Sketch the graph and Determine the domain and range of $h(x)=3+e^{-2x}$.

Question

How do I even start on this? How do I sketch the graph and find the domain and range? I am really lost on how to do this problem! Please walk me through this question!

Answer 1

Make it easier for yourself. You know that the $+3$ only shifts everything up by $3$. The $2$ just makes the graph go a bit "faster", but it won't change the shape at all, so for the purpose of a sketch or finding domain/range, it does nothing. The negative next to the $x$ will just flip everything horizontally, about the $y$ axis. The only thing left is $e^x$, which is a graph you should be familiar with.

So the sketch shouldn't be too hard: take a standard exponential curve and flip it about the $y$ axis. Then move it up by $3$ units. You should now see the domain and range already, but we can do a bit more work for that. I had some coming but it seems another answer has addressed that, give it a go first.

Answer 2

To find the domain, you need to consider if there are any restriction on which values of x the function can take. For example, $f(x) = \sqrt{x}$ cannot take any negative values, so the domain is restricted to the non-negative real numbers. Are there any similar problems with your function?

To find the range, you again need to look at the function. For example, $f(x) = x^2$ can only give a positive output, so the range is all the non-negative real numbers. If $f(x) = 4 + x^2$, then you again have that $x^2$ can only take non-negative values, but you are adding a constant $4$ to every output, so the range becomes $[4, \infty)$. Try to think about what is happening with the different parts of your function.

To sketch the graph, start by plotting some points (x = 0,1,-1,10,-10) and see what happens. Together with what you know about the shape of exponential graphs, you should be able to sketch it.