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