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the function changes at \(x=1\) right?
I have no idea :/ I'm really confused with what the questions are asking for Part 2 especially
erase the line to the left of \(x=1\) and erase the parabola to the right of \(x=1\)
Wait so just the entire left side?
your function is two different things right? one part is a parabola, the other is a line
it changes at 1
to the left of 1 it is a parabola to the right of 1 it is a line
here is a nice picture http://www.wolframalpha.com/widgets/view.jsp?id=5075763dba6ec763f31004272f8aa7fa
ok that didn't work, but you can fill it in for yourself to see what it should look like i can't send it with fields filled out
wait let me take a picture with the erased parts bc i'm not sure if i erased it correctly
yeah good job!
yayy that's c right? how would i do d
closed circle goes on the parabola, because of the \(x\leq 1\) open circle on the line because of \(x>1\)
get rid of the arrows
do i put circles at the end of each line and parabola
oh okay thank you! :)
i don't put it on the other side?
just at the break
okay thank you so much!!
I'm going to add some information. So basically, we were given a piecewise function \[x^2, x \leq 1 \] this means that when our x is less than or equal to 1 we have to graph \[x^2 \] since we we have \[\leq \] we draw a closed circle. This also applies when we have \[\geq \] ( we draw a closed circle) \[2x+1, x > 1\] this means that our x is greater than 1. So on the graph, we draw 2x+1 after x = 1 since we have > we draw an open circle. This also applies when we have < (we draw an open circle. So our graph will look at this |dw:1436843159513:dw|