## anonymous one year ago WILL FAN AND MEDAL :) Plz help me walk through this! Functions f(x) and g(x) are described as follows: f(x) = -2x^2 + 3 x g(x) 0 0 1 2 2 3 3 2 4 0

1. anonymous

Which statement best compares the maximum value of the two functions? It is equal for both functions. It is 2 units higher for f(x) than g(x). It is 2 units lower for f(x) than g(x). It is 1 unit higher for f(x) than g(x).

2. anonymous

@geerky42 @pooja195

3. anonymous

@Concentrationalizing @sammixboo

4. anonymous

@amistre64 @TheSmartOne

5. anonymous

@wio @IrishBoy123

6. anonymous

@Vocaloid @hartnn

7. anonymous

@Hero @perl

8. anonymous

@billj5 can you help me? :)

9. anonymous

Ok, so the part I'm having trouble with is finding the maximum of f(x)

10. anonymous

f(x) is a quadratic. The graph of a quadratic is a parabola, which would either have a finite minimum value or a finite maximum value. In this case you have a finite maximum value (because the question implies this and because the leading coefficient is negative). Given a quadratic with a general form of $$ax^{2} + bx + c = 0$$, the x-coordinate of the maximum or minimum value is equal to $$-b/2a$$.

11. anonymous

this might help http://www.purplemath.com/modules/fcntrans.htm

12. anonymous

Have you seen the -b/2a idea before, horse?

13. anonymous

Ok thank you @billj5 ;)

14. anonymous

Yes I have :) @Concentrationalizing

15. anonymous

Okay, awesome. So for our problem, what is -b/2a?

16. anonymous

I think it would be -3/2(-2)

17. anonymous

Actually, it would be 0. Note that the b in the general form is the coefficient of the x to the first power variable. Yet our equation has no x to the first power, we only have an x^2 and a constant term. Thus b is 0, which makes -b/2a = 0

18. anonymous

I meant those would be the values. So would the -3/2(-2) = 0?

19. anonymous

b is not -3, though, b is 0. Since a quadratic is $$ax^{2} + bx + c$$, a is what goes with the x^2 term. So a for us is -2. b is what goes with the x term. But our equation is -2x^2 + 3, we dont have an x term. Thus b has to be 0. Then our constant, c, is 3. So a = -2, b = 0, c = 3. Which means -b/2a = 0/2(-2) = 0/-4 = 0 Does that make sense?

20. anonymous

But f(x) only has two terms?:|

21. anonymous

So it's not a quadratic :|

22. anonymous

A quadratic is any function in which all the powers are integers and the highest power is a 2. So all of these would be quadratics: $$x^{2} + 5x + 6$$ $$\frac{-3x^{2}}{5} - 9$$ $$x^{2}$$ The first has 3 terms, the 2nd has 2, the last one has 1 term. But they're all quadratics because all of the exponents are integers and the highest one is a 2.

23. anonymous

Ok but I still don't get why b = 0.

24. anonymous

|dw:1433381836170:dw|

25. anonymous

|dw:1433381878344:dw|

26. anonymous

Ooooh! That makes sense ;)

27. anonymous

So it would all just equal 0. Got it :)

28. anonymous

Lol, glad the visual helped. But yes, b is 0 because that part is missing :3 So that means the x-coordinate of the maximum value is 0. So since we need x to be 0, we can just plug in 0. If we plug in 0, we get $$-2(0)^{2} + 3 = 0 + 3 = 3$$ Thus the maximum value is 3 for f(x)

29. anonymous

Got it! That makes total sense now. I guess I must be a visual person lol :3 Thanks again @Concentrationalizing !

30. anonymous

No problem :)