Answer:
Option I
Step-by-step explanation:
Let
[tex]f(x)=x^2+3x-28[/tex]
We know by subsitution,
[tex]f(-9)=(-9)^2+3(-9)-28=26>0[/tex]
[tex]f(0)=0^2+3(0)-28=-28<0[/tex]
[tex]f(5)=5^2+3(5)-28=12>0[/tex]
Since a quadratic equation only have 1 turning point, we know for some values a in (-9, 5) such that f(a)<0. And the option of the range of a must lie with in (-9, 5).
Therefore, only Option I is correct.
