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Medina
Mathematics
5 February, 06:40
20-90x-50x^2
find the vertex
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Shayla
5 February, 06:57
0
20-90x-50x^2, written in standard form, would be - 50x^2 - 90x + 20. Let's use the method of completing the square to determine the vertex:
Divide all terms in - 50x^2 - 90x + 20 by - 50, obtaining - 50 (x^2 + (9/5) x - 2/5)
The coefficient of x is 9/5. Divide this by 2, obtaining 9/10, and then square this result: (9/10) ^2 = 81/100. Add this to x^2 + (9/5) x - 2/5 and then subtract it, resulting in x^2 + (9/5) x - 2/5 + 81/100 - 81/100, or
x^2 + (9/5) x + 81/100 - 2/5 - 81/100, or
(x + 9/10) ^2 - 12100
Then the vertex location is (-9/10, - 12100).
A faster approach would be to use x = - b / (2a) to determine the x-coordinate of the vertex and then to evaluate the function at this x-value to determine the y-coordinate. Here x = - (-90) / (-100) = - 9/10 (same as before).
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Mara Powers
5 February, 07:01
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I think it's your answer
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