Answer:
The correct option is 3.
Step-by-step explanation:
The given equation is
![2x^2+12x-14=0](https://tex.z-dn.net/?f=2x%5E2%2B12x-14%3D0)
It can be written as
![(2x^2+12x)-14=0](https://tex.z-dn.net/?f=%282x%5E2%2B12x%29-14%3D0)
Taking out the common factor form the parenthesis.
![2(x^2+6x)-14=0](https://tex.z-dn.net/?f=2%28x%5E2%2B6x%29-14%3D0)
If an expression is defined as
then we add
to make it perfect square.
In the above equation b=6.
Add and subtract 3^2 in the parenthesis.
![2(x^2+6x+3^2-3^2)-14=0](https://tex.z-dn.net/?f=2%28x%5E2%2B6x%2B3%5E2-3%5E2%29-14%3D0)
![2(x^2+6x+3^2)-2(3^2)-14=0](https://tex.z-dn.net/?f=2%28x%5E2%2B6x%2B3%5E2%29-2%283%5E2%29-14%3D0)
![2(x+3)^2-18-14=0](https://tex.z-dn.net/?f=2%28x%2B3%29%5E2-18-14%3D0)
.... (1)
Add 32 on both sides.
![2(x+3)^2=32](https://tex.z-dn.net/?f=2%28x%2B3%29%5E2%3D32)
The vertex from of a parabola is
.... (2)
If a>0, then k is minimum value at x=h.
From (1) and (2) in is clear that a=2, h=-3 and k=-32. It means the minimum value is -32 at x=-3.
The equation
reveals the minimum value for the given equation.
Therefore the correct option is 3.