Answer:
d.(-2,0)
Step-by-step explanation:
The given parametric equation is:
[tex]x=4t[/tex]
[tex]y=12t^2+4t-1[/tex]
We make t the subject in the first equation;
[tex]t=\frac{x}{4}[/tex]
We substitute into the second equation to get:
[tex]y=12(\frac{x}{4})^2+4(\frac{x}{4})-1[/tex]
[tex]y=\frac{3}{4}x^2+x-1[/tex]
When x=4 , [tex]y=\frac{3}{4}(4)^2+4-1=15[/tex]
When x=-2 , [tex]y=\frac{3}{4}(-2)^2+-2-1=0[/tex]
Therefore the point (-2,0) lies on the given parametric curve.
