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# Urgent! Find the slope of the tangent to y=x^3 + 2x^2 - 8x at (2,0)?

asked Jan 18, 2013 in CALCULUS

y = x3 + 2x2 - 8x

Let f(x) = y = x3 + 2x2 - 8x

Now, you have to know the derivate of this function

(d/dx)f(x) = (d/dx)(x3 + 2x2 - 8x)

f' (x) = (d/dx)(x3) + (d/dx)(2x2) - (d/dx)(8x)

f' (x) = (d/dx)(x3) + 2(d/dx)(x2) - 8(d/dx)(x)

Formula  : (d/dx)(xn) = nxn-1

f' (x) = 3x3-1 + 2(2x2-1) - 8x1-1

Simplify

f' (x) = 3x2 + 2(2x1) - 8x0

Note : A0 = 1

Than

f' (x) = 3x2 + 2(2x) - 8(1)

f' (x) = 3x2 + 4x - 8

You know that the abscissa of the point is : 2

f' (2) = 3(22) + 4(2) - 8

f' (2) = 3(4)+ 8 - 8

f' (2) = 12

Therefore, the slope of the tangent of the curve f at point of abscissa 2 is :  12

You can say that the equation of the tangent is : y = 12x + b

You've got an additional information :  point (2, 0)

It means that f(2) = 0

y = 12x + b

0 = 12(2) + b

0 = 24 + b

Subtract 24 from each side.

- 24 = 24 + b - 24

- 24 = b

There fore     b = - 24

Now you can complete the equation, and it becomes : y = 12x - 24

The function is y = x3 + 2x2 - 8x and the point is (2, 0).

We can use implicit differentiation to find y '.

The implicit(or total) derivation of y = x3 + 2x2 - 8x is,

y ' = 3x2 + 4x - 8

Substitute the values of (x, y ) = (2, 0) in the above equation.

y ' = 3(2)2 + 4(2) - 8

= (3 * 4) + 8 - 8

y ' = 12.

This is the slope (m ) of the tangent line to y = x3 + 2x2 - 8x at (2, 0).