determine the numbers a, b, and c such that it satisfies all of the 3 conditions I have a equation y=x^3+ ax^2 + bx + c and I need to find the numbers a, b and c such that the graph of y=x^3+ ax^2 + bx + c satisfies all of the following 3 conditions:


*

*It has a horizontal tangent line at x = -1

*its tangent line at x = 1 is parallel to the line y = 8x - 5.

*it passes through the point (1,2)
for 2), parallel means they have to have the same slope. SO derivative of the equation is 3x^2-2ax+b. When I put 1 in for x, i get 3-2a+b. When I put 1 in for y=8x-5 , I get y=3. So 3-2a+b=3, so a should be 1 b should be 2.
I'm not sure i'm doing it correctly.. Any help would be very much appeciated
 A: Well,first consider (1).A horizontal tangent line means the derivative with respect to x at x=- 1 is 0. This can be seen geometrically by the fact this represents a horizontal line at y = c where c is a real constant. 
Now consider (2). A line is parallel to another line when their slopes are the same, which means the derivatives are the same at x= 1. So f'(1) = 8. You're lastly given that when x = 1, y =2 on the graph of the curve of the function. 
So now let's put this all together: 
$$1)\frac{dy}{dx}=0 = 3x^2 + 2ax + b when x = -1 ----->  0 = 3 - 2a +b$$ 
 $$2)  \frac{dy}{dx}=8 = 3x^2 + 2ax + b when x = 1 -------> 8= 3 + 2a + b $$
 $$3) At (1,2) , y= x^3 + ax^2 + bx + c -----> 2 = 1^{3}+ a*(1^{2}) +b + c = 1+a+b+c $$ 
You now have a series of 3 linear equations in 3 unknowns that you can solve by your favorite method. I'm a matrix guy myself. 
A: You don't get the slope of a line by substituting an $x$ value, the slope is the coefficient of $x$.  In your case that's $8$.  So the equations you have to solve are
$$\eqalign{
  &\frac{dy}{dx}=0\quad\hbox{when $x=-1$}\cr
  &\frac{dy}{dx}=8\quad\hbox{when $x=1$}\cr
  &y=2\quad\hbox{when $x=1$}\ .\cr}$$
See if you can do the problem from here.  BTW there is an error in your derivative, but it's probably just a typo, please check it carefully.
