Write the equation of a parabola with a vertex of (2,3) that opens downward and is congruent to y=1/3x^2.  

Expert Answers

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You need to use the vertex form of equation of parabola, such that:

`y = a(x - h)^2 + k`

`(h,k)` represents the vertex of parabola

Since the problem provides the coordinates of vertex, `(2,3)` , you need to substitute the coordinates in equation above, such that:

`y = a(x - 2)^2 + 3`

You need to find the leading coefficient a using the information provided by the problem, hence, since parabola you need to determine is congruent to `y = (1/3)x^2` , hence the magnitudes of the leading coefficients are equal, such that: `a = |1/3|`

Since the problem provides the information that parabola opens downward, hence `a = -1/3` .

`y = (-1/3)(x - 2)^2 + 3 `

Hence, evaluating the equation of parabola that opens downward, in vertex form, yields `y = (-1/3)(x - 2)^2 + 3.`

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