Answer
(a) $x = 2t-1$
$y = t^2+2$
We can see the graph below.
(b) $y = \frac{x^2+2x+9}{4}$
Work Step by Step
(a) $x = 2t-1$
$y = t^2+2$
When $t = -3$:
$x = 2(-3)-1 = -7$
$y = (-3)^2+2 = 11$
When $t = -2$:
$x = 2(-2)-1 = -5$
$y = (-2)^2+2 = 6$
When $t = -1$:
$x = 2(-1)-1 = -3$
$y = (-1)^2+2 = 3$
When $t = 0$:
$x = 2(0)-1 = -1$
$y = (0)^2+2 = 2$
When $t = 1$:
$x = 2(1)-1 = 1$
$y = (1)^2+2 = 3$
When $t = 2$:
$x = 2(2)-1 = 3$
$y = (2)^2+2 = 6$
When $t = 3$:
$x = 2(3)-1 = 5$
$y = (3)^2+2 = 11$
We can see the graph below.
(b) $x = 2t-1$
$t = \frac{x+1}{2}$
We can replace this expression for $t$ in the equation for $y$:
$y = t^2+2$
$y = (\frac{x+1}{2})^2+2$
$y = (\frac{x^2+2x+1}{4})+2$
$y = \frac{x^2+2x+9}{4}$
