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. $$x = - 4 t , \quad y = - 3 t + 1 , \quad t \in [ - 2,1 ]$$
A) $y = \frac { 3 } { 4 } x + 1 , \quad - 4 \leq x \leq 8 , \quad \text { left to right }$
B) $y = \frac { 3 } { 4 } x - 1 , \quad - 4 \leq x \leq 8 , \quad \text { left to right }$
C) $y = \frac { 3 } { 4 } x - 1 , \quad - 4 \leq x \leq 8 , \quad \text { right to left }$
D) $y = \frac { 3 } { 4 } x + 1 , \quad - 4 \leq x \leq 8 , \quad \text { right to left }$

Question

Quizplus · Accepted Answer

To eliminate the parameter $t$, we solve one of the equations for $t$ and substitute into the other. From $x = -4t$, we get $t = -\frac{x}{4}$. Substituting this into $y = -3t + 1$ gives $y = -3\left(-\frac{x}{4}\right) + 1 = \frac{3}{4}x + 1$. The range of $x$ is found by substituting the $t$ values into $x = -4t$, giving $x \in [-8, 4]$, not $[-4, 8]$ as stated in all options, which seems to be a mistake in the question. The direction is from right to left since as $t$ increases, $x$ decreases.

Find an Equation of Y as a Function of X x=−4t,y=−3t+1,t∈[−2,1]x = - 4 t , \quad y = - 3 t + 1 , \quad t \in [ - 2,1 ]x=−4t,y=−3t+1,t∈[−2,1]

Find an Equation of Y as a Function of X $x = - 4 t , \quad y = - 3 t + 1 , \quad t \in [ - 2,1 ]$