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A parabola opening up or down has vertex $(0,-3)$ and passes through $(-6,-12)$ . Write its equation in vertex form. $\newline$ Simplify any fractions.

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Write a quadratic function from its vertex and another point

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Q. A parabola opening up or down has vertex

(0,-3)

and passes through

(-6,-12)

. Write its equation in vertex form.

\newline

Simplify any fractions.

Vertex Form of Parabola: What is the vertex form of the parabola? $\newline$ The vertex form of a parabola is given by the equation $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Equation with Vertex: What is the equation of a parabola with a vertex at $(0, -3)$ ? $\newline$ Substitute $0$ for $h$ and $-3$ for $k$ in the vertex form. $\newline$ $y = a(x - 0)^2 + (-3)$ $\newline$ $y = ax^2 - 3$
Use Point to Find $a$ : Use the point $(-6, -12)$ to find the value of $a$ . $\newline$ Replace the variables with $(-6, -12)$ in the equation. $\newline$ Substitute $-6$ for $x$ and $-12$ for $y$ . $\newline$ $-12 = a(-6)^2 - 3$ $\newline$ $-12 = 36a - 3$
Solve for a: Solve for a. $\newline$ $-12 = 36a - 3$ $\newline$ Add $3$ to both sides. $\newline$ $-9 = 36a$ $\newline$ Divide both sides by $36$ . $\newline$ $-\frac{9}{36} = a$ $\newline$ Simplify the fraction. $\newline$ $-\frac{1}{4} = a$
Equation with $a=-\frac{1}{4}$ : What is the equation of the parabola if $a = -\frac{1}{4}$ ? $\newline$ Substitute $-\frac{1}{4}$ for $a$ in the equation $y = ax^2 - 3$ . $\newline$ $y = (-\frac{1}{4})x^2 - 3$ $\newline$ Vertex form of the parabola: $y = -\frac{1}{4}x^2 - 3$

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