It's a quadratic function.
[tex]f(x)=ax^2+bx+c[/tex]
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[tex]f(x)=3x^2-8[/tex]
a = 3, b = 0, c = -8
a = 3 > 0, therefore the parabola open up.
The range of this function is [k, ∞).
k - second coordinate of the vertex (h, k)
[tex]h=\dfrac{-b}{2a},\ k=f(h)[/tex]
Substitute:
[tex]h=\dfrac{-0}{2(3)}=0\\\\k=f(0)=3(0^2)-8=0-8=-8[/tex]
Answer: The range is [tex][8,\ \infty)[/tex]
