Consider the parametric equations
𝑥 of 𝑡 equals 𝑡 squared plus two and 𝑦 of 𝑡 equals three 𝑡 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 𝑡
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
𝑡 equals negative two and 𝑡 equals one, we know that 𝑡 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
Now we’re just looking to identify
which of the graphs is the sketch of our equation. So I’ve chosen seven values of
𝑡. If we were looking to sketch the
graph from scratch, we may choose more values of 𝑡, perhaps choosing to go up in
intervals of 0.25 as supposed to 0.5.
To find our 𝑥- and 𝑦-coordinates,
we’re going to substitute each value of 𝑡 into the parametric equations. For example, when 𝑡 is negative
two, 𝑥 is equal to negative two squared plus two, which is six. And when 𝑡 is equal to negative
two, 𝑦 is equal to three times negative two minus one, which is negative seven. When 𝑡 is equal to negative 1.5,
𝑥 is equal to negative 1.5 squared plus two, which is 4.25. And when 𝑡 is equal to negative
1.5, 𝑦 is equal to three times negative 1.5 minus one, which gives us a 𝑦-value of
We continue this process,
substituting 𝑡 equals negative one into 𝑥 to get three and into 𝑦 to get negative
four. When we substitute 𝑡 equals
negative 0.5, we get 𝑥 equals 2.25 and 𝑦 equals negative 2.5. When 𝑡 is zero, our coordinate
pair is two, negative one. And the final two values of 𝑡 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 𝑡 from negative two to 𝑡 equals one. The sketch representing the
parametric equations 𝑥 of 𝑡 equals 𝑡 squared plus two and 𝑦 of 𝑡 equals three
𝑡 minus one is C.