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21 Jan 2021, 04:18
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Difficulty:
35%
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Question Stats:
70% (01:56) correct
30%
(02:13)
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Find the least possible value of y when y = x^2 - 12x + 86
(1) 50
(2) 51
(3) 52
(4) 53
(5) 54
(1) 50
(2) 51
(3) 52
(4) 53
(5) 54
\(y = x^2 - 12x + 86\)
when substituting \(x=6\), then \(y=50\) (using the options)
Mathematically, we can use the Quadratic Formula
\(y \ge \dfrac{4ac - b^2}{4a} = \dfrac{4 \times 1 \times 86-12^2}{4\times 1} =\dfrac{200}{4} = 50\)
when substituting \(x=6\), then \(y=50\) (using the options)
Mathematically, we can use the Quadratic Formula
\(y \ge \dfrac{4ac - b^2}{4a} = \dfrac{4 \times 1 \times 86-12^2}{4\times 1} =\dfrac{200}{4} = 50\)
22 Oct 2021, 23:02
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The minimum value of y will be when x = -b/2a (minimum value for quadratic equation)
y = x^2 - 12x + 86
:. -b/2a = -(-12)/2*1 = 6
substitute x
y = 6^2 - 12*(6) + 86 = 50
Hope it helps...
y = x^2 - 12x + 86
:. -b/2a = -(-12)/2*1 = 6
substitute x
y = 6^2 - 12*(6) + 86 = 50
Hope it helps...
