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A parabola opening up or down has vertex $(0,5)$ and passes through $(6,-4)$ . 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

(0,5)

and passes through

(6,-4)

. 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 $(0, 5)$ ? $\newline$ Substitute $0$ for $h$ and $5$ for $k$ in the vertex form. $\newline$ $y = a(x - 0)^2 + 5$ $\newline$ $y = ax^2 + 5$
Use Point to Find 'a': Use the point $(6, -4)$ to find the value of 'a'. $\newline$ Replace the variables with $(6, -4)$ in the equation. $\newline$ Substitute $6$ for $x$ and $-4$ for $y$ . $\newline$ $-4 = a(6)^2 + 5$ $\newline$ $-4 = 36a + 5$
Solve for 'a': Solve for 'a'. $\newline$ $-4 = 36a + 5$ $\newline$ Subtract $5$ from both sides. $\newline$ $-4 - 5 = 36a$ $\newline$ $-9 = 36a$ $\newline$ Divide both sides by $36$ . $\newline$ $-\frac{9}{36} = a$ $\newline$ Simplify the fraction. $\newline$ $-\frac{1}{4} = a$
Write Equation with 'a': Write the equation of the parabola using the value of 'a'. $\newline$ Substitute $-\frac{1}{4}$ for $a$ in the equation $y = ax^2 + 5$ . $\newline$ $y = \left(-\frac{1}{4}\right)x^2 + 5$ $\newline$ Vertex form of the parabola: $y = -\left(\frac{1}{4}\right)x^2 + 5$

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