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