Draw the graph of `y = sin^2x.`

Explanation

We have `y = f(x) = sin 2^(x)`
Clearly, the domain of the function is R.
`sin 2^(x) = 0`
`therefore` `2^(x) = npi, n in Z^(+)`
`therefore` `x = log_(2) (npi), n in Z^(+)`
For `x lt 0, 0 lt 2^(x) lt 1, 0 lt sin (2^(x)) lt sin 1`
As the value of x increases, `2^(x)` increases more rapidly and `sin(2^(x))` oscillates between '-1' and '1' more frequenly.
So the graph of the function is as shown in the following figure.

When `x to - oo, sin (2^(x)) to 0.` So y = 0 is an asymptote.