Question

Find p(0),p(1),p(-2) when p(y)= y^2-y+6

Answer 1

Answer:

⏩ ⒶⓃⓈⓌⒺⓇ ⏪

Given polynomial, p(y) = y^2-y +6

(1) p(y) = p(0)

➡ p(0) = (0) ^2- (0) +6

➡ p(0) = 0 - 0 + 6

➡ p(0) = 6

Therefore, the value of p(0) =6

(2) p(y) =p(1)

➡ p(1) = (1) ^2 - (1) + 6

➡ p(1) = (1) - (1) +6

➡ p(1) = 6

Therefore, the value of p(1) = 6

(3) p(y) = p(-2)

➡ p(-2) = (-2) ^2 - (-2) + 6

➡ p(-2) = 4 + 2 + 6

➡ p(-2) = 12

Therefore, the value of p(-2) = 12

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Step-by-step explanation:

℘ℓḙᾰṧḙ ՊᾰԻк Պḙ ᾰṧ ♭Իᾰ!ℵℓ!ḙṧт✌✌