Asked by maria
Find the standard equation of a parabola with zeros at x=2 and x=15 passing through the point (6,β5).
Remember to write your answers as exact numbers (fractions), rather than a decimal approximation.
thank you
Remember to write your answers as exact numbers (fractions), rather than a decimal approximation.
thank you
Answers
Answered by
Anonymous
(x-2)(x-15) = x^2 - 17 x + 30 = 4a(y-k)
if x = 6, y = -5 and also if x = (2+15)/2 , y = 0
36 - 102 + 30 = 4 a (-5-k)
-36 = -20 a -4 k a
and
(17/2)^2 - 17(17/2) + 30 = -4ak
so
-36 = -20 a - 17^2/4 + 30
20 a = 66 + 17^2/4
20 a = (264 + 289)/4 = 553/4
a = 553 / 80
plug and chug
if x = 6, y = -5 and also if x = (2+15)/2 , y = 0
36 - 102 + 30 = 4 a (-5-k)
-36 = -20 a -4 k a
and
(17/2)^2 - 17(17/2) + 30 = -4ak
so
-36 = -20 a - 17^2/4 + 30
20 a = 66 + 17^2/4
20 a = (264 + 289)/4 = 553/4
a = 553 / 80
plug and chug
Answered by
maria
How do write that in standard dorm
y=...(x+-)^2+-.....
Answered by
anonymous maria
help anonymous
thank you
thank you
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