[tex]p(x)=x^2+2x-3[/tex]
The zeroes of p(x) are given by:
[tex]\begin{gathered} x=\frac{-2\pm\sqrt{2^2-4\cdot(-3)}}{2} \\ x_=\frac{-2\pm4}{2} \\ x_+=-3 \\ x_-=1 \end{gathered}[/tex]
The vertex of the polynomial is given by:
[tex](-\frac{b}{2a},-\frac{(b^2-4ac)}{4a})=(-1,-4)[/tex]
The y-intercept is (0,-3)
Therefore, we can graph the polynomial tracing a parabola passing through (-3,0), (-1,-4), (0,-3) and (1,0):
