Answer: Stationary point \((-1,-4)\). Intercepts: \((-3,0)\), \((1,0)\), \((0,-3)\).

(a) Complete the square:

\(y=x^2+2x-3.\)

Since

\(x^2+2x=(x+1)^2-1,\)

we get

\(y=(x+1)^2-1-3=(x+1)^2-4.\)

This is a positive quadratic, so the stationary point is a minimum. The minimum occurs when

\(x+1=0,\)

so \(x=-1\), and then \(y=-4\).

Therefore the stationary point is

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

(b) For the \(x\)-intercepts, put \(y=0\):

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

Factorise:

\((x+3)(x-1)=0.\)

So the \(x\)-intercepts are

\((-3,0)\quad\text{and}\quad(1,0).\)

For the \(y\)-intercept, put \(x=0\):

\(y=-3.\)

So the \(y\)-intercept is

\((0,-3).\)

The sketch should be an upward-opening parabola passing through these intercepts with minimum point \((-1,-4)\).