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A parabola opening up or down has vertex $(0,-6)$ and passes through $(12,12)$ . 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,-6)

and passes through

(12,12)

. Write its equation in vertex form.

\newline

Simplify any fractions.

Identify Vertex Values: We have: $\newline$ Vertex: $(0,-6)$ $\newline$ Identify the values of $h$ and $k$ . $\newline$ Vertex is $(0,-6)$ . $\newline$ $h = 0$ $\newline$ $k = -6$
Select Equation: We have: $\newline$ $y = a(x - h)^2 + k$ $\newline$ $h = 0$ and $k = -6$ $\newline$ Select the equation after substituting the values of $h$ and $k$ . $\newline$ Substitute $h = 0$ and $k = -6$ in $y = a(x - h)^2 + k$ . $\newline$ $y = a(x - 0)^2 - 6$ $\newline$ $y = ax^2 - 6$
Find Value of a: We have: $y = ax^2 - 6$
Point: $(12,12)$
Find the value of a.
$y = ax^2 - 6$
$12 = a(12)^2 - 6$
$12 + 6 = a \times 144$
$18 = 144a$
$\frac{18}{144} = a$
$a = \frac{1}{8}$
Write Vertex Form: We found: $\newline$ $a = \frac{1}{8}$ $\newline$ $h = 0$ $\newline$ $k = -6$ $\newline$ Write the equation of a parabola in vertex form. $\newline$ $y = a(x - h)^2 + k$ $\newline$ $y = \frac{1}{8}(x - 0)^2 - 6$ $\newline$ Vertex form: $y = \frac{1}{8}x^2 - 6$

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