Bash variable quoting in a function

I've defined a short helper function in a bash script. For legibility, I'd like to replace the boilerplate headers with a variable declared outside the function. I've tried quoting the declaration many different ways, but nothing's working. I want to replace this put() { local response=$(curl -H 'Content-Type: application/json' -H 'Accept: application/vnd.urbanairship+json; version=3;' ...) echo $response } with something like this headers=??? put() { local response=$(curl $headers ...) echo $response } i.e. $headers should expand to -H 'Content-Type: application/json' -H 'Accept: application/vnd.urbanairship+json; version=3;' $headers needs to expand to separate words to get passed properly as arguments, so I've even tried using an array. Note: I've pretty much given up. It's not worth fighting with for such a simple use. At this point the question is for edification.