Find an equation of y as a function of x for the parametric equations given below by eliminating the parameter. Also, indicate the direction of the curve. $$x = - 2 t , \quad y = 6 t ^ { 2 } - 1 , \quad t \in [ - 2,3 ]$$
A) $y = \frac { 3 } { 2 } x ^ { 2 } + 1 , x \in ( - \infty , \infty ) \text {, left to right }$
B) $y = \frac { 3 } { 2 } x ^ { 2 } + 1 , x \in ( - \infty , \infty ) \text {, right to left }$
C) $y = \frac { 3 } { 2 } x ^ { 2 } - 1 , x \in ( - \infty , \infty ) \text {, left to right }$
D) $y = \frac { 3 } { 2 } x ^ { 2 } - 1 , - 6 \leq x \leq 4 \text {, right to left }$

Question

Quizplus · Accepted Answer

To eliminate the parameter $t$, we solve the equation $x = -2t$ for $t$, getting $t = -\frac{x}{2}$. Substituting this into the equation for $y$, we get $y = 6\left(-\frac{x}{2}\right)^2 - 1 = \frac{3}{2}x^2 - 1$. The domain of $x$ is determined by the given range of $t$, which translates to $x = -2(-2) = 4$ and $x = -2(3) = -6$, so $x$ ranges from $-6$ to $4$. The direction is right to left because as $t$ increases, $x$ decreases.

Find an Equation of Y as a Function of X x=−2t,y=6t2−1,t∈[−2,3]x = - 2 t , \quad y = 6 t ^ { 2 } - 1 , \quad t \in [ - 2,3 ]x=−2t,y=6t2−1,t∈[−2,3]

Find an Equation of Y as a Function of X $x = - 2 t , \quad y = 6 t ^ { 2 } - 1 , \quad t \in [ - 2,3 ]$