SOLUTION: what is the vertex of parabola y=8x+1-4x^2? I need all the steps so I can understand the solution. Thank you

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
-----------------------
a = -4, b = 8, c = 1
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Vertex occurs at x = -b/2a = -8/(2*-4) = 1
---------
f(1) = 8 + 1 - 4 = 5
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Vertex: (1,5)
=======================
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)
====================================
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
.
There are two ways to do it. By "completing the square" OR by find the "axis of symmetry".
.
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.
.
"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
.
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
.
Solution:
Vertex is at (-1, 5)

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