None of Option are correct
Step-by-step explanation:
The Given function is f(x) = y = [tex]\frac{1}{2} x^{2} + x -9[/tex]
Here, f(x) is quadratic function and it asked that when function cross the Negative X-axis.
If any function cross the X-axis then, Point of intersection will have y component as zero.
Therefore, f(x) = y = [tex]\frac{1}{2} x^{2} + x -9 [/tex]
y = [tex]\frac{1}{2} x^{2} + x -9 = 0 [/tex]
Solving quadratric equation,
Comparing with [tex] ax^{2} + bx + c= 0 [/tex]
We get a=[tex]\frac{1}{2}[/tex], b=1, c=(-9)
Since,
[tex]x=\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]x=\frac{- 1+\sqrt{(1)^{2}-4(\frac{1}{2})(-9) }}{2(\frac{1}{2})}[/tex]
[tex]x=\frac{-1+\sqrt{1+(2)(9)}}{1}[/tex]
[tex]x=\frac{-1+\sqrt{19}}{1}[/tex]
[tex]x=-1+\sqrt{19}and x=-1-\sqrt{19}[/tex]
Thus, None of option are correct.
Verification.....
A) (–6, 0) and (–5, 0) give quadratic function : y = [tex]x^{2} + 11x +30 [/tex]
B) (–4, 0) and (–3, 0) give quadratic function : y = [tex]x^{2} + 7x +12 [/tex]
C) (–3, 0) and (–2, 0) give quadraric function : y = [tex]x^{2} + 5x +6 [/tex]
D) (–2, 0) and (–1, 0) give quadratic function : y = [tex]x^{2} + 3x +2 [/tex]
