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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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