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Using a directrix of y=-2 and a focus of (2,6), what quadratic function is created?

A. f(x)=-1/8(x-2)^2-2
B. f(x)=1/16(x-2)^2+2
C. f(x)=1/8(x-2)^2-2
D. f(x)=-1/16(x+2)^2-2

User Deeksy
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2 Answers

2 votes

Answer:

b). f(x) = 1/16 (x-2)^2 + 2

Explanation:

took the test

User Sod
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5.4k points
2 votes

Answer:

B. f(x)=1/16(x-2)^2+2

Explanation:

The equation can be written as ...

f(x) = 1/(4p)(x -h)^2 +(k-p)

where (h, k) is the focus, and p is half the distance between focus and directrix. Here, (h, k) = (2, 6) and p=(6-(-2))/2 = 4. So the equation is ...

f(x) = 1/16(x -2)^2 +2

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The graph shows the focus, directrix, and parabola. It also shows the parabola is the set of points that are the same distance from focus and directrix. (Dashed orange lines are the same length.)

Using a directrix of y=-2 and a focus of (2,6), what quadratic function is created-example-1
User Sharra
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