Question

1. Find the equation of the tangent line to the parabola y = x^3 at the point P(2,8).
Use x = 2.1, 2.01, 2.001, 2.0001, 2.00001; 1.9,1.99, 1.999,1.9999, 1.99999 to find slope.

Answer 1

Answer:

a) Equation of the tangent of the parabola

12 x - y -16=0

b)

slope of the tangent  at x=2.1

$(\frac{dy}{dx}) = 3(2.1)^{2} = 13.23$

slope of the tangent  at x=2.01

$(\frac{dy}{dx}) = 3(2.01)^{2} = 12.1203$

Step-by-step explanation:

Step(i):-

Given parabola  y = x³  ....(i)

Differentiating equation (i) with respective to 'x'

  $\frac{dy}{dx} = 3x^{2}$

slope of the tangent

$(\frac{dy}{dx})_{(2,8)} = 3x^{2} = 3(2)^{2} =12$

Step(ii):-

Equation of the tangent of the parabola

$y-y_{1} = m (x-x_{1} )$

y-8 = 12 (x -2)

12 x - 24 -y +8 =0

12 x - y -16=0

b)

slope of the tangent  at x=2.1

$(\frac{dy}{dx}) = 3(2.1)^{2} = 13.23$

slope of the tangent  at x=2.01

$(\frac{dy}{dx}) = 3(2.01)^{2} = 12.1203$