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Question 213455: what is the vertex of parabola y=8x+1-4x^2? I need all the steps so I can understand the solution.
Thank you
Found 2 solutions by stanbon, nerdybill:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! what is the vertex of parabola y=8x+1-4x^2? I need all the steps so I can understand the solution
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a = -4, b = 8, c = 1
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Vertex occurs at x = -b/2a = -8/(2*-4) = 1
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f(1) = 8 + 1 - 4 = 5
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Vertex: (1,5)
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If you work it by completing the square you get:
-4x^2 + 8x + ? = y - 1 + ?
-4(x^2 - 2x + (-1)^2) = y - 1 + -4*1
-4(x - 1)^2 = y-5
From this you can see that h=1 and k = 5
So vertex is at (1,5)
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Cheers,
Stan H.
Answer by nerdybill(7384)  (Show Source):
You can put this solution on YOUR website! Rewrite:
y=8x+1-4x^2
as:
y = -4x^2+8x+1
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There are two ways to do it. By "completing the square" OR by find the "axis of symmetry".
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I'll do the latter. The "axis of symmetry" defines a line which divides the parabola right down the middle. Once you know that, you can determine the vertex.
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"axis of symmetry" is found by plugging in the values to:
-b/(2a) (derives from a portion of the quadratic equation)
-8/(2*(-4))
-8/-8
-1
Our axis of symmetry is then at:
x = -1
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We now have half of our vertex: (-1, __)
to find the y coordinate, subsititute -1 for 'x' into the equation and solve for y:
y = -4x^2+8x+1
y = -4(1)^2+8(1)+1
y = -4+8+1
y = 5
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Solution:
Vertex is at (-1, 5)
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