Question 398670: This is my first time using this site, so I apologize ahead of time if I do something wrong. I'm having trouble with this problem in my Algebra II homework and I was wondering if you could help me out.
*For the graph of y = ax^2 + bx + c, show that the y-coordinate of the vertex is (I don't know how to write fractions on here) -b^2 over 4a + c.
I thought about this one for a long time and I just can't seem to understand it. Help please?
-thank you for your time.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! For the graph of y = ax^2 + bx + c, show that the y-coordinate of the vertex is (I don't know how to write fractions on here) -b^2 over 4a + c.
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y = a(x^2+(b/a)x+(b/(2a))^2) + c
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y = a(x+(b/2a))^2 + c
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Vertex: [(-b/(2a),?]
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The y-coordinate is f(-b/2a)
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y = a[-b/(2a)]^2 + b[-b/(2a)] + c
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y = a[b^2/(4a^2)] - [b^2/(2a) + c
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y = [b^2/4a] - [2b^2/4a] + 4ac/4a
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y = [b^2-2b^2 + 4ac]/(4a)
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y = [-b^2+4ac]/(4a)
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Cheers,
Stan H.
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