Question

p(y) = y^2 - y+1 answer fast pls

Answer 1

Step-by-step explanation:

given

p(y)=y^2-y+1

putting y=0

p(0)=(0)^2-0+1

=0-0+1

=1

putting y=1

p(1)=()^2-1+1

=1-1+1

=1

putting y=2

p(2)=(2)^2-2+1

4-2+1

=3

p(2)=(2)^2-2+1