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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.

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Your answer is correct. Does the site want the answer in the form

$x=\frac{5y^2}{4}+\frac{50y}{4}+\frac{133}{4}$?

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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.

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