For each polynomial function, identify its graph from choices ...

Question

For each polynomial function, identify its graph from choices A–F.

$ƒ (x) = - (x - 2)^{2} (x - 5)$

Six graphs labeled A to F show different polynomial curves with x-intercepts near 2 and 5 for matching a given function.

Accepted Answer

Hello everybody. I have everything right today. Today, we're going to be looking at this question that states consider the given polynomial function and plot its graph where our function is Efax is equal to negative open parentheses, X minus 17, closed parentheses, squared, multiplying in open parentheses, X minus 29 closed parentheses. Now we're looking at this function. The first thing I can think of is the fact that we have it in kind of a factored form. So we have two different parentheses, multiply multiplying each other with X values inside of them. So when they're in this form, I can go ahead and solve for the zeros of my function. And to do that, I just had to set the terms inside of each parentheses, equal to zero. So one parentheses had X minus 17. So we had that equal to zero and then we have the other parentheses which was X minus 29 equal zero. So now we just solve for X in both cases. So X minus 17 equals zero. So as positive 17 by moving the 17 over and if X minus 29 equals zero, if we add 29 to both sides, we have X is equal to 29 which means that we have a point at 17 comma zero and another point at 29 comma zero. So now, now that I solve for my zeros, I can go ahead and also solve for my Y intercept. And to do that, I just have to set my X equals to zero. So we'll have F of zero equals negative open parentheses, zero minus 17 closed parentheses, squared, multiplying open parentheses, zero minus 29 closed parentheses. And once I do that math and distribute everything, I will get an answer of or A Y intercept of zero comma 8381. No, the next thing I wanna do is look at the multiplicity of my, my zeros. So X equals 17 and X equals 29 have different multiplicities. So let's look at X equals 17. 1st X equals 17 has multiplicity of two, which is an even number. And to look at the multiplicity, you just look at the power that each of each, each parentheses is set to. So the parentheses that had X minus 17 is to the second power. So we have a multiplicity of two. So because we have an even multiplicity, we will have a tangent and change directions. And once we start drawing, we can see this better illustrated. Now for X equals 29 that has multiplicity of one which is odd, which means we will cross the X axis. And finally, I want to see what each end of my graph will look like. I, I want to see what the ends of my graph will look like. And to do that, I'm going to look back at our function and kind of distribute or see what the end goal is going to be. So I see that I have one of the parentheses with an X inside of it squared so that X will be squared at some point. And once we multiply that X squared with the other parentheses, which also has an X in it, we have X squared multiplied by an X which will be OK. So we'll have, which will be X Q. So we'll have ends of graph will be determined by an A, a positive or a negative execute in this case because we have a negative outside of all the parentheses. So whatever we get from distributing that we multiply it by a negative. So we'll have a negative X cubed. Therefore, our graph, the ends of our graph will be the left hand side will be facing upwards and the right hand side will be facing downwards. And now with all this information, I can draw a general idea of what my graph look like. So we have a point at 17 comma zero a point at 29 comma zero A Y intercept at zero point or a zero comma 8381 we know that we have a tangent at X equals and we cross the X axis as at X equals 29. And we know what the ends of my graph will look like. So we'll draw a Y axis and an X axis and we will draw a point at X equals 17 and another at X equals 29. And we'll also put a point at a Y of 8381. So between the point at 17 comma zero, that was one of our zeros between that point and the Y intercept that is in the left hand side of our graph. And we know it's going to be facing upwards. So from that point to the Y intercept, we are facing up. So we draw that to connect those points. And from the point at 29 we know we're gonna be facing downwards and how we, we set in their most one, we're crossing the X axis. So from that X, at 29 comma zero, our graph will be going downwards. Now, we're missing this space between 17 comma zero and 29 comma zero. So at 17 comma zero, we had a multiple city of two, which means that point serves as a tangent. Therefore, we don't cross that X axis at that point. So from the Y intercept to that point, we're moving downwards and then at that point, we change directions touch the X axis and change directions upward. And then we know at some point the graph will have to change direction again downwards to cross the X equals 29 or the 29 comma zero. Because at 29 comma zero, we have a multi of one. So we cross the X axis and move downwards. So in total, we, we go from the Y intercept, move downwards to the 17 comma zero and do a first turn, then do a second turn and moved to 29 comma zero, cross the X axis and continue moving downward. So that will give us a general idea of what our graph should look like. So I hope that was helpful. And until next time.

For each polynomial function, identify its graph from choices A–F.
$ƒ (x) = - (x - 2)^{2} (x - 5)$

Key Concepts

Polynomial Function and Degree

Zeros and Their Multiplicities

End Behavior of Polynomial Graphs

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