Using a directrix of y = −2 and a focus of (2, 6), what quadratic function is created?
Asked by maham237 @ in Mathematics viewed by 319 People
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
Asked by maham237 @ in Mathematics viewed by 319 People
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
Answered by maham237 @
Answer:
B. f(x)=1/16(x-2)^2+2
Step-by-step 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
__
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.)
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