Question

Find P(0), P(1) and P(-2) for P(y) = (y^2) - y + 1 ​

Answer 1

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P(y)=(Y^2)-y+1

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

= 0-0+1

=1

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

= 1-1+1

=1

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

= 4 +2+1

=7

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Answer 2

Answer:

P(0)=1,p(1)=1,p(-2)=7

Step-by-step explanation:

P(0)= 0-0+1 =1

P(1)= 1²-1+1 =1

P(-2)= (-2)²-(-2) +1 =4 +2 +1 =7