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I really need help with this question please

+1 vote

 

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

asked Jan 18, 2013 in CALCULUS by linda Scholar

2 Answers

+2 votes
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
answered Jan 18, 2013 by richardson Scholar
0 votes

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).

answered May 30, 2014 by lilly Expert

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