`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
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
