To graph the parametric equations [tex]\( x = t^2 + 2t \)[/tex] and [tex]\( y = -t \)[/tex] for [tex]\( -4 \leq t \leq 1 \)[/tex], we must follow a systematic approach.
Here's a step-by-step solution:
1. Identify the range of [tex]\( t \)[/tex]:
[tex]\[ -4 \leq t \leq 1 \][/tex]
2. Evaluate [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex] at several points within the given range:
We will calculate a few key points to understand the behavior of the parametric curve.
For [tex]\( t = -4 \)[/tex]:
[tex]\[
x = (-4)^2 + 2(-4) = 16 - 8 = 8
\][/tex]
[tex]\[
y = -(-4) = 4
\][/tex]
So, the point is [tex]\( (8, 4) \)[/tex].
For [tex]\( t = -3 \)[/tex]:
[tex]\[
x = (-3)^2 + 2(-3) = 9 - 6 = 3
\][/tex]
[tex]\[
y = -(-3) = 3
\][/tex]
So, the point is [tex]\( (3, 3) \)[/tex].
For [tex]\( t = -2 \)[/tex]:
[tex]\[
x = (-2)^2 + 2(-2) = 4 - 4 = 0
\][/tex]
[tex]\[
y = -(-2) = 2
\][/tex]
So, the point is [tex]\( (0, 2) \)[/tex].
For [tex]\( t = -1 \)[/tex]:
[tex]\[
x = (-1)^2 + 2(-1) = 1 - 2 = -1
\][/tex]
[tex]\[
y = -(-1) = 1
\][/tex]
So, the point is [tex]\( (-1, 1) \)[/tex].
For [tex]\( t = 0 \)[/tex]:
[tex]\[
x = 0^2 + 2(0) = 0
\][/tex]
[tex]\[
y = 0
\][/tex]
So, the point is [tex]\( (0, 0) \)[/tex].
For [tex]\( t = 1 \)[/tex]:
[tex]\[
x = 1^2 + 2(1) = 1 + 2 = 3
\][/tex]
[tex]\[
y = -1
\][/tex]
So, the point is [tex]\( (3, -1) \)[/tex].
3. Plot the points:
- (8, 4)
- (3, 3)
- (0, 2)
- (-1, 1)
- (0, 0)
- (3, -1)
4. Draw the smooth curve passing through these points:
Given the nature of parametric equations, the transition of [tex]\( t \)[/tex] from -4 to 1 will generate a continuous curve.
5. Analyze the general behavior:
- For [tex]\( t \)[/tex] becoming more positive, [tex]\( x \)[/tex] initially decreases, reaches a minimum at [tex]\( t = -1 \)[/tex], and then increases.
- The [tex]\( y \)[/tex]-coordinate is linearly decreasing.
The graph will have points from right to left first reaching downwards, especially around [tex]\( (-1, 1) \)[/tex] being the turning point where the slope of [tex]\( x \)[/tex] switches direction.
6. Conclusion:
The resultant graph is a curved path starting from a peak on the top-right (for larger negative [tex]\( t \)[/tex]), dips, and climbs back as [tex]\( t \)[/tex] goes from [tex]\(-4\)[/tex] to [tex]\( 1 \)[/tex]. It shows the inverse linear behavior in the y-direction and a parabola-like behavior in the x-direction until it climbs again towards a point sweeping from [tex]\((-x_0, y_1)\)[/tex] gradually turning upwards.
