## hugsandkisses Group Title What is the value of h in the function f(x)=1/2(x=5)2-9? A. 5 B. 1/2 C. 9 D. -5 2 years ago 2 years ago

1. across Group Title

I am very confused: Why do you have a second equals sign in the middle of the function, and where is h?

2. hugsandkisses Group Title

they didnt give me an h value :-/ and that was supposed to be x=5

3. across Group Title

I am now more confused. n_n

4. myininaya Group Title

what? I'm confused too. The way you typed your problem is that exactly the way it looks?

5. hugsandkisses Group Title

yeah. but the 2 is small and its raised

6. myininaya Group Title

and inside the parenthesis is x=5?

7. myininaya Group Title

$f(x)=\frac{1}{2}(x \pm 5)^2-9$ i think it should either be + or -

8. across Group Title

$f(x)=\frac{1}{2}(x=5)^2-9...$I can't seem to make any sense of this.

9. myininaya Group Title

and I'm assuming you want to find the x coordinate of the parabola

10. hugsandkisses Group Title

Ummm, I guess.. I dont know. Im really really bad at math

11. myininaya Group Title

can you write the function down one more time?

12. myininaya Group Title

and i meant x coordinate of the vertex of the parabola lol

13. hugsandkisses Group Title

f(x)=1/2(x+5)2-9

14. myininaya Group Title

ok great!

15. myininaya Group Title

now it makes more sense

16. hugsandkisses Group Title

oh. sorry! haha

17. myininaya Group Title

$f(x)=\frac{1}{2}(x+5)^2-9$ $f(x)=a(x-h)^2+k$ what is h?

18. myininaya Group Title

well we want -h=5 right?

19. myininaya Group Title

so if -h=5, then h=?

20. hugsandkisses Group Title

-5? lol sorry, im super weak in math

21. myininaya Group Title

-5 is tight!

22. myininaya Group Title

i mean right*

23. hugsandkisses Group Title

YAY!! :D

24. hugsandkisses Group Title

can you help me with more? Ive got a ton

25. myininaya Group Title

Ok

26. myininaya Group Title

But you have to give me a sprite or dr. pepper

27. hugsandkisses Group Title

I'll give you both!

28. myininaya Group Title

lol

29. myininaya Group Title

ok thanks for the imaginary soda