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A parabola opening up or down has vertex $(0,2)$ and passes through $(-6,5)$ . 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

(0,2)

and passes through

(-6,5)

. Write its equation in vertex form.

\newline

Simplify any fractions.

Vertex Form: 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 Equation: What is the equation of a parabola with a vertex at $(0, 2)$ ? $\newline$ Substitute $0$ for $h$ and $2$ for $k$ in the vertex form. $\newline$ $y = a(x - 0)^2 + 2$ $\newline$ $y = ax^2 + 2$
Find Value of $a$ : Use the point $(-6, 5)$ to find the value of $a$ . Replace the variables with $(-6, 5)$ in the equation. Substitute $-6$ for $x$ and $5$ for $y$ . $5 = a(-6)^2 + 2$ $5 = 36a + 2$
Solve for a: Solve for a. $\newline$ $5 = 36a + 2$ $\newline$ $5 - 2 = 36a$ $\newline$ $3 = 36a$ $\newline$ $a = \frac{3}{36}$ $\newline$ $a = \frac{1}{12}$
Equation with $a=\frac{1}{12}$ : What is the equation of the parabola if $a = \frac{1}{12}$ ? $\newline$ Substitute $\frac{1}{12}$ for $a$ in the equation $y = ax^2 + 2$ . $\newline$ $y = \left(\frac{1}{12}\right)x^2 + 2$ $\newline$ Vertex form of the parabola: $y = \left(\frac{1}{12}\right)x^2 + 2$

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Solve by completing the square.

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