`y=3(x+1)^2-4`, find Parity of a function

Solution:
`y=3(x+1)^2-4`

Siplyfing vertex form equation `y=3(x+1)^2-4`, we get

`y=3(x+1)^2-4`

`y=3(x^2+2x+1)-4`

`y=3x^2+6x+3-4`

`y=3x^2+6x-1`

Parity :
Even Function : A function is even if `f(-x)=f(x)` for all `x in R`

Odd Function : A function is odd if `f(-x)=-f(x)` for all `x in R`

`f(-x)=3(-x)^2+6(-x)-1`

`f(-x)=3x^2-6x-1`

`f(x)!=f(-x)`

`3x^2+6x-1` is not an even function

`-f(x)=-(3x^2+6x-1)`

`-f(x)=-3x^2-6x+1`

`f(x)!=-f(x)`

`3x^2+6x-1` is not an odd function

`:. 3x^2+6x-1` is neither even nor odd function

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