Answer: \(2(x+\frac54)^2-\frac18\), stationary point \((-\frac54,-\frac18)\), least \(p=-\frac54\).

(a) Complete the square:

\(2x^2+5x+3=2\left(x^2+\dfrac52x\right)+3.\)

Inside the bracket,

\(x^2+\dfrac52x=\left(x+\dfrac54\right)^2-\left(\dfrac54\right)^2.\)

Therefore

\(2x^2+5x+3=2\left[\left(x+\dfrac54\right)^2-\dfrac{25}{16}\right]+3.\)

Simplifying,

\(2x^2+5x+3=2\left(x+\dfrac54\right)^2-\dfrac{25}{8}+3 =2\left(x+\dfrac54\right)^2-\dfrac18.\)

Thus \(a=\dfrac54\) and \(b=-\dfrac18\).

(b) The completed square form shows that the minimum point of the parabola is

\(\left(-\dfrac54,-\dfrac18\right).\)

(c)(i) For \(\mathrm f^{-1}\) to exist, \(\mathrm f\) must be one-one on its domain. The parabola is one-one on either side of its vertex. Since the domain is \(x\geqslant p\), the least value of \(p\) that keeps only the right-hand branch is

\(p=-\dfrac54.\)

(c)(ii) With \(p=-\dfrac54\), the graph of \(y=\mathrm f(x)\) is the right-hand branch of the parabola with vertex

\(\left(-\dfrac54,-\dfrac18\right).\)

The \(y\)-intercept of \(y=\mathrm f(x)\) is found by putting \(x=0\):

\(y=3,\)

so the \(y\)-intercept is \((0,3)\).

For the \(x\)-intercepts, solve

\(2x^2+5x+3=0.\)

Factorising,

\((2x+3)(x+1)=0,\)

so \(x=-\dfrac32\) or \(x=-1\). The value \(-\dfrac32\) is not in the restricted domain \(x\geqslant-\dfrac54\), so the only \(x\)-intercept of \(\mathrm f\) is

\((-1,0).\)

The graph of \(y=\mathrm f^{-1}(x)\) is the reflection of \(y=\mathrm f(x)\) in the line \(y=x\). Therefore the intercept \((-1,0)\) reflects to \((0,-1)\), and the intercept \((0,3)\) reflects to \((3,0)\). The vertex reflects to

\(\left(-\dfrac18,-\dfrac54\right).\)