WEBVTT
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Which of the graphs shown represents inverse variation?
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And then we have four graphs (A), (B), (C), and (D) plotted in the first quadrant of the coordinate plane.
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In order to answer this question, weβll remind ourselves what we mean by the term βinverse variation.β
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Suppose we have the variables π¦ and π₯.
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If these are inversely proportional to one another, or if they represent inverse variation, we say that π¦ is proportional to one over π₯.
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The corresponding equation we use to describe this relationship is π¦ equals π over π₯.
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Now, of course, π¦ could also be inversely proportional to π₯ squared.
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In this case, we say that π¦ is proportional to one over π₯ squared.
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And the corresponding equation is π¦ equals π over π₯ squared.
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This holds for π¦ being inversely proportional to the square root of π₯, the cube of π₯, and so on.
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So essentially, weβre looking to find the graph of an equation of the form π¦ equals π over π₯ to the πth power, where π must be a positive number.
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And so there are some graphs that we can disregard straight away.
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Graph (C) is a straight line.
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That means its general form is π¦ equals ππ₯ plus π.
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But it passes through the origin, the point zero, zero, so, in fact, since its π¦-intercept is zero, we can write it as π¦ equals ππ₯ or π¦ equals ππ₯.
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This, in fact, is an example of two variables that are in direct proportion to one another.
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And so we disregard option (C).
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Similarly, option (D) looks like it could be of the form π¦ equals π times the square root of π₯.
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And looking at a couple of coordinates here, for instance, four, two and one, one, we can certainly verify that these coordinates satisfy this equation.
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This is an example of π being directly proportional to the square root of π₯.
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And so this does not represent inverse variation, and we disregard (D).
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Similarly, equation (B) could be of the form π¦ equals ππ₯ squared.
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It certainly looks like a quadratic.
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And itβs a reflection of our previous graph in the line π¦ equals π₯.
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Once again, this is a graph that represents direct variation.
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Itβs π¦ varies directly with π₯ squared, and so we disregard option (B).
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And so this must leave option (A) only.
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And this makes a lot of sense when we think about what we know about the graph of, say, π¦ equals one over π₯.
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It looks like this, and it has asymptotes given by the π¦- and π₯-axes.
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Similarly, the graph of π¦ equals one over π₯ squared looks like this.
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In fact, if we think about the shape of the curve purely in the first quadrant, the graph of π¦ equals π over π₯ to the πth power for positive values of π will always have this shape.
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And so the answer is (A).
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Graph (A) then represents inverse variation: π¦ is inversely proportional to some power of π₯.