Answer:
The intercepts of the parabola are at;
[tex](0,-12),(-2,0),(6,0)[/tex]Explanation:
Given the function:
[tex]f(x)=x^2-4x-12[/tex]The intercepts of the function would be at x=0 or y=0.
At x=0;
[tex]\begin{gathered} y=f(0)=0^2-4(0)-12 \\ y=-12 \\ (0,-12) \end{gathered}[/tex]at y=0;
[tex]\begin{gathered} f(x)=x^2-4x-12=0 \\ x^2-4x-12=0 \\ (x+2)(x-6)=0 \\ x+2=0 \\ x=-2 \\ or \\ x-6=0 \\ x=6 \\ (-2,0),(6,0) \end{gathered}[/tex]Therefore, the intercepts of the parabola are at;
[tex](0,-12),(-2,0),(6,0)[/tex]