JAMB Mathematics 2021 Paper

Concept:A quadratic expression with a positive coefficient of $x^2$ opens upward, so it has a minimum value, not a maximum value.Explanation:Given $y = 3x^2 + 5x - 3$.Differentiate: $\frac{dy}{dx} = 6x + 5$.Set $\frac{dy}{dx} = 0$ to find the turning point:$6x + 5 = 0$$x = -\frac{5}{6}$The second derivative is $\frac{d^2y}{dx^2} = 6$, which is positive.Therefore, the turning point is a minimum, not a maximum.At $x = -\frac{5}{6}$:$y = 3\left(-\frac{5}{6}\right)^2 + 5\left(-\frac{5}{6}\right) - 3$$y = \frac{75}{36} - \frac{25}{6} - 3$$y = \frac{25}{12} - \frac{50}{12} - \frac{36}{12} = -\frac{61}{12}$This is the minimum value of the function.Since the parabola opens upward, $y$ has no maximum value.Answer:No correct option.