Question

Which graph represents the parametric equations x = t^2 + 2t and y = –t, where –4 ≤ t ≤ 1?

Answer 1

The graph of the parametric equations x = t^2 + 2t and y = -t for -4 ≤ t ≤ 1 can be found by plotting points derived from these equations as t varies within the given range, resulting in a distinct path on the coordinate plane.

Step-by-step explanation:

The question is asking to identify the graph of the parametric equations x = t^2 + 2t and y = -t, with the parameter t ranging from -4 to 1. To find the graph, we can plug in values for t within the given range into both equations to get corresponding x and y values, which can then be plotted on a graph. The graph will illustrate how the point (x, y) moves as t changes. As this graph represents a set of parametric equations, it would not necessarily be a function in the traditional sense, where each x value is associated with only one y value. Instead, each value of t corresponds to a unique point (x, y) on the plane.

Answer 2

We have:

x = t^2 + 2*t

y = -t

where:

–4 ≤ t ≤ 1

Isolating t in the "y" equation, we get:

t = -y

Now we can use the limits of t, to find similar limits for y.

When t = -4

-4 = -y

4 = y

when t = 1

1 = -y

-1 = y

Then the limits for y are:

-1 < y < 4

Now, knowing that t = -y

we can replace that in the "x" equation:

x = t^2 + 2*t

x = (-y)^2 + 2*(-y)

x = y^2 - 2*y

Then the graph of the parametric equation is the one defined by:

x = y^2 - 2*y

in the range

-1 < y < 4

The graph of this can be seen below, where the two black dots are the points where the graph should start/end.