## anonymous one year ago can I please get some help for this Algebra question?

1. anonymous

2. anonymous

I don't know why I am so confused with this question! Can someone please point me in the right direction?

3. misty1212

HI!!

4. misty1212

look at the graph what is $$f(-2)$$?

5. anonymous

$(f+g)(a)=f(a)+g(a)$$(f-g)(a)=f(a)-g(a)$$(fg)(a)=f(a) \times g(a)$$(\frac{f}{g})(a)=\frac{f(a)}{g(a)} \space ; \space g(a)\neq 0$ Use these formulae and find the values from the graph

6. anonymous

so to find for your first part $(f+g)(-2)=f(-2)+g(-2)$ you do it like that and the value of f(-2) and g(-2) may be obtained from the graph

7. anonymous

Okay but I how do I find that?

8. anonymous

F=-2,4? G=-2,0??

9. anonymous

yep! what are the y values for these points

10. anonymous

Think of your f(x) as y-values for the blue curve and g(x) as the y-values for the pink straight line so to find f(-2), we find the corresponding y value for the x value -2 on the blue curve and to find g(-2), we find the corresponding y value for the x value -2 on the pink straight line

11. anonymous

Hmm..okay..Now what? :/ still kind of confused

12. anonymous

you've found your $f(-2)=4$$g(-2)=0$ Because -2,4 is a point lying on the blue curve and -2,0 is a point lying on the pink straight line

13. anonymous

basically u look at the x value, look for what value of y for that x we have a point lying on the curve and the line

14. anonymous

Ok do I add or subtract the numbers now? I think I remember learning that.

15. anonymous

$(f+g)(-2)=f(-2)+g(-2)$ add them up :)

16. anonymous

-4 ?

17. anonymous

|dw:1444071199043:dw|

18. anonymous

Sorry got confused there. Now I get it...Lol. Ok for the next one, f(1)= 1 right?

19. anonymous

yep!

20. anonymous

and g(1)=3...so answer would be 4 also?

21. anonymous

No! this time you are subtracting! $(f-g)(1)=f(1)-g(1)$

22. anonymous

Ok I see that now lol So it would be 2

23. anonymous

Nevermind, -2!

24. anonymous

Good!

25. anonymous

Remember, for the next one you have to multiply!

26. anonymous

Yes I got 0 on that one..Now for D.

27. anonymous

yep!

28. anonymous

I got 1/3. Thanks SO much for your help. Have a great day!!

29. anonymous

correct!!

30. anonymous

An interesting thing to note here, if by chance your g(1) was 0, you would say that that your (f/g)(1) is not defined, as you can't divide by 0 $(\frac{f}{g})(1)=\frac{f(1)}{g(1)}=\frac{f(1)}{0}$