Question

If y = 2 then find the value of p(y) = 4y2 - 8y + 1​

Answer 1

Answer:

16-16+1

=1 is the value of p(y).

Answer 2

$If \: y = 2 \:then \: the \: value \:of \\p(y) = 4y^{2}-8y+1 \:is \:p(2)$

$p(2) = 4\times 2^{2} - 8 \times 2 + 1 \\= 4 \times 4 - 16 + 1 \\= 16 - 16 + 1 \\= 1$

Therefore.,

$\red{ If \: y = 2 \:then \: the \: value \:of \:\p(y)}\green { = 1 }$

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