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]}

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