Question

1. you get this: fill in this: factored form of the equation standard form of the equation

Explanation:

Step1: Determine the vertex and a point on the parabola

The vertex of the parabola (since it opens to the left) is \((h,k)=(10, - 1)\). Let's take a point \((x,y)=(0, - 4)\) on the parabola. The general form of a parabola that opens to the left is \(x=a(y - k)^{2}+h\).
Substitute \(h = 10,k=-1,x = 0,y=-4\) into \(x=a(y - k)^{2}+h\):
\(0=a(-4+1)^{2}+10\)
\(0 = 9a+10\)
\(9a=-10\)
\(a=-\frac{10}{9}\)
The equation in vertex - form is \(x=-\frac{10}{9}(y + 1)^{2}+10\)

Step2: Expand to standard form

$$
LATEXBLOCK0
$$

Step3: Find the factored form (assuming we rewrite the vertex - form in a different factored - like way for a parabola opening left. If we consider the original vertex - form \(x=-\frac{10}{9}(y + 1)^{2}+10=\frac{-10(y + 1)^{2}+90}{9}=\frac{-10((y + 1)^{2}-9)}{9}=\frac{-10(y + 1 - 3)(y + 1+3)}{9}=\frac{-10(y - 2)(y+4)}{9}\)

Answer:

• Standard form: \(10y^{2}+20y + 9x-80 = 0\)
• Factored form: \(x=\frac{-10(y - 2)(y + 4)}{9}\)