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A parabola opening up or down has vertex $(0,0)$ and passes through $(-8,16)$ . 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

(0,0)

and passes through

(-8,16)

. Write its equation in vertex form.

\newline

Simplify any fractions.

Vertex Form Explanation: What is the vertex form of the parabola? $\newline$ 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 at Vertex: 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. $\newline$ $y = a(x - 0)^2 + 0$ $\newline$ $y = ax^2$
Value of 'a' Calculation: Determine the value of 'a' using the point $(-8, 16)$ that lies on the parabola. $\newline$ We substitute $x = -8$ and $y = 16$ into the equation $y = ax^2$ to find 'a'. $\newline$ $16 = a(-8)^2$ $\newline$ $16 = 64a$
Solving for 'a': Solve for 'a'. $\newline$ Divide both sides of the equation by $64$ to isolate 'a'. $\newline$ $a = \frac{16}{64}$ $\newline$ $a = \frac{1}{4}$
Final Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of 'a' found in Step $4$ . $\newline$ Substitute $a = \frac{1}{4}$ into the equation $y = ax^2$ . $\newline$ $y = \left(\frac{1}{4}\right)x^2$ $\newline$ This is the equation of the parabola in vertex form.

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