we need the quadratic to be one-to-one in order for its inverse to be a function. we can accomplish this by splitting the quadratic into two parts divided by the vertex, and then picking one of those parts.

so, if, for example, the x coordinate of the vertex is , then the largest restricted domain for which f has an inverse function is or

we find the inverse function by switching x and y and solving for y2nd part is to find the inverse using the original function and the domain i found in part 1 and state its domain and range.

that is, take

then switch x and y to obtain

now solve for y

the domain and range of your inverse function depends on what interval you chose for the first question, but i will say this, the domain of your original function is the range of your inverse function, and the range of your inverse function is the domain of your original function

do you know how to find the domain and range?

can you continue?