Find the gradient of the tangent to the curve $y = x^{3} - 2 x^{2} + 5$ at the point $(2, 5)$ .

1

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Solution

The correct option is A $4$
$y = x^{3} - 2 x^{2} + 5$
$\frac{d y}{d x} = 3 x^{2} - 4 x$

Thus,
$\frac{d y}{d x}$ at $x = 2$ = $3 (2)^{2} - 4 (2) = 12 - 8 = 4$

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Similar questions

y = x^{3} - 2 x^{2} - x

y = 3 x - 2

y = x^{3} - 2 x^{2} - x

3 x - y + 1 = 0

y = x^{3} - 2 x^{2} + 4

x = 2

y = x^{3} - 2 x^{2} + 4

x = 2

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