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JAMB Mathematics 2014 Paper

Question : 31

Total: 47

Find the minimum value of y = x² - 2x - 3

Solution:

Concept:

For a quadratic function, the minimum value occurs where the derivative is zero.

Explanation:

Given

y = x^2 - 2x - 3

Differentiate with respect to

x

\frac{dy}{dx} = 2x - 2

At the turning point, set

\frac{dy}{dx} = 0

2x - 2 = 0

Solve for

x

2x = 2

x = 1

Substitute

x = 1

into the original equation to find the minimum value:

y_{\text{min}} = (1)^2 - 2(1) - 3

y_{\text{min}} = 1 - 2 - 3

y_{\text{min}} = -4

Answer:

The minimum value is

-4

, which is option D.

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