Asked by George
Find a formula for the following function.
A parabola opening downward with its vertex at (6,1) and y -intercept equal to -35.
(y-1)^2=4(-35)(x-6) Please help
A parabola opening downward with its vertex at (6,1) and y -intercept equal to -35.
(y-1)^2=4(-35)(x-6) Please help
Answers
Answered by
N'aicha
It means that y is equal to 3-1 times 2=4 times-35 times x-6.Use PEMDASLR to find your answer.
Answered by
Damon
I agree that it is of form
(y-1)^2 = 4a (x-6)
but then when x = 0, y must be -35 so put that in
(-36)^2 = 4a (-6)
4a = - 216
so I think it is
(y-1)^2 = -4*54 (x-6)
(y-1)^2 = 4a (x-6)
but then when x = 0, y must be -35 so put that in
(-36)^2 = 4a (-6)
4a = - 216
so I think it is
(y-1)^2 = -4*54 (x-6)
Answered by
Reiny
Damon I respectfully disagree
Your parabola opens sideways
I think you meant to say:
y-1 = a(x-6)^2
then at (0,-35)
-36 = a(36)
a = -1
so y-1 = -(x-6)^2
y = -(x-6)^2 + 1
Your parabola opens sideways
I think you meant to say:
y-1 = a(x-6)^2
then at (0,-35)
-36 = a(36)
a = -1
so y-1 = -(x-6)^2
y = -(x-6)^2 + 1
Answered by
Damon
You are right, I was not thinking!
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