Answer
$x = 4t-3$
$y = t^2$
$t$ in $[-3,4]$
We can see the graph below:
Work Step by Step
$x = 4t-3$
$y = t^2$
$t$ in $[-3,4]$
When $t = -3$:
$x = 4(-3)-3 = -15$
$y = (-3)^2 = 9$
When $t = -2$:
$x = 4(-2)-3 = -11$
$y = (-2)^2 = 4$
When $t = -1$:
$x = 4(-1)-3 = -7$
$y = (-1)^2 = 1$
When $t = 0$:
$x = 4(0)-3 = -3$
$y = (0)^2 = 0$
When $t = 1$:
$x = 4(1)-3 = 1$
$y = (1)^2 = 1$
When $t = 2$:
$x = 4(2)-3 = 5$
$y = (2)^2 = 4$
When $t = 3$:
$x = 4(3)-3 = 9$
$y = (3)^2 = 9$
When $t = 4$:
$x = 4(4)-3 = 13$
$y = (4)^2 = 16$
We can see the graph below:
