How do you graph #Y=2^x+3# by plotting points?

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Answer 1 · 2017-04-21T00:19:16

Aysmptote is already given. Sub in #x#-values and solve for #y#.

This graph is slightly easier to graph than the other exponential functions.

The base function is #2#, describing the increase rate as the function's #x#-value increases.

The only transformation that's done is a translation #3# units up. This gives us the asymptote of #y>3#.

graph{(2^x) + 3 [-10, 10, -5, 5]}

Now, we need to plot the points. We can simply just sub #x# as #-1#, #0#, and #1#, and solve for #y#.

Doing so gives us #7/2 ->3.5#, #4#, and #5# respectively.

Connect the dots in a curving motion. Make sure when you extend the function in the 2nd quadrant, that it never touches #3#.

Hope this helps :)