can I please get some help for this Algebra question?

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- anonymous

can I please get some help for this Algebra question?

- jamiebookeater

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- anonymous

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- anonymous

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

- misty1212

HI!!

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## More answers

- misty1212

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

- 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

- 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

- anonymous

Okay but I how do I find that?

- anonymous

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

- anonymous

yep! what are the y values for these points

- 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

- anonymous

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

- 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

- 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

- anonymous

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

- anonymous

\[(f+g)(-2)=f(-2)+g(-2)\]
add them up :)

- anonymous

-4 ?

- anonymous

|dw:1444071199043:dw|

- anonymous

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

- anonymous

yep!

- anonymous

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

- anonymous

No! this time you are subtracting!
\[(f-g)(1)=f(1)-g(1)\]

- anonymous

Ok I see that now lol So it would be 2

- anonymous

Nevermind, -2!

- anonymous

Good!

- anonymous

Remember, for the next one you have to multiply!

- anonymous

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

- anonymous

yep!

- anonymous

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

- anonymous

correct!!

- 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
\[(\frac{f}{g})(1)=\frac{f(1)}{g(1)}=\frac{f(1)}{0}\]

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