Eliminate the Parameter
-$x = \sqrt { t } , y = 2 t + 4 ; 0 \leq t \leq 4$
A) $y = 2 x ^ { 2 } + 4 ; 0 \leq x \leq 2$
B) $y = 4 x ^ { 2 } + 2 ; - 1 \leq x \leq 2$
C) $y = - 4 x ^ { 2 } + 17 ; 0 \leq x \leq 2$
D) $\mathrm { y } = - 4 \mathrm { x } + 17 ; 0 \leq \mathrm { x } \leq 2$

Question

Quizplus · Accepted Answer

Given $x = \sqrt{t}$, we can square both sides to get $t = x^2$. Substituting $t = x^2$ into $y = 2t + 4$ gives $y = 2x^2 + 4$. The domain of $x$ is from 0 to 2 because $x = \sqrt{t}$ and $t$ ranges from 0 to 4, so $x$ ranges from $\sqrt{0} = 0$ to $\sqrt{4} = 2$.

Eliminate the Parameter - x=t,y=2t+4;0≤t≤4x = \sqrt { t } , y = 2 t + 4 ; 0 \leq t \leq 4x=t​,y=2t+4;0≤t≤4

Eliminate the Parameter
- $x = \sqrt { t } , y = 2 t + 4 ; 0 \leq t \leq 4$