Thread: Calculus Graph Help - Interpretation

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Please see link below:
http://oi48.tinypic.com/flgfpy.jpg

I know how to get 3.1. a & b but not c and d.

I also need help with questions 3.2. , 3.6. , 3.7. and 3.8.

I can do the rest myself

3.1.
(x+1) (x) (x-3)
=x^2 + x +x^2 -3x
= 2x^2 - 2x
a=2 & b= -2

That's all I can get.

(0,-6) will help you attain d (i.e. find f(0) ). It's a good attempt at 3.1, but you need to substitute points A, B, and C into f(x). Since you can obtain d, then you only have a,b, and c left. This means that you need 3 equations (hint: substitute A and C to get two). In order to get a 3rd equation, note that A is a critical point (read: do something with the derivative). This will get you a 3rd equation, and then use these three to find a,b, and c.

3.2 is asking about the critical point B. Note that you find these with the first derivative.

I only see up to 3.6 with the image link you provided.

Ok, I've figured out how to get d. All I need is C and question 3.2.

I've been trying for quite a while today but I can't seem to get it.

Please give me a detailed explanation on how to get C and then 3.2.

Don't worry about the other questions cos I've sorted them out already, just these two are bothering me.

Please explain how you found a,b, and d but not c. Whatever you did to find these, should be used to find c as well.

Once you have found all a,b,c, and d, then 3.2 is just a critical point (i.e. take the derivative and solve for x).

(x+1) (x) (x-3)
=x^2 + x +x^2 -3x
= 2x^2 - 2x
a=2 & b= -2

that's how I got A & B

d= 6x -6
6(0) -6
= -6 (I substituted the 0 from the graph into the equation)

If this is the wrong method please explain the correct one.

Update:
I think I have 3.2. now

ax^3 +bx^2 + c
= 2x^3 - 2x^2 - 10
= 6x^2 -4x -10
=3x^2 - 2x - 5
=(3x -5) (x+1)
x = 5/3 or -1

I know I take 5/3 because it is positive and B is on that side of the graph but how do I get the Y value? is it -10 because of c?

Originally Posted by VegetaZA

(x+1) (x) (x-3)
=x^2 + x +x^2 -3x
= 2x^2 - 2x
a=2 & b= -2

This is incorrect since is not in the form of f(x).

that's how I got A & B

d= 6x -6
6(0) -6
= -6 (I substituted the 0 from the graph into the equation)

If this is the wrong method please explain the correct one.

Update:
I think I have 3.2. now

ax^3 +bx^2 + c
= 2x^3 - 2x^2 - 10
= 6x^2 -4x -10
=3x^2 - 2x - 5
=(3x -5) (x+1)
x = 5/3 or -1

I know I take 5/3 because it is positive and B is on that side of the graph but how do I get the Y value? is it -10 because of c?

I don't follow your reasoning in the line of expressions above. You have three lines of differing polynomials, but I cannot determine where these came from. Again, you must use the fact that A is also a critical point, and also perhaps, that it is a double root (if you are going to take the factorization to f(x) route).