Asked by Jessie Dee
given three points: (0,30) (30,45) (60,30) and need to find a parabola (ie. ax^2+bx+c function) from these three points for a curve. Please help! :)
Answers
Answered by
drwls
It is an inverted parabola symmetric about the x=30 line.
Use the form
y = -a(x-30)^2 + C
When x = 30, (x-30) = 0 and
45 = C
When x = 0, 30 = -900a + 45
900 a = 15
a = 1/60
y = -(1/60)(x-30)^2 + 45
Multiply it out if you want it in
ax^2+bx+c form
Use the form
y = -a(x-30)^2 + C
When x = 30, (x-30) = 0 and
45 = C
When x = 0, 30 = -900a + 45
900 a = 15
a = 1/60
y = -(1/60)(x-30)^2 + 45
Multiply it out if you want it in
ax^2+bx+c form
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