Question

$find the eqution of tangent and normal to curve y = x {}^{3} - 2x {}^{2} + 4 : at x = 4$
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Answer 1

Step-by-step explanation:

We have,

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

$\frac{dy}{dx} = 3{x}^{2} - 4 x$

$\implies \frac{dy}{dx} _{x = 4} = 3.{4}^{2} - 4 \times 4 = 32$

Now, at x=4, y=36

Equation of Tangent:

$(y - 36) = 32(x - 4)$

$\implies \: y - 36 = 32x - 128$

$\implies \: y -32x - 36 +128 = 0$

$\implies \: y -32x +92 = 0$

Equation of Normal:

$(y - 36)32 + (x - 4) = 0$

$\implies32y - 1152 + x - 4 = 0$

$\implies32y + x -1156 = 0$