I really need help with this question please

Question

 

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

Answer 1

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) = nx^n-1

 

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

 

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

Answer 2

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).