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A parabola opening up or down has vertex $(0,7)$ and passes through $(4,5)$ . 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

(0,7)

and passes through

(4,5)

. Write its equation in vertex form.

\newline

Simplify any fractions.

Vertex Form of Parabola: 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 $(0, 7)$ ? $\newline$ Substitute $0$ for $h$ and $7$ for $k$ in the vertex form. $\newline$ $y = a(x - 0)^2 + 7$ $\newline$ $y = ax^2 + 7$
Use Point to Find 'a': Use the point $(4, 5)$ to find the value of 'a'. $\newline$ Replace the variables with $(4, 5)$ in the equation. $\newline$ Substitute $4$ for $x$ and $5$ for $y$ . $\newline$ $5 = a(4)^2 + 7$ $\newline$ $5 = 16a + 7$
Solve for 'a': Solve for 'a'. $\newline$ $5 = 16a + 7$ $\newline$ Subtract $7$ from both sides. $\newline$ $5 - 7 = 16a$ $\newline$ $-2 = 16a$ $\newline$ Divide both sides by $16$ . $\newline$ $-\frac{2}{16} = a$ $\newline$ $-\frac{1}{8} = a$
Write Equation with 'a': Write the equation of the parabola with the found value of 'a'. $\newline$ Substitute $-\frac{1}{8}$ for $a$ in the equation $y = ax^2 + 7$ . $\newline$ $y = \left(-\frac{1}{8}\right)x^2 + 7$ $\newline$ Vertex form of the parabola: $y = -\left(\frac{1}{8}\right)x^2 + 7$

More problems from Write a quadratic function from its vertex and another point

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Solve by completing the square.

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m^2 - 10m - 29 = 0

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\newline

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