I need to graph y=x^2+3. Do I just plug in #'s for x to find y then graph the points??

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I need to graph y=x^2+3. Do I just plug in #'s for x to find y then graph the points??

Mathematics
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Yes :)
K... Im just nervous Im doing something wrong. Are these right?? (0,0) (1,4) (2,7) (-1,4) (-2,7)
Yes, that's correct, well done :)

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My next one I have to graph (LAST ONE!) is y=lx+3l +2 what does that absolute value sign mean??/
Basically, what ever value is inside the absolute value sign, when you remove the sign, it is always positive. So if you use x = -45, the value inside the absolute bars is -42, but when you remove the bars, it's 42.
K..So I got the points (0,5) (1,6) (2,7) (-1,6) (-2,7) are they right?
Not quite, when you use negative x values, you first need to compute the value inside the absolute bars, then remove them. So in the case of x=-1 you get |x+3|=|-1+3|=|2|=2
So are they (0,5) (1,6) (2,7) (-1,4) (-2,3) are any of them right?
All are correct. :)
Thank YOU!!!
for \[y =x ^{2}+3\] , the y-intercept should be (0, 3) rather than (0,0).
Obviously I shouldnt help people when I'm busy with other things, too many mistakes today. ^^ Thanks for finding it.
i know the feeling - also i'm feeding my 3 month old and typing one handed:)
That's a challenge..! :)
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