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A parabola opening up or down has vertex $(0,1)$ and passes through $(12,-11)$ . 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

(0,1)

and passes through

(12,-11)

. 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 $(0, 1)$ ? $\newline$ Substitute $0$ for $h$ and $1$ for $k$ in the vertex form. $\newline$ $y = a(x - 0)^2 + 1$ $\newline$ $y = ax^2 + 1$
Use Point to Find $a$ : Use the point $(12, -11)$ to find the value of $a$ . Replace the variables with $(12, -11)$ in the equation. Substitute $12$ for $x$ and $-11$ for $y$ . $-11 = a(12)^2 + 1$ $-11 = 144a + 1$
Solve for a: Solve for a. $\newline$ $-11 = 144a + 1$ $\newline$ Subtract $1$ from both sides. $\newline$ $-12 = 144a$ $\newline$ Divide both sides by $144$ . $\newline$ $a = -\frac{12}{144}$ $\newline$ $a = -\frac{1}{12}$
Write Equation with $a$ : Write the equation of the parabola in vertex form using the value of $a$ . Substitute $-\frac{1}{12}$ for $a$ in the equation $y = ax^2 + 1$ . $y = \left(-\frac{1}{12}\right)x^2 + 1$ Vertex form of the parabola: $y = -\frac{1}{12}x^2 + 1$

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