Eliminating the parameter?

Question

How would you eliminate the parameter where the x coordinate is in terms of t, but the t is squared.

x= 3t - $t^2$
y= t + 1

I know to solve for y as a function of x, but I'm not sure how to do so with powers.

Answer 1

$y = t + 1 \implies t = y-1 \implies x = 3t-t^2 = 3(y-1)-(y-1)^2$.

This is a quadratic in $y$ and can be solved using the quadratic formula.

Answer 2

$y = t + 1$ and so $t = y - 1$

then you substitute this value in the 1st equation to get $$x= 3t -t^2 = t(3 - t) = (y-1)(3-(y-1)) = (y-1)(4-y) = 4y + y -y^2 - 4 = 5y -y^2 - 4$$