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Algebra
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Identify the vertex, focus, and directrix.

y= 1/2 x^2

May 18th, 2015

  y = (1/12)x^2

For an equation of the form y = a(x - b)^2 + c, where a, b, c are all real numbers, the vertex is at (b , c).
In this case, we have y = (1/12)(x - 0)^2 + 0, thus the vertex is (0 , 0).

Rearranging y = (1/12)x^2, we have x^2 = 12y

Suppose we have an equation of the form (x - a)^2 = b(y - c), where a, b, c are all real numbers.
Suppose the vertex is (d , e), where d, e are some real numbers.
The focus will be (d , e + (b/4))
The directrix will be y = e - (b/4)

Notice that in our case, b = 12
Thus, b / 4 = 12 / 4 = 3
Our vertex is (0 , 0).
Hence, the focus is (0 , 0 + 3) = (0 , 3).

The directrix will be:
y = 0 - 3
y = -3

Hence, the answer is c).

May 18th, 2015

5 star always give me

May 18th, 2015

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May 18th, 2015
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May 18th, 2015
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