Eliminate the Parameter
-$$x = 2 t , y = t + 1 ; - 2 \leq t \leq 3$$
A) $y = \frac { 1 } { 2 } x + 1 ; - 4 \leq x \leq 6$
B) $y = - 2 x + 1 ; - \infty

Question

Eliminate the Parameter
-$$x = 2 t , y = t + 1 ; - 2 \leq t \leq 3$$
A) $y = \frac { 1 } { 2 } x + 1 ; - 4 \leq x \leq 6$
B) $y = - 2 x + 1 ; - \infty < x < \infty$
C) $y = \frac { 1 } { 2 } x - 1 ; - \infty < x < \infty$
D) $\mathrm { y } = \mathrm { x } ^ { 2 } + 1 ; - 2 \leq \mathrm { x } \leq 2$

Quizplus · Accepted Answer

To eliminate the parameter $t$, we solve the first equation for $t$: $t = \frac{x}{2}$. Substituting this into the second equation gives $y = \frac{x}{2} + 1$, which simplifies to $y = \frac{1}{2}x + 1$. The range of $x$ is found by substituting the limits of $t$ into $x = 2t$, giving $x$ values from $-4$ to $6$, matching option A.

Eliminate the Parameter - x=2t,y=t+1;−2≤t≤3x = 2 t , y = t + 1 ; - 2 \leq t \leq 3x=2t,y=t+1;−2≤t≤3

Eliminate the Parameter
- $x = 2 t , y = t + 1 ; - 2 \leq t \leq 3$