Answer
Visually, the parabola intersects the x-axis in two points: $x-1$ and $x=3$. The x-intercepts are: $(-1,0)$ and $(3,0)$
Algebraically: the x-intercepts are $(3,0)$ and $(-1,0)$
Work Step by Step
$f(x)=0$
$x^2-2x-3=0~~$ ($a=1,b=-2,c=-3$):
$x=\frac{-(-2)±\sqrt {(-2)^2-4(1)(-3)}}{2(1)}=\frac{2±\sqrt {16}}{2}=\frac{2±4}{2}=1±2$
$x=3$ or $x=-1$
x-intercepts: $(3,0)$ and $(-1,0)$
