Chapter 10 - Parametric Equations and Polar Coordinates - 10.5 Exercises - Page 700: 36

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

$(y-k)^2=4p(x-h)$ which can be re-written as: $x=ay^2+by+c$ Since $(-1,0)$ is a point on the parabola, we can write $c=-1$ and $2=a-b$ ....(1) Since $(3,1)$ is a point on the parabola, we can write $4=a+b$ ....(2) Add equations (1) and (2), we get $a=3$ then from equation (2), we have $b=1$ Thus, $x=3y^2+y-1$