what is the asymptote of `y=2^x-4`

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`y=2^x- 4`

An exponential function in the form `y=a^x` , the `a^x` is always greater than zero. Since `a^xgt0` , y is always greater than zero too (y>0). Because of this, `y=a^x` has a horizontal asymptote at y=0.

Applying this, let's solve for the horizontal asymptote.
Set `2^x` greater than zero.

`2^x gt 0`

Then, subtract both sides by 4, since the given function is `y=2^x-4` .

`2^x - 4 gt -4`

And replace `2^x - 4` by y.

`y gt -4`

This means that the function is always greater than -4.

Hence, the horizontal asymptote is `y=-4` .

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