Question

How do you graph `y = 4(2)^x-3`?

Answer 1

graph{4*2^x-3 [-10, 10, -5, 5]}
See explanation below.

Explanation:

Start from exponential function `y=2^x`.

It is defined for all real numbers: `-oo < x < +oo`.
It is always positive: `y=2^x > 0`.
It is monotonically increases as `x` runs from `-oo` to `+oo`.
As `x -> -oo`, function asymptotically approaches `0`: `y=2^x -> 0` (while staying positive).
At `x=0` function `y=2^0=1`.
As `x -> +oo`, function `y=2^x -> +oo`.
Here is the graph of `y=2^x`:
graph{2^x [-10, 10, -5, 5]}

Next, let's transform this graph into `y=4*2^x`.
All we do is stretch this graph vertically from X-axis upward.
At `x=0` the function will have a value `y=4*2^0=4`.
Moving to the left, the function will still asymptotically go to `0`.
Moving to the right, the function will still go to `+oo`.
Here is the graph of `y=4*2^x`:
graph{4*2^x [-10, 10, -5, 5]}

Finally, to transform the graph to `y=4*2^x-3`, we have to shift the whole graph down by `3`.
The result is:
graph{4*2^x-3 [-10, 10, -5, 5]}
It asymptotically approaches `-3` from above, when `x->-oo`.
At `x=0` the function `y=4*2^x-3=1`
To the right of `x=0` the function is growing to `+oo`..