SOLUTION: Determine the equation of parabola which has turning point (3;4) and passes trough the point (4;2)

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Question 972032: Determine the equation of parabola which has turning point (3;4) and passes trough the point (4;2)
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
parabolas are symmetric around the vertex x-value. From (3,4) to (4,2) is one point; another is (2,2) The vertex is higher than the point, so this is convex upward, and a will be negative.
Vertex form of a parabola is a (x-h)^2 +k ; h is 3 and k is 4
a (x-3)^2 + 4=y
substitute for x and y
a(4-3)^2+4=2 ; a(2-3)^2 +4=2
a +4=-2 ; a=-2;;;; a+4=2, a= -2
y=-2(x-3)^2 +4 = -2(x^2-6x+9)+4
y= -2x^2 +12x -14,
Check: vertex x-value should be -b/2a; -12/-4 =3.

graph+%28300%2C300%2C-10%2C10%2C-10%2C10%2C-2x%5E2%2B12x-14%29