1.

Find the value of the polynomial f(y) = 6y – 3y2 + 3 at (i) y = 1 (ii) y = -1 (iii) y = 0

Answer»

(i) At y = 1, 

f(1) = 6(1) – 3(1)2 + 3 = 6 – 3 + 3 = 6 

(ii) At y = -1, 

f(-1) = 6(-1) – 3(-1)2 + 3 = -6 – 3 + 3 = -6 

(iii) At y = 0, 

f(0) = 6(0) – 3(0)2 + 3 = 0 – 0 + 3 = 3



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