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Mathematics 10 Online
OpenStudy (anonymous):

can anyone help me? http://i.imgur.com/E86ZTKE.jpg

OpenStudy (anonymous):

You want both ends to meet.

OpenStudy (anonymous):

They have to be equal when \(x=1\) and \(x=3\)

OpenStudy (anonymous):

@ineptAtMath Need more help?

OpenStudy (anonymous):

let me try some stuff first. xD

OpenStudy (anonymous):

is it, b=-4 and c =7?

OpenStudy (anonymous):

I'm not sure how you got that answer, but it doesn't seem right.

OpenStudy (anonymous):

awh ;'< then do u have anymore clues?

OpenStudy (anonymous):

When \(x=1\) then you want: \[ (1)+1=(1)^2+b(1)+c \]

OpenStudy (anonymous):

This gives you the equation \[ 2-1+b+c \]

OpenStudy (anonymous):

Use \(x=3\) to get the other equation.

OpenStudy (anonymous):

ooopss i see what i did wrong

OpenStudy (anonymous):

i accidently put (2) + 1

OpenStudy (anonymous):

also, im too sure but how do u know ur suppose to use 3 and 1?

OpenStudy (anonymous):

im not too sure*

OpenStudy (anonymous):

Because that is where the function switches between pieces.\[ 1<x<3 \]

OpenStudy (anonymous):

i think i get it. thank you. :>

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