How do you sketch the curve reprsented by the parametric equations : x=t2-6
y=1/2t-1?
pick some values for t, calculate x and y, and plot the points. Nothing hard about that.
You can tell it will be a parabola, since you already figured out that
x = (2y+2)^2 - 6
I guess you might mean:
x = t^2-6
y = .5t - 1
?????
if so, just substitute
2y = t -2
t = 2 y +2
then
x = 4(y^2+2y+1) -6
x = 4y^2 + 8y - 2
sketch that parabola
To sketch the curve represented by the parametric equations x = t^2 - 6 and y = 1/2t - 1, follow these steps:
1. First, let's find the range for the parameter t. Since there are no restrictions mentioned, we can assume t can take any real value.
2. Next, we'll examine the behavior of the equations for extreme values of t.
a. As t approaches positive infinity, both x and y will also tend to positive infinity. This means that the curve will go off towards the top-right with a steep incline.
b. As t approaches negative infinity, both x and y will tend to negative infinity. This implies that the curve will go off towards the bottom-left with a steep decline.
3. To get a clear understanding of the shape of the curve, we can plot some points. Choose a few values for t, substitute them into the equations, and compute the corresponding (x, y) coordinates.
Let's calculate the (x, y) coordinates for t = -3, -2, -1, 0, 1, 2, and 3:
- For t = -3, x = (-3)^2 - 6 = 9 - 6 = 3 and y = 1/2(-3) - 1 = -1.5 - 1 = -2.5. So, we have the point (3, -2.5).
- For t = -2, x = (-2)^2 - 6 = 4 - 6 = -2 and y = 1/2(-2) - 1 = -1 - 1 = -2. So, we have the point (-2, -2).
- For t = -1, x = (-1)^2 - 6 = 1 - 6 = -5 and y = 1/2(-1) - 1 = -0.5 - 1 = -1.5. So, we have the point (-5, -1.5).
- For t = 0, x = (0)^2 - 6 = 0 - 6 = -6 and y = 1/2(0) - 1 = 0 - 1 = -1. So, we have the point (-6, -1).
- For t = 1, x = (1)^2 - 6 = 1 - 6 = -5 and y = 1/2(1) - 1 = 0.5 - 1 = -0.5. So, we have the point (-5, -0.5).
- For t = 2, x = (2)^2 - 6 = 4 - 6 = -2 and y = 1/2(2) - 1 = 1 - 1 = 0. So, we have the point (-2, 0).
- For t = 3, x = (3)^2 - 6 = 9 - 6 = 3 and y = 1/2(3) - 1 = 1.5 - 1 = 0.5. So, we have the point (3, 0.5).
4. Now, plot these points on a graph making sure to include the behavior described in step 2. Once all the points are plotted, draw a smooth curve that passes through them.
The resulting graph should resemble a curved line that slopes upwards to the right, passes through the point (0, -1), and flattens out towards the top right.