Determine whether the function f(x)=3x2-12x-24 has a minimum or maximum value?

Question

Problem is f(x)= 3x(squared) - 12x - 24

Choices for answer:

minimum
maximum
neither
cannot be determined

I put the problems up here that I don't understand. I'm very thankful for the ones who are trying to help. I'm sorry if I'm pissing you off but I'd just like to fully understand everything!

I do not know calculus, sorry. I'm in Algebra 2 at the moment!

Answer 1

take 3 as a common factor
=3(x^2-4x-8)
=3((x-2)^2-12) putting in form of square

this has got a minimum

Answer 2

Ok, do you know calculus? I will assume not, but if you do, please tell me.
3x^2 - 12x - 24 is a U shape. At very negative values it is very positive, and it just keeps getting more positive for lower and lower values. At very positive values it is also positive and it just keeps getting most positive at higher and higher values, so it has no maximum. But it does have a lowest value. Unfortunately, I don't know if you can prove this without calculus, but if you do know calculus, you can see that the maximum and minimum values will occur when the slope of the graph at point (x,y)= 0, so the derivative of the graph, 6x -12, must equal 0. 6x = 12, or x = 1/2. Plugging in (3 * ((1 / 2)^2)) - (12 * (1 / 2)) - 24, we see it corresponds to y = -29.25. This is indeed a minimum as lower values of x have higher y values, and so do higher x values.