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A parabola opening up or down has vertex $(0,-4)$ 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

Full solution

Q. A parabola opening up or down has vertex

(0,-4)

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? $\newline$ Vertex form of a parabola: $y = a(x - h)^2 + k$
Equation at Vertex: What is the equation of a parabola with a vertex at $(0, -4)$ ? $\newline$ Substitute $0$ for $h$ and $-4$ for $k$ in vertex form. $\newline$ $y = a(x - 0)^2 + (-4)$ $\newline$ $y = ax^2 - 4$
Substitute Values: $y = ax^2 - 4$ $\newline$ Replace the variables with $(-6, 5)$ in the equation. $\newline$ Substitute $-6$ for $x$ and $5$ for $y$ . $\newline$ $5 = a(-6)^2 - 4$ $\newline$ $5 = 36a - 4$
Solve for $a$ : $5 = 36a - 4$ Solve for $a$ . $5 + 4 = 36a$ $9 = 36a$ $\frac{9}{36} = a$ $\frac{1}{4} = a$
Equation with $a=\frac{1}{4}$ : $y = ax^2 - 4$ $\newline$ What is the equation of the parabola if $a = \frac{1}{4}$ ? $\newline$ Substitute $\frac{1}{4}$ for $a$ . $\newline$ $y = (\frac{1}{4})x^2 - 4$ $\newline$ Vertex form of the parabola: $y = (\frac{1}{4})x^2 - 4$

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Solve by completing the square.

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