The given equation of quadratic :
[tex]f(x)=x^2-x-6[/tex]for the x - intercept :
The y coordinates are equal to 0
Substitute f(x) = 0
[tex]\begin{gathered} f(x)=x^2-x-6 \\ x^2-x-6=0 \end{gathered}[/tex]Simplify the expression for x :
[tex]\begin{gathered} x^2-x-6=0 \\ x^2-3x+2x-6=0 \\ x(x-3)+2(x-3)=0 \\ (x+2)(x-3)=0 \end{gathered}[/tex]Thus : x + 2= 0 and x - 3 = 0
so, x = -2 and x = 3
X intercept are ( -2, 0) & ( 3, 0)
