Conditions for Cubic Equation to Represent Curve

[elitewarr]

Homework Statement

2) Given a cubic equation y = (x+5)(ax^2 + bx - 2). Give conditions on a and b for the equation to represent the following curve. The curve is attached to the email.
http://img35.imageshack.us/img35/8246/question2o.png

Homework Equations

The Attempt at a Solution

I know that a>0.
Since there are 3 intersections with the x-axis, then for (ax^2 + bx - 2), b^2 - 4ac > 0.
b^2 - 4(a)(-2) > 0
b^2 + 8a > 0
b^2 > -8a

If i differentiate it, I get
3ax^2 + 2bx + 10ax + 5b - 2
Since there are 2 stationary points,
B^2 - 4ac = (2b + 10a)^2 - 4(3a)(5b-2) > 0
b^2 - 5ab + 25a^2 + 6a > 0
and I'm stucked again.
Which approach is correct?

Thanks.

 

[Mark44]

Homework Statement

2) Given a cubic equation y = (x+5)(ax^2 + bx - 2). Give conditions on a and b for the equation to represent the following curve. The curve is attached to the email.
http://img35.imageshack.us/img35/8246/question2o.png

Homework Equations

The Attempt at a Solution

I know that a>0.
Since there are 3 intersections with the x-axis, then for (ax^2 + bx - 2), b^2 - 4ac > 0.
b^2 - 4(a)(-2) > 0
b^2 + 8a > 0
b^2 > -8a

It seems to me that you are just about finished here. As you already said, a > 0, which you can tell from the behavior of the graph for very negative or very positive x.

The inequality b² > -8a is true for all real b, as long as a > 0. You were finding conditions on the discriminant so that there would be two real, distinct roots. If the discriminant had been equal to zero, there would have been a repeated root.

If i differentiate it, I get
3ax^2 + 2bx + 10ax + 5b - 2
Since there are 2 stationary points,
B^2 - 4ac = (2b + 10a)^2 - 4(3a)(5b-2) > 0
b^2 - 5ab + 25a^2 + 6a > 0
and I'm stucked again.
Which approach is correct?

Thanks.

 

[elitewarr]

eh.. Why did I never thought of that?? Thank you! Haha.