Video: Graphing a Pair of Parametric Equations

Consider the parametric equations π‘₯(𝑑) = 𝑑² + 2 and 𝑦(𝑑) = 3𝑑 βˆ’ 1, where βˆ’2 < 𝑑 < 1. Which of the following is the sketch of the given equation? [A] Sketch A [B] Sketch B [C] Sketch C [D] Sketch D [E] Sketch E

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Video Transcript

Consider the parametric equations π‘₯ of 𝑑 equals 𝑑 squared plus two and 𝑦 of 𝑑 equals three 𝑑 minus one, which is greater than negative two and less than one. Which of the following is the sketch of the given equation?

Here we’ve been given a pair of parametric equations and asked to sketch the curve over the open interval for 𝑑 from negative two to one. We’re going to need to begin by finding some coordinate pairs that satisfy our parametric equations. And whilst we don’t want to include 𝑑 equals negative two and 𝑑 equals one, we know that 𝑑 approaches both of these values. So we’ll use direct substitution to find the coordinate pair that our curve approaches at the end points of our open interval.

Now we’re just looking to identify which of the graphs is the sketch of our equation. So I’ve chosen seven values of 𝑑. If we were looking to sketch the graph from scratch, we may choose more values of 𝑑, perhaps choosing to go up in intervals of 0.25 as supposed to 0.5.

To find our π‘₯- and 𝑦-coordinates, we’re going to substitute each value of 𝑑 into the parametric equations. For example, when 𝑑 is negative two, π‘₯ is equal to negative two squared plus two, which is six. And when 𝑑 is equal to negative two, 𝑦 is equal to three times negative two minus one, which is negative seven. When 𝑑 is equal to negative 1.5, π‘₯ is equal to negative 1.5 squared plus two, which is 4.25. And when 𝑑 is equal to negative 1.5, 𝑦 is equal to three times negative 1.5 minus one, which gives us a 𝑦-value of negative 5.5.

We continue this process, substituting 𝑑 equals negative one into π‘₯ to get three and into 𝑦 to get negative four. When we substitute 𝑑 equals negative 0.5, we get π‘₯ equals 2.25 and 𝑦 equals negative 2.5. When 𝑑 is zero, our coordinate pair is two, negative one. And the final two values of 𝑑 give us coordinate pairs of 2.25, 0.5 and three, two.

We could now plot each ordered pair on a pair of axes. But of course, we’re looking to find which of the given sketches represents our pair of parametric equations. The first coordinate that we’re going to look to find is six, negative seven. That’s here. We then look to find 4.25, negative 5.5, which is around here. And is in fact, as we would expect, the remaining five coordinates do indeed lie on this line sketched.

Notice how the arrows describe the direction in which the graph is sketched. So our curve is sketched through increasing values of 𝑑 from negative two to 𝑑 equals one. The sketch representing the parametric equations π‘₯ of 𝑑 equals 𝑑 squared plus two and 𝑦 of 𝑑 equals three 𝑑 minus one is C.

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