math question that's telling us for the following equation identify three order pairs and show that the graph passes through these points with the equation being Y is equal to the absolute value of X plus 13. Now the answer choices that they provide us are a, a graph in the shape of a V with a vertex at zero comma 13 and passing through the points negative 10 comma 23 positive 10 comma B A graph in the shape of A V with a vertex as zero comma negative and passing through the points negative 13 comma zero and positive 13 comma zero C A graph in the shape of a V with a vertex at negative 13 comma zero, passing through the points negative 20 comma seven and zero comma and D A graph in the shape of a V with a vertex at 13 comma zero, passing through the points zero, comma 13 and 20 comma seven. Now, the first thing we got to look at when approaching this question is the fact that this is an absolute value equation. When we have an absolute value equation, that means that whatever answer you get in within the absolute value of bars, that answer will be positive doesn't matter if it is negative within it. Once you take it out of the absolute value bars, it will become positive. So that means that we will have no negative Ys, we will have no negative answers for Y. Therefore, we can't have any points below the X axis. So we can eliminate answer B right away because it has just that some points below the X axis and we can't have those negative points. So with one answer choice eliminating now we can try to see if we can solve for a vertex equation. We see that the answer choices remaining have different vertexes. So we can try to solve for a different vertex. In this case, we have inside the absolute value bars X plus 13. So we know that if we plug in a negative 13 for X, this will give us a zero Y equals zero. So this could function as the vertex. So let's do just that. So we would have at X equals negative 13, Y equals absolute value of negative plus 13, which equals zero. So this is a point of negative 13 comma zero at an X of 13 will have a Y value of zero. So we see that a has a point of zero comma 13 which is not the vertex that we want. So we look at a choice C that has negative 13 comma zero. And that is the choice that we do want. And we look at a choice D has positive 13 comma zero. So we can theoretically eliminate answer choice A and D and go to sea. So I'll write the equation once again, we had the absolute value of X plus in a vertex of comma zero. So we focus on answer choice C now, now that we see that they had two other points that they provide us and they asked us to look for those two extra points. So let's plug those in and see what we get. Well, one point that they gave us was negative 20 comma seven, which means that we have to plug in negative 24 X and see if we get a positive seven as our answer. So let's do just that. So we'll have Y is equal to negative 20 plus 13 and this is the absolute value of negative 20 plus 13. And remember this is that X equals negative 20. So this will be equal to the absolute value of negative seven, which is positive seven leading to the point negative 20 comma seven. And that checks out. So now we have two points already checked with negative 13 comma zero and negative 20 comma seven. So now let's check the last point they gave us. So we have at X equals zero. We would have Y equal to the absolute value of zero plus 13. This is equal to the absolute value of 13, which is just positive 13 or the 0.0 comma 13. And once again, that checks out with the answer, so that confirms that answer choice. C will be the right answer for this question. So I hope that was helpful. And until next time.

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For each equation, (a) give a table with at least three ordered pairs that are solutions, and (b) graph the equation. See Examples 7 and 8. y=|x+4|