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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

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  1. anonymous
    • 5 years ago
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    Yes, it's a piece-wise 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!

  2. radar
    • 5 years ago
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    use the link provided previously. It worked well for the first one.

  3. anonymous
    • 5 years ago
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    @kaiyne what are the points? to graph it

  4. anonymous
    • 5 years ago
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    Well, 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?

  5. anonymous
    • 5 years ago
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    alittle i have one more can i ask you that one?

  6. anonymous
    • 5 years ago
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    Sure, what is it?

  7. anonymous
    • 5 years ago
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    F(x) 2x+1 x =>1 x^2+3 x<1 F(-2)= F(6)= F(1)=

  8. anonymous
    • 5 years ago
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    f(-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)

  9. anonymous
    • 5 years ago
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    is it a function?

  10. anonymous
    • 5 years ago
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    thanks so much i get it now :)

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