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@jim_thompson5910
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Using a directrix of y = -2 and a focus of (1, 6), what quadratic function is created? f(x) = one eighth (x - 1)2 - 2 f(x) = -one eighth (x + 1)2 - 2 f(x) = -one sixteenth (x + 1)2 - 2 f(x) = one sixteenth (x - 1)2 + 2
@Nnesha
@misty1212
@timo86m
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no one likes these conic section problems, they are not that hard
|dw:1433727883608:dw|
parabola opens up, and the vertex is half way between \((1,6)\) and \(y=-2\) so it is at \((1,3)\)
general form will be \[f(x)=\frac{1}{4p}(x-h)^2+k\] so you have \[f(x)=\frac{1}{4p}(x-1)^2+6\] all you need is \(p\)
\(p\) is the distance between the vertex and the focus (or the vertex and the directrix) which is 3, so final answer is \[f(x)=\frac{1}{12}(x-1)^2+6\]
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