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A parabola opening up or down has vertex $(0,0)$ and passes through $(6,-3)$ . 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

(0,0)

and passes through

(6,-3)

. Write its equation in vertex form.

\newline

Simplify any fractions.

Identify Vertex Form: Identify the vertex form of a parabola. $\newline$ Vertex form: $y = a(x - h)^2 + k$ $\newline$ Here, $h = 0$ and $k = 0$ , so the equation simplifies to $y = ax^2$ .
Substitute Point: Substitute the point ( $6$ , $-3$ ) into the equation to find $a$ . $\newline$ Using $y = ax^2$ and substituting $x = 6$ and $y = -3$ : $\newline$ $-3 = a(6)^2$ $\newline$ $-3 = 36a$ $\newline$ $a = -3/36$ $\newline$ $a = -1/12$
Write Final Equation: Write the final equation using the value of $a$ . $\newline$ Substitute $a = -1/12$ back into the vertex form: $\newline$ $y = -\frac{1}{12}x^2$ $\newline$ This is the equation of the parabola in vertex form.

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

Question

Solve by completing the square.

\newline

m^2 - 10m - 29 = 0

\newline

Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.

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Posted 8 months ago

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g(x)

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Posted 5 months ago

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Write the equation of the parabola that passes through the points

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Question

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\newline

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