What is the equation of the quadratic graph with a focus of (8, −8) and a directrix of y = −6?
f(x) = −one fourth (x − 7)2 + 1
f(x) = −one fourth (x − 8)2
f(x) = −one fourth (x − 8)2 − 7
f(x) = one fourth (x − 7)2
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The general equation for a vertical parabola is (x-h)^2 = 4p(y-k), where (h,k) is the vertex and p is the directed distance from the vertex to the focus.
For the parabola in question (h,k) = (8,-7) and p = -1, so the equation of the quadratic graph is
(x-8)^2 = -4(y+7).
The vertex is halfway between the focus and directrix, so will be (8, -7). The only selection among the answers that has the correct vertex is
.. f(x) = (-1/4)(x - 8)^2 - 7
As a check, you can find a point on the parabola that is both 2 units from the directrix and 2 units from the vertex. Such points are (6, -8) and (10, -8). Both of these points will satisfy the equation above.
4 p [ y - (-8)] = [ x - 8 ]² where p = distance from directrix to focus...{ - 2 }...
The general equation for a vertical parabola is (x-h)^2 = 4p(y-k), where (h,k) is the vertex and p is the directed distance from the vertex to the focus.
For the parabola in question (h,k) = (8,-7) and p = -1, so the equation of the quadratic graph is
(x-8)^2 = -4(y+7).
The vertex is halfway between the focus and directrix, so will be (8, -7). The only selection among the answers that has the correct vertex is
.. f(x) = (-1/4)(x - 8)^2 - 7
As a check, you can find a point on the parabola that is both 2 units from the directrix and 2 units from the vertex. Such points are (6, -8) and (10, -8). Both of these points will satisfy the equation above.
4 p [ y - (-8)] = [ x - 8 ]² where p = distance from directrix to focus...{ - 2 }...
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