Answer:
See below
Step-by-step explanation:
[tex]f(x)=3x^2+6x-24\\\\f(x)=3(x^2+2x-8)\\\\f(x)+3(9)=3(x^2+2x-8+9)\\\\f(x)+27=3(x^2+2x+1)\\\\f(x)=3(x+1)^2-27[/tex]
Here, we can see that the vertex of the parabola is [tex](-1,-27)[/tex] when we compare to the equation [tex]y=(x-h)^2+k[/tex] with vertex [tex](h,k)[/tex].
A second point you could plot is the y-intercept, when [tex]x=0[/tex]:
[tex]f(0)=3(0+1)^2-27\\\\f(0)=3-27\\\\f(0)=-24[/tex]
So, your y-intercept is [tex](0,-24)[/tex].