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You can even 'message' them if you need to on occasion....

Nope....real (smart) folks !

Welcome, Huggy ! 

Hello

I like cat and i like taylor swift too!!!!!!

You are really good at photography I love it !

Yah, I looked around the 'net and there seems to be a lot of controversy re: this.....boils down to what your teacher's expectations are and what they have taught in class.....    I guess.  

Thank you for explaining more. I talked with some classmates and no one knows what the teacher wants! She isn't very strict so as long as I explain in detail why I chose what I chose, she should be happy with the answer. If not, I may only get a couple of points deducted but not a lot.

Mmmm....I don't know....I suspect that the answrer  might be "True".....they are probably not as nit-picky as Calculus is....LOL!!!

Thanx for the clarification, Chris!    I wonder what her teacher wants for an answer......

Thanks, Omi....I missed the "positive integer" part    !!!

Technically.....f(x)  does NOT have a minimum at x  = -8   [  if I remember my Calculus correctly  ]

A local minimum  cannot occur at an endpoint of a function......only an absolute  max or min can occur at an endpoint

So..... at x = -8 we cannot have an absolute min since we  have an absolute min  at  x   = 6   and no local minimum at any point on the graph

This might help  : http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx

The equation x^2 - kx - 12 = 0 has only integer solutions for certain positive integers k. What is the sum of all such values of k?

1+4+11=15

1. (a,b) = (4,-6).

2. The y-intercept is 5.

3. n = -3.

x^2  - kx  - 12

The sum of the possibile solutions  =  k

So.....the possibilities  are

 Solutions                             k

-1     12                                11

1     -12                               -11

-4      3                                -1

4      -3                                 1

6      -2                                 4

-6      2                                -4

So.....the sum  of the k's   =   0