How do you graph #y = 2\sqrt{2x+3}+1#?

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Answer 1 · 2014-10-29T21:09:27

#y=2sqrt{2x+3}+1#

by factoring out #2#,

#Rightarrow y=2sqrt{2(x+3/2)}+1#

by pulling #sqrt{2}# out,

#Rightarrow y=2sqrt{2}sqrt{x+3/2}+1#

Now, let us sketch #y=sqrt{x}#, which looks like:

enter image source here

By stretching vertically by a factor of #2sqrt{2]#, we have #y=2sqrt{2}sqrt{x}#, which looks like:

enter image source here

By shifting left by #3/2#, we have #y=2sqrt{2}sqrt{x+3/2}#, which looks like:

enter image source here

By shifting up by 1 unit, we have #y=2sqrt{2}sqrt{x+3/2}+1#, which looks like:

enter image source here

Above is the graph of #y=2sqrt{2x+3}+1#.

I hope that this was helpful.