Answer with Step-by-step explanation:
We are given that function f(x) which is quadratic function.
x -intercept of function f(x) at (-1,0) and (-3,0)
x-Intercept of f means zeroes of f
x=-1 and x=-3
Range of f =[-4,
)
g(x)=![2x^2+8x+6=2(x^2+4x+3)](https://tex.z-dn.net/?f=2x%5E2%2B8x%2B6%3D2%28x%5E2%2B4x%2B3%29)
![g(x)=0](https://tex.z-dn.net/?f=g%28x%29%3D0)
![2(x^2+4x+3)=0](https://tex.z-dn.net/?f=2%28x%5E2%2B4x%2B3%29%3D0)
![x^2+4x+3=0](https://tex.z-dn.net/?f=x%5E2%2B4x%2B3%3D0)
![x^2+3x+x+3=0](https://tex.z-dn.net/?f=x%5E2%2B3x%2Bx%2B3%3D0)
![x(x+3)+1(x+3)=0](https://tex.z-dn.net/?f=x%28x%2B3%29%2B1%28x%2B3%29%3D0)
![(x+1)(x+3)=0](https://tex.z-dn.net/?f=%28x%2B1%29%28x%2B3%29%3D0)
![x+1=0\implies x=-1](https://tex.z-dn.net/?f=x%2B1%3D0%5Cimplies%20x%3D-1)
![x+3=0\implies x=-3](https://tex.z-dn.net/?f=x%2B3%3D0%5Cimplies%20x%3D-3)
Therefore, x-intercept of g(x) at (-1,0) and (-3,0).
Substitute x=-2
![g(-2)=2(-2)^2+8(-2)+6=8-16+6=-2](https://tex.z-dn.net/?f=g%28-2%29%3D2%28-2%29%5E2%2B8%28-2%29%2B6%3D8-16%2B6%3D-2)
![g(x)=2(x^2+4x)+6](https://tex.z-dn.net/?f=g%28x%29%3D2%28x%5E2%2B4x%29%2B6)
![g(x)=2(x^2+2\times x\times 2+4-4)+6=2(x^2+2\times x\times 2+4)-8+6](https://tex.z-dn.net/?f=g%28x%29%3D2%28x%5E2%2B2%5Ctimes%20x%5Ctimes%202%2B4-4%29%2B6%3D2%28x%5E2%2B2%5Ctimes%20x%5Ctimes%202%2B4%29-8%2B6)
![g(x)=2(x+2)^2-2](https://tex.z-dn.net/?f=g%28x%29%3D2%28x%2B2%29%5E2-2)
By comparing with the equation of parabola
![y=a(x-h)^2+k](https://tex.z-dn.net/?f=y%3Da%28x-h%29%5E2%2Bk)
Where vertex=(h,k)
We get vertex of g(x)=(-2,-2)
Range of g(x)=[-2,
)
Zeroes of f and g are same .
But range of f and g are different.
Range of f contains -3 and -4 but range of g does not contain -3 and -4.
f and g are both quadratic functions.