Find the equation in standard form of the parabola that has vertex (2, −5), has its axis of symmetry parallel to the x-axis, and passes through the point (7, −3).
I got $f(y)=\frac{5y^2}{4}+\frac{50y}{4}+\frac{133}{4}$ My Analytic geometry skills are very rusty, though I'm sure of my answer the problem is correct though the system of the online quiz keeps rejecting this.
Your answer is correct. Does the site want the answer in the form
$x=\frac{5y^2}{4}+\frac{50y}{4}+\frac{133}{4}$?
The parabola, with its vertex at $(2,-5)$, has the form
$$x-2=c(y+5)^2$$
Plug the point $(7,-3)$ in to find $c$ to be $c = \frac{5}{4}$. Thus, its form is
$$x-2=\frac{5}{4}(y+5)^2$$
or, in its standard form
$$x=\frac 54 y^2+\frac {25}{2}y + \frac{133}{4}$$
So, you just need to simplify the middle coefficient.
