The answer is 2/-2
Using the formula y^2-y^1/x^2-x^1
An even function can be reflected about the y axis and map onto itself
example: y=x^2
an odd function can be reflected about the origin and map onto itself
example: y=x^3
a simple test is the following
if f(x) is even then f(-x)=f(x)
if f(x) is odd then f(-x)=-f(x)
so
even function
subsitute -x for each and see if we get the same function
remember to fully expand these
g(x)=(x-1)^2+1=x^2-2x+1+1=x^2-2x+2 is the original one
g(x)=(x-1)^2+1
g(-x)=(-x-1)^2+1
g(-x)=(1)(x+1)^2+1
g(-x)=x^2+2x+1+1
g(-x)=x^2+2x+2
not same because the original has -2x
not even
g(x)=2x^2+1
g(-x)=2(-x)^2+1
g(-x)=2x^2+1
same, it's even
g(x)=4x+2
g(-x)=4(-x)+2
g(-x)=-4x+2
not the same, not even
g(x)=2x
g(-x)=2(-x)
g(-x)=-2x
not same, not even
g(x)=2x²+1 is the even function
The answer is ten hope this helps :-)
Answer :The first one
Step-by-step explanation:
