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Write the equation in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola. $$x+y^2-8y=-20$$

I have seen some students answer it, but I just don't understand. I know the basics, but I think that isn't enough for this question apparently.

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Hishamsack
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The standard form of the parabola (which opens left or right in this case) is $(y-k)^2=4p(x-h)$, where $(h,k)$ is the vertex and $p$ is the distance between the vertex and the focus.

For your equation, we want to get $x+y^2-8y=-20$ into standard form:

\begin{align} x+y^2-8y&=-20 \\ y^2-8y&=-x-20 & \text{subtract $x$ both sides} \\ y^2-8y+16&=-x-20+16 & \text{complete the square on left side} \\ (y-4)^2&=-x-4 & \text{factor left side} \\(y-4)^2 &= -(x+4) & \text{factor out a negative on right side} \\ (y-4)^2 &= 4\left(-\frac{1}{4} \right)(x-(-4)) \end{align}

The last equation is standard form of $(y-k)^2=4p(x-h)$. From this we see that:

  • the vertex $(h,k)=(-4,4)$
  • the axis of symmetry $y=4$
  • the parabola opens to the left (because $p$ is negative; $p=-\frac{1}{4}$).
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