First slide

Tangents and normals

Question

The tangent line at (2, 4) to the curve y = $x^{3} - 3 x + 2$ meets the x-axis at

Moderate

A

(2, 0)
B

$(\frac{7}{2}, 0)$
C

$(\frac{11}{9}, 0)$
D

$(\frac{14}{9}, 0)$

Solution

$\frac{d y}{d x} = 3 x^{2} - 3 = 3 (x^{2} - 1)$
${\frac{d y}{d x}|}_{x = 2} = 3 (4 - 1) = 9$
Equation of tangent at (2, 4) is
$Y - 4 = 9 (X - 2)$
This meets x-axis if Y = 0
$\Rightarrow \frac{- 4}{9} = X - 2 \Rightarrow X = 2 - \frac{4}{9} = \frac{14}{9}$

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