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gradu8 · May 20, 2009, 02:46 PM

I tried to do it multiple times but fractions always gets me frustrated :confused:

P(y)=y^2-4y+3
P(-/4)

This is as far as I got

P(y)=(-1/4)^2-4(-1/4)+3

Perito · May 21, 2009, 05:18 AM

I tried to do it multiple times but fractions always gets me frustrated

P(y)=y^2-4y+3
P(-/4)

This is as far as I got

P(y)=(-1/4)^2-4(-1/4)+3

$P(y)=y^2-4y+3$

What is P(-4)?

$P(y)=(-4)^2-4(-4)+3 = (-4 \times -4) -(4 \times -4) + 3$

$P(y)=16 - (-16) + 3 = 16+16+3=35$

What is P(-1/4)?

$P(y)=(-\frac 14)^2-4(-\frac 14)+3 = \frac {1}{16} + 1 + 3$

$P(y)=4\,\frac {1}{16}=4.0625$