anonymous one year ago Using the completing-the-square method, find the vertex of the function f(x) = –2x2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.

1. anonymous

it is a maximum because the leading coefficient|dw:1434844163162:dw| is negative so the parabola opens down

2. campbell_st

|dw:1434843957618:dw|

3. anonymous

Ok:)

4. anonymous

$f(x) = –2x^2 + 12x + 5$to find the vertex, the first coordinate of the vertex is always $-\frac{b}{2a}$ which in your case is $-\frac{12}{2\times (-2)}$

5. anonymous

the second coordinate of the vertex is what you get when you replace $$x$$ by the first coordinate

6. anonymous

3

7. dmndlife24

Graph opens downward so it would be a maximum point Vertex formula|dw:1434844392512:dw|

8. anonymous

yeah that is the first coordinate the second coordinate if $$f(3)$$

9. anonymous

Now what do I do?

10. anonymous

have a beer you are done

11. campbell_st

well completing the square $f(x) = (-2x^2 + 12x) + 5$ factor out -2 $f(x) = -2(x^2 - 6x) + 5$ now complete the square in x

12. anonymous

So is it maximum or minimum??

13. anonymous

, find the vertex of the function $(3,23)$ it is a max have a nice day

14. anonymous

Solve x2 + 12x + 6 = 0 using the completing-the-square method.

15. anonymous

Thanks friend. I also need help with that one

16. anonymous

|dw:1434844578090:dw|

17. campbell_st

so inside the brackets you need to add (-6/2)^2 so the function is $f(x) = -2(x^2 - 6x + 9) + 5 -18$ which becomes f(x) = -2(x -3)^2 - 13 now you have completed the square have the equation in vertex form and can answer the question using the required method

18. campbell_st

|dw:1434844393236:dw|

19. anonymous

$x^2 + 12x + 6 = 0 \\ x^2+12x=-6$ is a start

20. anonymous

hahahah you're funny @satellite73 !!!

21. anonymous

I see, I see

22. anonymous

|dw:1434844713212:dw|

23. anonymous

24. campbell_st

|dw:1434844498447:dw|

25. anonymous

$x^2+12x=-6$takes two steps to complete the square what is half of 12?

26. anonymous

6

27. anonymous

right, and what is $$6^2\_? 28. anonymous oops what is \(6^2$$?

29. anonymous

36

30. anonymous

so go right from $x^2+12x=-6$ to $(x+6)^2=-6+36$ or $(x+6)^2=30$

31. anonymous

then take the square root of both sides, don't forget the $$\pm$$ and get $x+6=\pm\sqrt{30}$

32. anonymous

subtract 6 and you are done

33. anonymous

so is it 6 plus or negative radical 30?

34. anonymous

no

35. anonymous

SUBTRACT 6 to solve for $$x$$

36. anonymous

|dw:1434844993983:dw|

37. anonymous

nope

38. anonymous

|dw:1434845032667:dw|

39. anonymous

|dw:1434845026665:dw|

40. anonymous

yeah that one