WEBVTT
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What function is represented in the figure below?
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We have the graph of some function that we want to know what function that that graph is of.
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Letβs take a look at the features of this graph.
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Thereβs a vertical asymptote, the π¦-axis, and a horizontal asymptote with equation π¦ equals negative three.
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Can we think of a function whose graph has very similar features?
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Well, here is a graph of the reciprocal function π of π₯ equals one over π₯.
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It doesnβt match the graph in our question exactly.
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But thereβs some strong family resemblance.
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Weβre going to use graph transformations to transform this reciprocal graph into the graph in our question.
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Like the graph in the question, the reciprocal graph has a vertical asymptote and a horizontal asymptote.
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But unlike the graph in our question, which to the left to the π¦-axis approaches the vertical asymptote by going upward, the reciprocal graph approaches this asymptote from the left by going downward.
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The graphs are different to the right of the π¦-axis too.
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The function represented on the left is increasing when π₯ is greater than zero.
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And so its graph comes up from below.
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Whereas, the reciprocal function is decreasing when π₯ is greater than zero.
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And so its graph comes down from above to the right of the π¦-axis.
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How can we transform the reciprocal graph on the right to make it more like the graph on the left in our question?
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One thing we could do is to reflect this graph in the π₯-axis.
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Hereβs a quick sketch of what this reflected graph looks like.
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We can see that the reflected graph does the right thing to the left and right of the π¦-axis, going up to approach the vertical asymptote to the left of the π¦-axis and coming up from below to the right of it.
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What is the function associated with this reflected graph?
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If you reflect the graph of π of π₯ in the π₯-axis, you get to the graph of negative π of π₯.
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As we started off with the graph of π of π₯ equals one over π₯, the reflected graph represents π of π₯ equals negative one over π₯.
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Before tidying up and moving to the next step, weβll just note that instead of reflecting in the π₯-axis, we could have chosen to reflect in the π¦-axis.
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Looking at the symmetry of this picture, you might guess that reflecting in the π¦-axis gives you the graph of π of π₯ equals negative one over π₯ as well.
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And youβd be right.
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Reflection in the π¦-axis takes the graph of π of π₯ to π of negative π₯.
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As we started with π of π₯ equals one over π₯, weβd end up with π of π₯ equals one over negative π₯ which can be rewritten as π of π₯ equals negative one over π₯.
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Now that weβve tidied up, we can more easily compare the graph of our improved guess, π of π₯ equals negative one over π₯, with the graph in our question.
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The vertical asymptote matches the graph in the question.
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Itβs the π¦-axis with equation that π₯ equals zero.
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However, the horizontal asymptote is not where it should be.
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It is the π₯-axis with equation that π¦ equals zero and not the line with equation that π¦ equals negative three, as in the graph in our question.
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We need to transform this graph by translating it down by three units to get it to match the graph in our question.
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We can draw a quick sketch of what this translated graph would look like.
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And it seems to match up well.
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The question is, what function does this translated graph represent?
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Translating the graph of π of π₯ by π units in the positive π¦-direction takes this graph to the graph of π of π₯ plus π.
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Weβre translating it down by three units.
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So thatβs three units in the negative π¦-direction or negative three units in the positive π¦-direction.
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And so we subtract three from our function getting π of π₯ equals negative one over π₯ minus three.
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And perhaps by trying a few points on the graph in our question, we can convince ourselves that this really is the graph of the function π of π₯ equals negative one over π₯ minus three.
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There is no additional scaling or stretching required.
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Letβs just recap how we solved this question.
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We started off with the reciprocal graph.
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That is the graph of the function π of π₯ equals one over π₯ which we thought was a good starting point because it had a vertical and horizontal asymptote like the graph in our question.
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Certainly, it was a better starting point than a straight line graph or a parabola.
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And we then applied successive graph transformations.
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First, a reflection and then a translation in the π¦-direction until we had a perfect match.
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One transformation fact that we didnβt have to use is that translation by π units in the positive π₯-direction takes the graph of π of π₯ to the graph of π of π₯ minus π.
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We would have had to use this fact had the vertical asymptote of the graph in our question not been the π¦-axis.