Answer: \(\mathrm{f}^{-1}\) exists because \(\mathrm{f}\) is one-one for \(x\geq-1\). The vertices are \((-1,-4)\) and \((-4,-1)\).

(a) Complete the square:

\(\mathrm{f}(x)=x^2+2x-3=(x+1)^2-4.\)

The vertex of \(y=\mathrm{f}(x)\) is \((-1,-4)\). Since the domain is restricted to \(x\geq-1\), only the right-hand branch of the parabola is used. On this interval the function is one-one, so \(\mathrm{f}^{-1}\) exists.

(b) The graph of \(y=\mathrm{f}(x)\) has vertex

\((-1,-4).\)

Its \(y\)-intercept is

\(\mathrm{f}(0)=-3,\)

and its \(x\)-intercept in the domain is

\(x^2+2x-3=0 \quad\Rightarrow\quad (x+3)(x-1)=0,\)

so \(x=1\).

The graph of \(y=\mathrm{f}^{-1}(x)\) is the reflection of \(y=\mathrm{f}(x)\) in the line \(y=x\). Therefore its vertex is

\((-4,-1).\)

The intercepts are also swapped: it has \(x\)-intercept \(-3\) and \(y\)-intercept \(1\).