SOLUTION: Can u please show me the table and graph of the parabola for this equation. I don't understand how to find the axis of symmetry and how to pick the points on my graph if the vertix

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Can u please show me the table and graph of the parabola for this equation. I don't understand how to find the axis of symmetry and how to pick the points on my graph if the vertix      Log On


   



Question 304414: Can u please show me the table and graph of the parabola for this equation. I don't understand how to find the axis of symmetry and how to pick the points on my graph if the vertix isnt a whole number..... here is the problem... y < 4X^2-2X+1
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Can u please show me the table and graph of the parabola for this equation. I don't understand how to find the axis of symmetry and how to pick the points on my graph if the vertix isnt a whole number..... here is the problem...
y < 4X^2-2X+1
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Graph the boundary line, y = 4x^2-2x+1, of the inequality.
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The axis of symmetry is x = -b/2a = 2/(2*4) = 1/4
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Pick a couple of x-values to the left and to the right of x = 1/4
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If x = -2, y = 4(-2)^2-2(-2)+1 = 21
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If x = -1, y = 4(-1)^2 -2(-1) + 1 = 7
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If x = 0, y = 1
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If x = 2 y = 4(2)^2-2(2)+1 = 13
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Plot the points and draw a smooth curve thru them.
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graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C4x%5E2-2x%2B1%29
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Cheers,
Stan H.
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