Need help finding 2 more points in a quadratic
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Need help finding 2 more points in a quadratic

[From: ] [author: ] [Date: 11-11-14] [Hit: ]
5) and Im trying to figure out how to get 2 more points so I can graph more accurately. I know its a downwards parabola, Im just stuck with finding the other points. Any help would be greatly appreciated.How I can find points on both sides of the parabola?f( - 2) = - 2*( - 2)^2 + 4*( - 2) + 3 = - 13,......
The math problem f(x) = -2x^2 + 4x + 3[I need to find the vertex and graph]

So I found the vertex using -b/(2a) which gave me x = 1 and substituted the X value on the formula then got that y = 5. I have the vertex (1, 5) and I'm trying to figure out how to get 2 more points so I can graph more accurately. I know it's a downwards parabola, I'm just stuck with finding the other points. Any help would be greatly appreciated.

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f(x) = - 2*x^2 + 4*x + 3

How I can find points on both sides of the parabola?

f( - 2) = - 2*( - 2)^2 + 4*( - 2) + 3 = - 13, point( - 2, - 13)

f(2) = - 2*(2)^2 + 4*(2) + 3 = 3, point (2, 3)

f( - 3) = - 2*( - 3)^2 + 4*( - 3) + 3 = - 27, point ( - 3; - 27)

f(3) = - 2*(3)^2 + 4*(3) + 3 = - 3, point (3; - 3), and so on!

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when f(x)=0...
so on you x axis, the two points are:
(-0,58;0)
(2.58;0)

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Take the y-intercept
f(0) = 3
One point : (0,3)
take a point to the right of a turning point ( any point )

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You know that the y-intercept is 3, so one point is (0,3). (3, -3), (2, 3)
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