Question
How to determine the interval for the domain of an inverse function of a quadratic function
For an example the graph of $y=x^2$ 's inverse has the domain of x is larger or equal to 0 using algebraic methods. (if we were to use graphic it would be obvious)
I was wondering how we determine such interval of the inverse of a quadratic
On
A function is invertable on an interval for which it is injective. A function is injective when $\forall a,b,f(a)=f(b)\Rightarrow a=b$ (think about why this would guarantee an inverse!).
It's not too hard to show that a quadratic equation is injective on an interval if and only if the interval is entirely on one side of its local min/max. The $x$ coordinate of the min/max is given by $x=\frac{-b}{2a}$, so the quadratic is invertable on the intervals $(-\infty, \frac{-b}{2a}]$ and $[\frac{-b}{2a},\infty)$, as well as any smaller interval that is properly contained in one of those intervals.
The domain of the inverse function is obtained by just applying $f$ to these intervals. That gives $(-\infty, c-\frac{b^2}{4a}]$ and $[c-\frac{b^2}{4a},\infty)$ as the corresponding domains of the inverse function.
On
Hint: If we have a quadratic function with image $X$ then the inverses' domain would be $X$.
Example: $f(x) = x^2+100x-271$. Clearly, this function's end behavior is going to infinity, and its lowest point is at its vertex. So its image is $[-2771,\infty)$, as $-2771$ is the y value of its vertex (Verification left to the O. P. as an exercise). As such, the domain of its inverse is $[-2771,\infty)$
A quadratic of the form $y=ax^2+bx+c$ with $a\ne0$, then it has a vertex at $x=-\frac b{2a}$, hence, the domain of the inverse will fall into two cases
If $a>0$, then $x\ge-\frac b{2a}$.
If $a<0$, then $x\le-\frac b{2a}$