How do you graph #y=6(1/2)^(x+5)-2#?

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Answer 1 · 2016-12-02T02:34:29

The exponential-decay Graph for #y=3/16 (1/2)^x-2# is inserted.

#y=6 (1/2)^(x+5)-2=6(1/2)^5(1/2)^x-2=3/16(1/2)^x-2#.

y-intercept ( x = 0 ) is #-29/16 > -2#.

As # x to oo, y to 0 and # as # x to -oo, y to oo#.

#y = -2# is the ( horizontal ) asymptote.

In #Q_4#, note that the initial vertical space 3/16, in between the

curve and the asymptote, tends to 0, as #x to oo.#.

graph{y-3/16(1/2)^x+2=0 [-40, 40, -20, 20]}