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A parabola opening up or down has vertex $(0,1)$ and passes through $(-8,-15)$ . 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

(-8,-15)

. 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 $(0, 1)$ ? $\newline$ Substitute $0$ for $h$ and $1$ for $k$ in the vertex form. $\newline$ $y = a(x - 0)^2 + 1$ $\newline$ $y = ax^2 + 1$
Find Value of $a$ : Use the point $(-8, -15)$ to find the value of $a$ . Replace the variables with $(-8, -15)$ in the equation. Substitute $-8$ for $x$ and $-15$ for $y$ . $-15 = a(-8)^2 + 1$ $-15 = 64a + 1$
Solve for a: Solve for a. $\newline$ $-15 = 64a + 1$ $\newline$ Subtract $1$ from both sides. $\newline$ $-16 = 64a$ $\newline$ Divide both sides by $64$ . $\newline$ $-\frac{16}{64} = a$ $\newline$ Simplify the fraction. $\newline$ $-\frac{1}{4} = a$
Equation with $a = -\frac{1}{4}$ : What is the equation of the parabola if $a = -\frac{1}{4}$ ? $\newline$ Substitute $-\frac{1}{4}$ for $a$ in the equation $y = ax^2 + 1$ . $\newline$ $y = (-\frac{1}{4})x^2 + 1$ $\newline$ Vertex form of the parabola: $y = -\frac{1}{4}x^2 + 1$

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