JAMB 1995 · UME · Q39

Calculus (JAMB bridge)

For what value of $x$ is the tangent to the curve $y = x^2 - 4x + 3$ parallel to the $x$ -axis?

Worked solution (try it first)

A tangent parallel to the $x$ -axis has gradient 0.
$\frac{dy}{dx} = 2x - 4 = 0$ , so $x = 2$ , option B.

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