WEBVTT
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Which of the following functions is graphed in the given figure?
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Is it A: π¦ equals negative π₯ plus one squared times π₯ plus seven, B: π¦ equals π₯ minus one squared times π₯ minus seven, C: π¦ equals π₯ minus seven squared times π₯ minus one, D: π¦ equals negative π₯ plus seven squared times π₯ plus one, or is it E: π¦ equals π₯ plus seven squared times π₯ plus one?
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We can see that the graph has π₯-intercepts at one and seven.
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This means that all functions must have the form π¦ equals some number π΄ times π₯ minus one to the power of some number π times π₯ minus seven to the power of some number π.
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We can eliminate several of the options which donβt have this form.
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We have eliminated A, D, and E.
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And so weβre left to choose between options B and C.
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How are we going to choose between these two options?
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Well in option B, the value of π is two and the value of π is one.
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But in option C, itβs the other way around.
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The value of π is two and the value of π is one.
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On the graph, you can see that these two π₯-intercepts are quite different.
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The π₯-intercept at π₯ equals one has the graph crossing the π₯-axis, whereas the π₯-intercept at π₯ equals seven has the graph touching but not crossing the π₯-axis.
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The fact that the graph touches the π₯-axis at π₯ equals seven tells us that there must be a repeated root here.
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And so the exponent of π₯ minus seven, π, cannot be equal to one.
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Actually, it turns up we can say more than that.
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We actually can say that π is even, but we donβt need that for this problem.
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We already have enough information to eliminate option B, because π there was one.
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And weβre left only with option C: π¦ equals π₯ minus seven squared times π₯ minus one.
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In this question, we didnβt need to know anything about the value of π΄, but we can actually see from the graph that π΄ would be positive which makes sense given our answer where π΄ is one.
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Had π΄ been negative, then the graph of the function would have moved from the second quadrant to the fourth quadrant, as π₯ increases rather than from the third quadrant to the first quadrant as it does.
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If π΄ had been negative one, then the graph you would have seen would be the graph that we have reflected in the π₯-axis.
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And Iβve sketched in that graph using a dotted line.
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If π¦ equals negative π₯ minus seven squared times π₯ minus one had been one of the options, then we could have eliminated it by expanding that option and seeing that that implied that the π¦-intercept was positive whereas clearly on our graph π¦-intercept is negative.