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anonymous
 5 years ago
graph the following. Identify wheather or not the graph is a funtion. Then evalutute the graph at my specified domain.
F(x)= x+5 x<2
X^2+2x+3 x=>2
F(x) (3)=
F(x) (4)=
F(x) (2)=
Please help me...i will give you a medal if you can help
anonymous
 5 years ago
graph the following. Identify wheather or not the graph is a funtion. Then evalutute the graph at my specified domain. F(x)= x+5 x<2 X^2+2x+3 x=>2 F(x) (3)= F(x) (4)= F(x) (2)= Please help me...i will give you a medal if you can help

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, it's a piecewise function, but still a function. For f(3) you would evaluate it at 3, with whichever equation is at 3. In this case, it's x^2+2x+3. So you plug in 3. 3^2+2(3)+3 = 18 f(4) you would use x+5 so it would be 4+5=1 f(2) would be x^2+2x+3 = 3 Hope that helps!

radar
 5 years ago
Best ResponseYou've already chosen the best response.0use the link provided previously. It worked well for the first one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@kaiyne what are the points? to graph it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, value given is the x value and you solve for the y value. So it ends up being (3,18), (4,1) and (2,3) Does it make sense how I got those?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alittle i have one more can i ask you that one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0F(x) 2x+1 x =>1 x^2+3 x<1 F(2)= F(6)= F(1)=

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(2) is less than one, so its x^2+3, or (2)^2+3 = 7. (2,7) f(6) is greater than one, so you use 2x+1 or 2(6)+1 = 13. (6,13) f(1) is equal to 1, so its still on 2x+1 (since that is x=>1). It ends up being 2(1)+1 = 3. (1,3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks so much i get it now :)
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