Find the maximum value of y in the equation y = 1 - 2x - 3x\(^2\)
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Correct Answer: Option D
Explanation:
\(y = 1 - 2x - 3x^2\)
At maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)
\(\frac{\mathrm d y}{\mathrm d x} = -2 - 6x\)
\(-2 - 6x = 0 \implies -2 = 6x\)
\(x = -\frac{1}{3}\)
At x = \(-\frac{1}{3}\), y = \(1 - 2(-\frac{1}{3}) - 3(-\frac{1}{3})^{2}\)
= \(1 + \frac{2}{3} - \frac{1}{3}\)
= \(\frac{4}{3}\)
\(y = 1 - 2x - 3x^2\)
At maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)
\(\frac{\mathrm d y}{\mathrm d x} = -2 - 6x\)
\(-2 - 6x = 0 \implies -2 = 6x\)
\(x = -\frac{1}{3}\)
At x = \(-\frac{1}{3}\), y = \(1 - 2(-\frac{1}{3}) - 3(-\frac{1}{3})^{2}\)
= \(1 + \frac{2}{3} - \frac{1}{3}\)
= \(\frac{4}{3}\)