Question

P(0),p(1),p(2)forp(y)=y²-3y+1

Answer 1

Answer :

p(y) = y²-3y+1

p(0) = (0)²-3(0)+1

=> p(0) = 0-0+1

=>p(0) = 1

p(1) = (1)²-3(1)+1

=> p(1) = 1-3+1

=> p(1) = 2-3

=> p(1) = -1

Here we just need to substitute the values in p(y).

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Answer 2

Explanation:

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• p(y) = y²-3y+1

p(0) = (0)²-3(0)+1

• => p(0) = 0-0+1

=>p(0) = 1

• p(1) = (1)²-3(1)+1

=> p(1) = 1-3+1

• => p(1) = 2-3

=> p(1) = -1

Here we just need to substitute the values in p(y).

hope this helps you