Eliminate the Parameter
-$$\mathrm { x } = \mathrm { t } ^ { 3 } + 1 , \mathrm { y } = \mathrm { t } ^ { 3 } - 10 ; - 2 \leq \mathrm { t } \leq 2$$
A) $\mathrm { y } = \mathrm { x } - 11 ; - 7 \leq \mathrm { x } \leq 9$
B) $y = - x - 11 ; - 7 \leq x \leq 9$
C) $\mathrm { y } = - \mathrm { x } ^ { 2 } ; - 4 \leq \mathrm { x } \leq 4$
D) $y = x ^ { 3 } ; - 3 \leq x \leq 1$

Question

Quizplus · Accepted Answer

Since both $x$ and $y$ are defined in terms of $t^3$, we can eliminate $t$ by expressing $y$ in terms of $x$. From the given equations, $y = x - 11$ is obtained by subtracting the equation for $x$ from the equation for $y$. The range for $x$ is found by substituting the limits of $t$ into the equation for $x$, resulting in $x$ ranging from $-7$ to $9$.

Eliminate the Parameter - x=t3+1,y=t3−10;−2≤t≤2\mathrm { x } = \mathrm { t } ^ { 3 } + 1 , \mathrm { y } = \mathrm { t } ^ { 3 } - 10 ; - 2 \leq \mathrm { t } \leq 2x=t3+1,y=t3−10;−2≤t≤2

Eliminate the Parameter
- $\mathrm { x } = \mathrm { t } ^ { 3 } + 1 , \mathrm { y } = \mathrm { t } ^ { 3 } - 10 ; - 2 \leq \mathrm { t } \leq 2$