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. $$x = 2 t + 1 , \quad y = 3 t , \quad t \in [ - 1,3 ]$$
A) $y = \frac { 3 } { 2 } x + 1 , \quad - 1 \leq x \leq 7 , \text { left to right }$
B) $y = \frac { 3 } { 2 } x - 1 , \quad - 1 \leq x \leq 7 , \text { left to right }$
C) $y = \frac { 3 } { 2 } ( x - 1 ) , \quad - 1 \leq \mathrm { x } \leq 7 , \text { left to right }$
D) $y = \frac { 3 } { 2 } ( x - 1 ) , \quad - 1 \leq \mathrm { x } \leq 7 , \quad \text { right to left }$

Question

Quizplus · Accepted Answer

To eliminate the parameter $t$, solve the first equation for $t$: $t = \frac{x - 1}{2}$. Substitute this into the second equation to get $y = 3\left(\frac{x - 1}{2}\right)$, which simplifies to $y = \frac{3}{2}(x - 1)$. The range of $x$ is found by substituting the limits of $t$ into $x = 2t + 1$, giving $x$ values from $-1$ to $7$. The direction is left to right since as $t$ increases, $x$ increases.

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

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