FInd the equation of the tangent to the curve defined by y= x(x2) at the point where x = 2?

Answer 1

First of all, learn to write an equation properly. "x(x2)" is incomprehensible.

Assuming "x(x2)" is really "x*x*2", which is my best guess at what you meant to write, you'd do it in the following way.

The equation tangent to the curve at that point will be a line equation, in the form m*x+b. You need to solve for m (the slope) and b (the y-intercept).

To find the slope, you take the derivative of the equation with respect to x
dy/dx = d/dx*(2x^2) = 4x
Then input the point the tangent line passes through to find the slope:
m = dy/dx = 4x = 4(2) = 8
You know point the line passes through by the original equation, inputting our x coordinate of 2:
y = 2x^2 = 2*(2)^2 = 8
Now we solve y = m*x+b to find b:
y = m*x+b
8 = m*2+b
8 = 8*2+b
8 = 16+b
-8 = b
This yields the final equation for the tangent line, which is
y = 8*x-8

Answer 2

Assume y = f(x) = x³
f ` (x) = 3x²
f ` (2) = 12 = m
f (2) = 8
y - 8 = 12.(x - 2)
y = 12x - 16 is equation of required tangent.

Answer 3

y = x(x^2) = x^3
dy/dx = 3x^2
dy/dx|x=2 = 3(2^2) = 12

When x=2, y=2^3 = 8

y=mx+c
8=(12)(2)+c
8-24=c
c=-16

Equation of tangent: y = 12x - 16