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A parabola opening up or down has vertex $(0,-2)$ and passes through $(-6,7)$ . 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

(-6,7)

. Write its equation in vertex form.

\newline

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.
Equation with Vertex: What is the equation of a parabola with a vertex at $(0, -2)$ ? $\newline$ Since the vertex is $(0, -2)$ , we substitute $h = 0$ and $k = -2$ into the vertex form equation. $\newline$ $y = a(x - 0)^2 - 2$ $\newline$ $y = ax^2 - 2$
Value of 'a' Calculation: Determine the value of 'a' using the point $(-6, 7)$ . $\newline$ We know the parabola passes through the point $(-6, 7)$ , so we substitute $x = -6$ and $y = 7$ into the equation to find 'a'. $\newline$ $7 = a(-6)^2 - 2$ $\newline$ $7 = 36a - 2$
Solving for 'a': Solve for 'a'. $\newline$ To find the value of 'a', we solve the equation from the previous step. $\newline$ $7 = 36a - 2$ $\newline$ $7 + 2 = 36a$ $\newline$ $9 = 36a$ $\newline$ $a = \frac{9}{36}$ $\newline$ $a = \frac{1}{4}$
Final Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of $a$ . $\newline$ Now that we have found $a$ to be $\frac{1}{4}$ , we substitute it back into the equation $y = ax^2 - 2$ . $\newline$ $y = \left(\frac{1}{4}\right)x^2 - 2$ $\newline$ This is the equation of the parabola in vertex form.

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