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A parabola opening up or down has vertex $(0,4)$ and passes through $(8,-12)$ . Write its equation in vertex form. $\newline$ Simplify any fractions.

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Write a quadratic function from its vertex and another point

Full solution

Q. A parabola opening up or down has vertex

(0,4)

and passes through

(8,-12)

. Write its equation in vertex form.

\newline

Simplify any fractions.

Vertex Form Explanation: What is the vertex form of the parabola? $\newline$ 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 $(0, 4)$ ? $\newline$ Substitute $0$ for $h$ and $4$ for $k$ in the vertex form. $\newline$ $y = a(x - 0)^2 + 4$ $\newline$ $y = ax^2 + 4$
Finding Value of 'a': Use the point $(8, -12)$ to find the value of 'a'. $\newline$ Replace the variables with $(8, -12)$ in the equation. $\newline$ Substitute $8$ for $x$ and $-12$ for $y$ . $\newline$ $-12 = a(8)^2 + 4$ $\newline$ $-12 = 64a + 4$
Solving for 'a': Solve for 'a'. $\newline$ $-12 = 64a + 4$ $\newline$ Subtract $4$ from both sides. $\newline$ $-12 - 4 = 64a$ $\newline$ $-16 = 64a$ $\newline$ Divide both sides by $64$ . $\newline$ $-\frac{16}{64} = a$ $\newline$ $-\frac{1}{4} = a$
Final Parabola Equation: Write the equation of the parabola using the value of $a$ . $\newline$ Substitute $-\frac{1}{4}$ for $a$ in the equation $y = ax^2 + 4$ . $\newline$ $y = \left(-\frac{1}{4}\right)x^2 + 4$ $\newline$ Vertex form of the parabola: $y = -\left(\frac{1}{4}\right)x^2 + 4$

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