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March 29, 2015

March 29, 2015

Posted by **Unknown** on Thursday, November 29, 2012 at 11:23pm.

25. f(x)= x^2-1 ; -3, -2, -1, 0, 1, 2 ,3

=

f(x); -3^2-1=-10

f(x) ; -2^2-1=-5

f(x) ; -1^2-1=-2

f(x) ; 0^2-1=-1

f(x) ; 1^2-1=0

f(x) ; 2^2-1=3

f(x) ; 3^2-1=8

26. g(x); √(x+1)-2 ; x+-1, 0, 3, 8

g(x) ; √(-1+1)-2=-2

g(x) ; √(0+1)-2=-1

g(x) ; √(3+1)-2=4

g(x) ; √(8+1)-2=1

27. h(x)= |x-3| ; x=0,1,2,3,4,5,6

h(x); |0-3| =3

h(x); |1-3|=4

h(x); |2-3|=5

h(x); |4-3|=7

h(x); |5-3|=8

h(x); |6-3|=9

- Algebra2 -
**Steve**, Friday, November 30, 2012 at 12:25am25.

all squares are positive

thus f(-3) = (-3)^2-1 = 9-1 = 8

-3^2 = -9, but (-3)^2 = +9

when substituting in for x, to avoid mistakes, always enclose the value in ().

26.

f(3) = √(3+1)-2 = √4 - 2 = 2-2 = 0

27.

|x| = -x if x<0

|x| = x if x >= 0

so, |1-3| = |-2| = 2, not 4

do the addition first, then the ||.

Treat the || like special parentheses, or treat |x| like a function abs(x), so all the inside stuff is done first

|1-3| = 2

|1| + |-3| = 1+3 = 4

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