Question

p(y)=y square-y+1 find p(0)
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Answer 1

$p(y) = {y}^{2} - y + 1$

$p(0) = {0}^{2} - 0 + 1$

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

$p(0) = + 1$

Hope it is helpful :-)

Answer 2

Answer.

Given,

p(y) = y² – y + 1

∴ p(0) = 0² – 0 + 1 = 1

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