Respuesta :
The correct parametric equations are:
[tex]x=4t-1 \\ \\ y=2t^{2}+3t-4 [/tex]
From first equation, we can write the value of t in terms of x to be:
[tex]t= \frac{x+1}{4} [/tex]
Using this value of t in equation of y, we get:
[tex]y=2( \frac{x+1}{4} )^{2}+3( \frac{x+1}{4} )-4 \\ \\ y=2( \frac{ x^{2} +2x+1}{16} )+ \frac{3(x+1)}{4} -4 \\ \\ y= \frac{ x^{2} +2x+1}{8}+ \frac{3(x+1)}{4}-4 \\ \\ y= \frac{ x^{2} +2x+1+6x+6-32}{8} \\ \\ y= \frac{ x^{2} +8x-25}{8} [/tex]
Thus, the curve in option C is described by the given parametric equations.
[tex]x=4t-1 \\ \\ y=2t^{2}+3t-4 [/tex]
From first equation, we can write the value of t in terms of x to be:
[tex]t= \frac{x+1}{4} [/tex]
Using this value of t in equation of y, we get:
[tex]y=2( \frac{x+1}{4} )^{2}+3( \frac{x+1}{4} )-4 \\ \\ y=2( \frac{ x^{2} +2x+1}{16} )+ \frac{3(x+1)}{4} -4 \\ \\ y= \frac{ x^{2} +2x+1}{8}+ \frac{3(x+1)}{4}-4 \\ \\ y= \frac{ x^{2} +2x+1+6x+6-32}{8} \\ \\ y= \frac{ x^{2} +8x-25}{8} [/tex]
Thus, the curve in option C is described by the given parametric equations.
1. Counter clockwise
2. (30,401)
3. t=2(x-3)
4. She should have taken both the positive and negative square root
5. y= x^2+8x-25/8
6. Hyperbola
7. (A) the ellipse is positive meaning it is above the x axis it is going counter clock wise so the arrow on top of the ellipse is pointing left and the arrow on the bottom of the ellipse is pointing right
