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

(0,0)

and passes through

(-6, -3)

. Write its equation in vertex form. Simplify any fractions.

Vertex Form Explanation: 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.
Vertex at Origin: What is the equation of a parabola with a vertex at $(0, 0)$ ? $\newline$ Since the vertex is at the origin $(0, 0)$ , we substitute $h = 0$ and $k = 0$ into the vertex form equation. $\newline$ $y = a(x - 0)^2 + 0$ $\newline$ $y = ax^2$
Determine 'a': Determine the value of 'a' using the point $(-6, -3)$ . $\newline$ We know the parabola passes through the point $(-6, -3)$ . We substitute $x = -6$ and $y = -3$ into the equation $y = ax^2$ to find 'a'. $\newline$ $-3 = a(-6)^2$ $\newline$ $-3 = 36a$
Solve for 'a': Solve for 'a'. $\newline$ Divide both sides of the equation by $36$ to solve for 'a'. $\newline$ $a = \frac{-3}{36}$ $\newline$ $a = \frac{-1}{12}$
Write Equation: Write the equation of the parabola in vertex form using the value of $a$ . Substitute $a = -\frac{1}{12}$ into the equation $y = ax^2$ . $y = \left(-\frac{1}{12}\right)x^2$ This is the equation of the parabola in vertex form.

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