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December 21, 2014

December 21, 2014

Posted by **Jimmy** on Thursday, May 5, 2011 at 4:13pm.

1/3x - 4 if -6 </= x < - 3

f(x) = 2 if -3 < x < 3

x^2 - 4 if x > 3

I found out the domain, range, and points of disontinuity. I think i did them right:

Domain= [-6,infinity]

Range= [-6,infinity]

and points of discontinuity = x=-3,x=3

How do i graph these functions?

- Algebra -
**Reiny**, Thursday, May 5, 2011 at 5:09pmYour first segment runs from (-6,-6) to (-3,-5)

for -6 ≤ x < -3

so draw a solid dot for (-6,-6) and an "open" dot for (-3,-5)

the 2nd segment runs from (-3,2) to (3,2).

draw that line leaving the end points as open dots.

the 3rd graph is part of a parabola, which has its vertex at (0,-4) and which would open upwards.

I would pass through (1,-3), (2,0) and (3,5)

draw it only from (3,5) on, leaving the (3,5) open.

you are correct with your domain and range, and your value of x that have a discontinuity.

- Algebra -
**Jimmy**, Thursday, May 5, 2011 at 5:36pmthanks

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