Download Homework

18 Days
Four Days

Definition of the nth root:
For any real numbers a and b, and any positive integer n, if a n  b,
then a is an nth root of b.
(2)3  8  2 is the 3rd root of 8
(2) 4  16 and (2) 4  16  2 and - 2 are 4th roots of 16
Radicand - the number under the root (a)
Index - the degree of the root
n
(n)
a

Lets start with a few familiar examples:
16
24
4x2
36 x 4
100 x 3 y 2
3
27 x 6
3
16 x 4
.25 x 2
4
3
81x 6
 64 x 3 y 6 z 4
1
x2
x3
x4
x5
2
3
4
5
6
7
8
9
10
11
12
13
14
15

pg 372 (# 1-27 odd)
Three Days
8  2  24 
2 5  125  36 
3
16  3 54 

For a Radical Expression to be in simplest for
the following conditions must be met:
◦ No perfect nth power factors, other than 1.
◦ No fractions in the radicand.
◦ No radicals in the denominator.
If n a and n b are real numbers, then n a  n b  n a  b
n
n
n
If a and b are real numbers, then
n
a n a

b
b
3 12 
3
33 9 
3
4 x 2 y  3 16 xy5 
4
27 x 3 y  4 3 xy6 
47

36
3
3
216

3
8
27 x 6 y 3
3
8

12 x 4

3x

Rationalizing the denominator of an
expression is the process of re-writing so
that there are no radicals in any denominator
and there are no denominators in any radical.
2
3
1
3x
x3
5 xy
3
3
4
6x

pg 377 (# 1-35 odd)

Practice 7.2 WS (1-33 odd, 34)

Practice 6-5 (#1-35 odd) - Glencoe
Three Days
Three Days
First of all, what is a rational number?
It’s a number that can be written as a fraction
of integer values.


Definition of Rational Exponents
If the nth root of a is a real number and m and n are integers, then
n
xx
1
n
and
m
n
x  x 
n
m
 x
n
m
3
2
x 
y  3 .5 
a5 
3
z 
4
5
4
16 
4  2.5 
2
7
5
14
x x 
27 x 
6
1
3


pg 388 (# 1-25 odd, 39-49 odd)
Three Days

Solve the following radical expression:
2  2x  4  6
2  2x  4  6
2  2 x  4  6
1
2

1. Convert the radical to rational exponents.

2. Isolate the “radical” part of the equation.


3. Raise both sides of the equation to the
reciprocal power of the rational exponent.
4. You’ve now “cleared” the radical, solve
using the appropriate method for the
resulting equation.
2x  2  4  22
2
3
x 7 5  x


The maximum flow of water in a pipe is modeled
by the formula Q=Av, where A is the cross
sectional area of the pipe, v is the velocity of the
water, and Q is the maximum volume of water than
can flow through the pipe per minute.
Find the diameter of a pipe that allows a maximum
flow of 50 cubic feet per minute at a velocity of
600ft/min. Round to the nearest inch.

pg 394 (# 1-25 odd)

Practice 7.5 WS (#2-32 Even)

Practice 6-7 WS (# 2-22 even) - Glencoe
Three Days



Parent:
Shift up k units:
Shift down k units:

Shift right h units:
Shift left h units

Combined Shift:

y  x
y  f ( x)
y  x k
y  f ( x)  k
y  x k
y  f ( x)  k
y  xh
y  f ( x  h)
y  xh
y  f ( x  h)
y  xh k
◦ (right h units, up k units)
y  f ( x  h)  k




Parent:
Reflection in x-axis:
Vertical Stretch a>1
Vertical Shrink 0<a<1
y  x
y  f ( x)
y  a  x
y  a  f ( x)
y  a x
y  a  f ( x)

Horizontal Stretch 0<c<1 :
Horizontal Compression c>1:

Combined Transformation:

y  a x h  k
y  c x
y  f (c  x )
y  a  f ( x  h)  k
-Name of Family
f x  =
x
8
Square Root
6
-Parent Equation
4
y x
-General Equation
2
-15
-10
-5
5
10
-2
y  a xh k
-Locator Point
Endpoint : (h, k )
-4
-6
-8
-Domain
-Range
[0, )
[0, )
x
0
1
4
9
16
y
0
1
2
3
4
-Name of Family
1
f x  = x
3
8
Cube Root
6
-Parent Equation
4
y x
3
2
-General Equation
y  a3 x  h  k
-Locator Point
-15
-10
-5
5
-2
-4
-6
Inflection : (h, k )
-8
-Domain
-Range


x y
-8 -2
-1 -1
0 0
1 1
8 2
10
y  x 3
y  x 1
y  x2
y  x4
y  x 1  3
y3 x
y 3 x4
y 3 x4 2

https://www.desmos.com/calculator/renedj48tv

pg 417 (# 1-23 odd)

Practice 7.8 WS (# 1-15 odd, 28, 29, 31, 35,
37)