## anonymous 4 years ago 2x^-3

1. anonymous

$2/\sqrt[3]{x}$

2. anonymous

oops 2/x^3

3. anonymous

4. anonymous

when you have a negative exponent you just drop the value to the denominator

5. anonymous

x^-3 is the same as 1/x^3

6. anonymous

so 2(1/x^3)=2/x^3

7. anonymous

i guess thats what you are looking for since you give no further directions

8. anonymous

what is it you are trying to do?

9. anonymous

horizontal translation equation

10. anonymous

sec

11. anonymous

did they give you a point?

12. anonymous

no I just need to make up an equation and graphing

13. anonymous

I am confused because normally for a horizontal translation a point is given

14. anonymous

I have to create a horizontal function coming up with the points

15. anonymous

so in other words i can plug 0 in for x and find my own point?

16. anonymous

yes

17. anonymous

well x cant be 0 so lets do 1 then

18. anonymous

ok

19. anonymous

f(1)=2/(1)^3=2

20. anonymous

so if x = 1 y = 2

21. anonymous

2=2/(1-a)^3 2=2/1-a^3 2(1-a^3)=2 1-a^3=1 -a^3=0 a=0 which this would not change the equation, not having a point given confuses me >.<

22. anonymous

23. anonymous

y=1/4 so 1/4= 2/(2-a)^3

24. anonymous

14=2/(8-a^3)

25. anonymous

1/4(8-a^3)=2

26. anonymous

(1/4(8-a^3)=2)4

27. anonymous

8-a^3=8

28. anonymous

wait wait ok, I'm confused now too without an point given. Ok so example : f(X)= 4^3-x is horizontal equation. How would that be solved?

29. anonymous

-a^3=0

30. anonymous

so a agaon would = 0

31. anonymous

meaning it would touch the line?

32. anonymous

can't you plug in x value's ?

33. anonymous

meaning the new equation which would be y=2(x-a)^3 would not be a change

34. anonymous

y=2(x-0)^-3

35. anonymous

if a is 0 there was no shift

36. anonymous

ooh ok

37. anonymous

this is why i dont like it when they do not give me a point to work with >.<

38. anonymous

O.o me either