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Name
Class
6-1
Date
Reteaching
Roots and Radical Expressions
For any real numbers a and b and any positive integer n, if a raised to the nth
power equals b, then a is an nth root of b. Use the radical sign to write a root. The
following expressions are equivalent:
index
power
an 5 b
g
radicand
n
!b
!
b5a
radical sign
Problem
What are the real-number roots of each radical expression?
3
a. !343
b.
4 1
Å625
3
Because (7)3 5 343, 7 is a third (cube) root of 343.
3
Therefore, !343
5 7.
(Notice that (27)3 5 2343, so 27 is not a cube root of 343.)
1
1
1
Because Q 15 R 4 5 625
and Q 215 R 4 5 625
, both 15 and 215 are real-number fourth roots of 625
.
c. !20.064
Because (20.4)3 5 20.064,20.4 is a cube root of 20.064 and is, in fact, the only one.
3
So, !20.064
5 20.4.
d. !225
Because (5)2 5 (25)2 5 25, neither 5 nor 25 are second (square) roots of 225. There are
no real-number square roots of 225.
Exercises
Find the real-number roots of each radical expression.
3
1. !169 213, 13
4.
7.
3
4
2. !729 9
218 212
5.
4
225
no real sq root
8. !0.1296 20.6, 0.6
Å
Å
4
10. !20.0001
no real 4th root
4
3. !0.0016 20.2, 0.2
Å121
2 2
211
, 11
4
11.
5 1 1
Å 243 3
6.
3 125 5
Å 216 6
3
9. !20.343 20.7
12.
3 8
Å 125
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2
5
Name
Class
6-1
Date
Reteaching (continued)
Roots and Radical Expressions
n
You cannot assume that "an 5 a. For example, "(26)2 5 !36 5 6, not 26.
This leads to the following property for any real number a:
n
If n is odd
"an 5 a
If n is even
"an 5 u a u
n
Problem
What is the simplified form of each radical expression?
3
a. " 1000x3y9
3
3
"
1000x3y9 5 "
103x3(y3)3
Write each factor as a cube.
3
5"
(10xy3)3
Write as the cube of a product.
5 10xy3
Simplify.
8
256g
b. 4 4 16
Åh k
4
256g8
Å h4k16
5
4
44(g2)4
Å h4(k4)4
Write each factor as a power of 4.
2 4
4g
5 4 a 4b
Å hk
5
4g2
u hu k4
Write as the fourth power of a quotient.
Simplify.
The absolute value symbols are needed to ensure the root is positive when
h is negative. Note that 4g2 and k4 are never negative.
Exercises
Simplify each radical expression. Use absolute value symbols when needed.
3
13. "36x2 6»x…
16.
"x20
"y8
14. " 216y3 6y
(x 1 3)3
17. 3
Å (x 2 4)6
x10
y4
27z3
19. 3
Å (z 1 12)6
(y 2 4)8
22. 4
Å (z 1 9)4
3z
(z 1 12)2
(y 2 4)2
»z 1 9…
15.
x 1 3
(x 2 4)2
4
20. " 2401x12 7»x3…
6 6
a b
23. 3 3
Å c
a 2b 2
c
1
Å 100x2
5
18. " x10y15z5 x2y3z
1331
21. 3
Å x3
11
x
3
24. " 2x3y6 2xy2
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1
10»x…
Name
Class
6-2
Date
Reteaching
Multiplying and Dividing Radical Expressions
You can simplify a radical if the radicand has a factor that is a perfect nth power
and n is the index of the radical. For example:
n
n
!xynz 5 y !xz
Problem
What is the simplest form of each product?
3
3
a. !
12 ? !
10
n
n
n
Use !a ? !b 5 !ab.
3
3
3
!
12 ? !
10 5 !
12 ? 10
3
5 "22 ? 3 ? 2 ? 5
Write as a product of factors.
3 3
5"
2 ?3?5
Find perfect third powers.
3
3
5 "23 ? "3 ? 5
n
Use !ab
5 !a ? !b.
3
5 2!
15
Use "an 5 a to simplify.
n
n
n
b. "7xy3 ? "21xy2
n
n
n
Use !a
? !b
5 !ab.
"7xy3 ? "21xy2 5 "7xy3 ? 21xy2
5 "7xy2y ? 3 ? 7xy2
Write as a product of factors.
5 "72x2(y2)2 ? 3y
Find perfect second powers.
5 7xy2"3y
Use !an 5 a to simplify.
n
Exercises
Simplify each product.
3
3
1. !15x ? !35x 5x"21
2. " 50y2 ? " 20y 10y
4. 5"7x3y ? "28y2
5. 2" 9x5y2 ? " 2x2y5
70xy"xy
3
3
3
2x2y2 !18xy
3
6. !3 Q !12 2 !21 R
6 2 3"7
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3
3
3. " 36x2y5 ? " 26x2y 26xy2 !x
Name
Class
Date
Reteaching (continued)
6-2
Multiplying and Dividing Radical Expressions
Rationalizing the denominator means that you are rewriting the expression so
that no radicals appear in the denominator and there are no fractions inside the
radical.
Problem
What is the simplest form of
!9y
!2x
?
Rationalize the denominator and simplify. Assume that all variables are positive.
!9y
!2x
5
9y
Å 2x
Rewrite as a square root of a fraction.
5
9y ? 2x
Å 2x ? 2x
Make the denominator a perfect square.
5
18xy
Å 4x2
Simplify.
5
!18xy
Write the denominator as a product of perfect squares.
"22 ? x2
5
"18xy
2x
Simplify the denominator.
5
"32 ? 2 ? x ? y
2x
Simplify the numerator.
5
3"2xy
2x
Use !an 5 a to simplify.
n
Exercises
Rationalize the denominator of each expression. Assume that all variables
are positive.
3
"6ab2
8. 3
"2a4b
!5 "5x
7.
x
!x
3
11.
4"k9
3
16"k5
3
k "k
4
12.
4
4
"9y "
9x3y
9. 4
x
"x
3
"3b
a
4
"10
13. 4
"z2
3x5 x2"15xy
Ä 5y
5y
4
"10z2
z
10.
3
19a2b "
19ac2
14. 3
4
c2
Ä abc
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"10xy3 "30xy
6
"12y2
Name
Class
Date
Reteaching
6-3
Binomial Radical Expressions
Two radical expressions are like radicals if they have the same index and the same
radicand.
Compare radical expressions to the terms in a polynomial expression.
Like terms:
4x3
11x3
Unlike terms:
4y3
11x3
Like radicals:
4!6
3
3
11 !
6
Unlike radicals:
4!5
3
3
11 !
6
The power and the variable are the same
4y2
Either the power or the variable are not the same.
The index and the radicand are the same
2
4 "6
Either the index or the radicand are not the same.
When adding or subtracting radical expressions, simplify each radical so that you
can find like radicals.
Problem
What is the sum? !63 1 !28
!63 1 !28 5 !9 ? 7 1 !4 ? 7
Factor each radicand.
5 "3 2 ? 7 1 "2 2 ? 7
Find perfect squares.
5 "3 2 "7 1 "2 2 "7
n
n
n
Use !ab
5 !a ? !b.
5 3 !7 1 2 !7
Use "an 5 a to simplify.
5 5 !7
Add like radicals.
n
The sum is 5!7.
Exercises
Simplify.
1. !150 2 !24 3 !6
3
3
3
4. 5 !2 2 !54 2 !2
3
3
3
2. !135 1 !40 5 !5
3. 6 !3 2 !75 !3
3
3
3
5. 2!48 1 !147 2 !27 0 6. 8 !3x 2 !24x 1 !192x
3
10 !3x
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Name
Class
Date
Reteaching (continued)
6-3
Binomial Radical Expressions
• Conjugates, such as !a 1 !b and !a 2 !b, differ only in the sign of the
second term. If a and b are rational numbers, then the product of conjugates
produce a rational number:
2
2
Q !a 1 !b R Q !a 2 !b R 5 Q !a R 2 Q !b R 5 a 2 b.
• You can use the conjugate of a radical denominator to rationalize the
denominator.
Problem
What is the product? Q 2 !7 2 !5 R Q 2 !7 1 !5 R
Q 2 !7 2 !5 R Q 2 !7 1 !5 R These are conjugates.
2
5 Q 2 !7 R 2 Q !5 R
2
Use the difference of squares formula.
5 28 2 5 5 23
Simplify.
Problem
How can you write the expression with a rationalized denominator?
4 !2
1 1 !3
4 !2
1 1 !3
5
4 !2
1 2 !3
?
1 1 !3 1 2 !3
Use the conjugate of 1 1 !3 to rationalize the denominator.
5
4 !2 2 4 !6
123
Multiply.
5
A4 !2 2 4 !6B
4 !2 2 4 !6
5
2
22
2
Simplify.
5
24 !2 1 4 !6
5 22 !2 1 2 !6
2
Exercises
Simplify. Rationalize all denominators.
7. A3 1 !6B A3 2 !6B 3
10.
2 2 !7
2 1 !7
211 1 4"7
3
8.
2 !3 1 1 "3 1 1
2
5 2 !3
11. A2 !8 2 6B A !8 2 4B
40 2 28 !2
9. Q 4 !6 2 1 R Q !6 1 4 R
12.
!5
2 1 !3
2 !5 2 !15
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20 1 15 !6
Name
Class
Date
Reteaching
6-4
Rational Exponents
You can simplify a number with a rational exponent by converting the expression
to a radical expression:
1
1
2
92 5 !9
53
n
xn 5 !x, for n . 0
1
3
83 5 !
852
You can simplify the product of numbers with rational exponents m and n by
raising the number to the sum of the exponents using the rule
am ? an 5 am1n
Problem
What is the simplified form of each expression?
1
1
1
1
a. 364 ? 364
1
1
364 ? 364 5 364 1 4
Use am ? an 5 am1n .
1
5 362
Add.
2
5 !36
Use xn 5 !x.
56
Simplify.
1
n
3
2
b. Write Q 6x3 R Q 2x4 R in simplified form.
3
2
Q 6x3 R Q 2x4 R
3
2
5 6 ? 2 ? x3 ? x4
2
Commutative and Associative
Properties of Multiplication
3
5 6 ? 2 ? x3 1 4
Use xm ? xn 5 xm1n.
17
5 12x12
Simplify.
Exercises
Simplify each expression. Assume that all variables are positive.
1
2
1
2 1
1
7
2. Q 2y4 R Q 3y3 R 6y 12
1. 53 ? 53 5
13
4. 2y3 y5 2y 15
1
1
1
1
5. 54 ? 54 !5
1
2
6. Q 23x 6 R Q 7x 6 R 221 !x
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1
3. (211)3 ? (211)3 ? (211)3 211
Name
Class
Date
Reteaching (continued)
6-4
Rational Exponents
To write an expression with rational exponents in simplest form, simplify all
exponents and write every exponent as a positive number using the following
rules for a 2 0 and rational numbers m and n:
1
a2n 5 n
a
1
a2m
5 am
(am)n 5 amn
(ab)m 5 ambm
Problem
2
What is A8x9y23 B 23 in simplest form?
2
(8x9y23)23
2
5 A23 x9 y23 B 23
2
2
Factor any numerical coefficients.
2
5 A23 B 23 Ax9 B 23 Ay23 B 23
Use the property (ab)m 5 ambm.
5 222x26y2
Multiply exponents, using the property (am)n 5 amn.
y2
5 2 6
2 x
y2
5 6
4x
Write every exponent as a positive number.
Simplify.
Exercises
Write each expression in simplest form. Assume that all variables
are positive.
1
1
7. A16x2 y8 B 22 4xy4
1
10. A25x26 y2 B 2 5y
x3
1
8. Az23 B 9
1
1 4
9. Q 2x4 R
1
z3
16z4
12. a
b
25x8
2
11. A8a23 b9 B 3 4b6
a2
1
x2 5 2 1
13. a 21 b x 5y 5
y
3
16. A9z10 B 2 27z15
16x
1
22
1
2
14. A27m9 n23 B 23
1
17. (2243)25 21
3
n2
9m6
1
32r2 4
15. a 4 b 2r 2
2s
s
2
x5 10
18. a 1 b
y2
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5x4
4z2
x4
y5
Name
Class
Date
Reteaching
6-5
Solving Square Root and Other Radical Equations
Equations containing radicals can be solved by isolating the radical on one side of
the equation, and then raising both sides to the same power that would undo the
radical.
Problem
What is the solution of the radical equation? 2!2x 1 2 2 2 5 10
2 !2x 1 2 2 2 5 10
2 !2x 1 2 5 12
Add 2 to each side.
!2x 1 2 5 6
Divide each side by 2.
(!2x 1 2)2 5 62
Square each side to undo the radical.
2x 1 2 5 36
Simplify.
2x 5 34
Subtract 2 from each side.
x 5 17
Divide each side by 2.
Check the solution in the original equation.
Check
2 !2x 1 2 2 2 5 10
2 !2(17) 1 2 2 2 0 10
2 !36 2 2 0 10
Write the original equation.
Replace x by 17.
Simplify.
12 2 2 0 10
10 5 10
The solution is 17.
Exercises
Solve. Check your solutions.
1
2. 3 !2x 5 12 8
1. x 2 5 13 169
1
4. (3x 1 4)2 2 1 5 4 7
1
7. (x 1 2) 2 2 5 5 0 23
1
3. !3x 1 5 5 11 12
5. (6 2 x) 2 1 2 5 5 23
6. !3x 1 13 5 4 1
8. !3 2 2x 2 2 5 3 211
3
9. !5x 1 2 2 3 5 0 5
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Name
Class
6-5
Date
Reteaching (continued)
Solving Square Root and Other Radical Equations
An extraneous solution may satisfy equations in your work, but it does not
make the original equation true. Always check possible solutions in the original
equation.
Problem
What is the solution? Check your results. !17 2 x 2 3 5 x
!17 2 x 2 3 5 x
!17 2 x 5 x 1 3
Add 3 to each side to get the radical alone on one side
of the equal sign.
A !17 2 xB 2 5 (x 1 3)2
Square each side.
17 2 x 5 x2 1 6x 1 9
0 5 x2 1 7x 2 8
Rewrite in standard form.
0 5 (x 2 1)(x 1 8)
Factor.
x 2 1 5 0 or x 1 8 5 0
x 5 1 or
Set each factor equal to 0 using the Zero Product Property.
x 5 28
Check
!17 2 x 2 3 5 x
!17 2 1 2 3 0 1
!16 2 3 0 1
151
!17 2 x 2 3 5 x
!17 2 (28) 2 3 0 28
!25 2 3 0 28
2 2 28
The only solution is 1.
Exercises
Solve. Check for extraneous solutions.
10. !5x 1 1 5 !4x 1 3 2
11. !x2 1 3 5 x 1 1 no
12. !3x 5 !x 1 6 3
solution
13. x 5 !x 1 7 1 5 9
14. x 2 3 !x 2 4 5 0 16
15. !x 1 2 5 x 2 4 7
16. !2x 2 10 5 x 2 5 5, 7
17. !3x 2 6 5 2 2 x 2
18. !x 2 1 1 7 5 x 10
19. !5x 1 1 5 !3x 1 15 7 20. !x 1 9 5 x 1 7 25
21. x 2 !x 1 2 5 40 47
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Name
6-6
Class
Date
Reteaching
Function Operations
When you combine functions using addition, subtraction, multiplication, or division, the
domain of the resulting function has to include the domains of both of the original functions.
Problem
Let f (x) 5 x2 2 4 and g(x) 5 !x. What is the solution of each function
operation? What is the domain of the result?
a. ( f 1 g)(x) 5 f (x) 1 g(x) 5 (x2 2 4) 1 (!x) 5 x2 1 !x 2 4
b. ( f 2 g)(x) 5 f (x) 2 g(x) 5 (x2 2 4) 2 (!x) 5 x2 2 !x 2 4
c. (g 2 f )(x) 5 g(x) 2 f (x) 5 (!x) 2 (x2 2 4) 5 2x2 1 !x 1 4
d. ( f ? g)(x) 5 f (x) ? g(x) 5 (x2 2 4)(!x) 5 x2 !x 2 4 !x
The domain of f is all real numbers. The domain of g is all x $ 0. For parts a2d, there are
no additional restrictions on the values for x, so the domain for each of these is x $ 0.
f
f (x)
(x2 2 4) !x
x2 2 4
e. g (x) 5
5
5
x
g(x)
!x
As before, the domain is x $ 0. But, because the denominator cannot be zero, eliminate
any values of x for which g(x) 5 0. The only value for which !x 5 0 is x 5 0. Therefore,
f
the domain of g is x . 0.
f.
g
g(x)
!x
(x) 5
5 2
f
f(x)
x 24
Similarly, begin with x $ 0 and eliminate any values of x that make the denominator
g
f(x) zero: x2 2 4 5 0 when x 5 22 and x 5 2. Therefore, the domain of f is x $ 0
combined with x 2 22 and x 2 2. In other words, the domain is x $ 0 and x 2 2, or all
nonnegative numbers except 2.
Exercises
Let f (x) 5 4x 2 3 and g(x) 5 x2 1 2. Perform each function operation and then
find the domain of the result.
1. ( f 1 g)(x)
2. ( f 2 g)(x)
x2
1 4x 2 1;
all real numbers
4. ( f · g)(x)
4x3 2 3x2 1 8x 2 6;
all real numbers
2x2
1 4x 2 5; all
real numbers
f
5. g (x)
4x 2 3
;
x2 1 2
all real numbers
3. (g 2 f )(x)
x2 2 4x 1 5; all
real numbers
g
6. (x)
f
x2 1 2
4x 2 3 ;
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x u 34
Name
Class
6-6
Date
Reteaching (continued)
Function Operations
• One way to combine two functions is by forming a composite.
• A composite is written (g + f ) or g( f (x)). The two different functions are g and f.
• Evaluate the inner function f (x) first.
• Use this value, the first output, as the input for the second function, g(x).
Problem
What is the value of the expression g( f (2)) given the inner function, f (x) 5 3x 2 5
and the outer function, g(x) 5 x2 1 2?
1st input
x ⫽ 2
f(x)
1st output, 1,
becomes
2nd input
3x ⫺ 5
3(2) ⫺ 5
g(x)
x2
2nd output
⫹ 2
3
12 ⫹ 2
6 ⫺ 5
1 ⫹ 2
1
1st output
3
2nd output
Exercises
Evaluate the expression g( f (5)) using the same functions for g and f as in the
Example. Fill in blanks 7–14 on the chart.
Use one color highlighter to highlight the first input. Use a second color to
highlight the first output and the second input. Use a third color to highlight the
second output, which is the answer.
1st input
x ⫽ 5
f(x)
1st output,
3x ⫺ 5
10.
5) ⫺ 5
7. 3(—
10 ,
———
becomes
2nd input
g(x)
x2
2nd output
⫹2
14.
102
———
2
11. 10
— ⫹2
8. 15
— ⫺ 5
12. 100
—⫹2
10
9. ———
102
13. ———
Given f (x) 5 x2 1 4x and g(x) 5 2x 1 3, evaluate each expression.
15. f (g(2))
77
16. g( f (2.5))
35.5
17. g( f (25))
13
18. f (g(25))
21
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Name
Class
Date
Reteaching
6-7
Inverse Relations and Functions
•
Inverse operations “undo” each other. Addition and subtraction are
inverse operations. So are multiplication and division. The inverse of
cubing a number is taking its cube root.
•
If two functions are inverses, they consist of inverse operations performed
in the opposite order.
Problem
What is the inverse of the relation described by f (x) 5 x 1 1?
f (x) 5 x 1 1
y5x11
Rewrite the equation using y, if necessary.
x5y11
Interchange x and y.
x215y
Solve for y.
y5x21
The resulting function is the inverse of the original function.
So, f 21 (x) 5 x 2 1.
Exercises
Find the inverse of each function.
2. y 5 3x3 1 2
1. y 5 4x 2 5
f 21 5
x 1 5
4
f 21 5
4. y 5 0.5x 1 2
f 21
5 2x 2 4
x
7. f (x) 5 5
f 21(x) 5 5x
3. y 5 (x 1 1)3
3
2
"x 2
3
3
f 21 5 "x 2 1
5. f (x) 5 x 1 3
f 21(x) 5 x 2 3
6. f (x) 5 2(x 2 2)
8. f (x) 5 4x 1 2
9. y 5 x
f 21(x)
4
f 21(x) 5 x 1
2
x22
4
5
x21
10. y 5 x 2 3
11. y 5 2
f 21 5 x 1 3
f 21 5 2x 1 1
2
13. f (x) 5 !x 1 2
14. f (x) 5 3 x 2 1
f 21(x) 5 32(x 1 1)
f21 (x) 5 x2 2 2 for x L 22
16. f (x) 5 2(x 2 5)2
f 21(x)
556
"x2
17. y 5 !x 1 4
f 21
5 (x 2
4)2
f 21 5 x
12. y 5 x3 2 8
3
f 21 5 "x 1 8
x13
5
f 21(x) 5 5x 2 3
15. f (x) 5
18. y 5 8x 1 1
for x L 0
1
f 21 5 x 2
8
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Name
6-7
Class
Date
Reteaching (continued)
Inverse Relations and Functions
Examine the graphs of f (x) 5 !x 2 2 and its inverse,
f 21(x) 5 x2 1 2, at the right.
6
5
4
3
2
1
Notice that the range of f and the domain of f 21 are the same:
the set of all real numbers x $ 0.
Similarly, the domain of f and the range of f 21 are the same:
the set of all real numbers x $ 2.
y
f ⫺1
f
x
O
1 2 3 4 5 6
This inverse relationship is true for all relations whenever both f and
f 21 are defined.
Problem
What are the domain and range of the inverse of the function f (x) 5 !3 2 x?
f is defined for 3 2 x $ 0 or x # 3.
Therefore, the domain of f and the range of f 21 is the set of all x # 3.
The range of f is the set of all x $ 0. So, the domain of f 21 is the set
of all x $ 0.
f
6
5
4
3
2
1
23 22 21 O
Exercises
y
x
1 2 3
Name the domain and range of the inverse of the function.
1
19. y 5 2x 2 1
20. y 5 2 2 x
The domain and the range is
domain: x u 2;
the set of all real numbers.
range: y u 0
21. y 5 !x 1 5
domain: x L 0;
range: y L 25
22. y 5 !2x 1 8
domain: x L 8;
range: y K 0
23. y 5 3 !x 1 2
domain: x L 2;
range: y L 0
24. y 5 (x 2 6)2
domain: x L 0;
range: all real numbers
25. y 5 x2 2 6
1
26. y 5 x 1 4
27. y 5
domain: x L 26;
range: all real numbers
domain: x u 0;
range: y u 24
1
(x 1 4)2
domain: x S 0;
range: y u 24
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Name
Class
6-8
Date
Reteaching
Graphing Radical Functions
The graph of y 5 a!x 2 h 1 k is a translation h units horizontally and k units
vertically of y 5 a!x. The value of a determines a vertical stretch or compression
of y 5 !x.
Problem
6
What is the graph of y 5 2 !x 2 5 1 3?
y
4
y 5 2 !x 2 5 1 3
2
x
a52
O
h55 k53
2
4
6
Translate the graph of y 5 2 !x right five units and up three units. The graph of
y 5 2 !x looks like the graph of y 5 !x with a vertical stretch by a factor of 2.
Exercises
Graph each function.
1. y 5 !x 2 4 1 1 4 y
2. y 5 !x 2 4
2
O
2
3. y 5 !x 1 1
5. y 5 2 !x 2 1
7. y 5 2!x 1 1
6. y 5 22 !x 1 3 1 4
y
2
2
2 y
8. y 5 !x 1 3 2 4
x
y
⫺4 ⫺2 O
⫺2
4
⫺4
⫺4
6 y
10. y 5 2!x 2 2
4
2
O
x
⫺4 ⫺2 O
O
⫺2
9. y 5 3 !x 1 2
6 y
4
x
⫺2 O
⫺2
2
4
x
⫺4 ⫺2 O
⫺2
2
2
y
4. y 5 2!x 1 2 2 3
x
⫺2 O
⫺2
2
⫺4
y
2
x
⫺2 O
⫺2
x
6
4
y
x
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x
Name
6-8
Class
Date
Reteaching (continued)
Graphing Radical Functions
Graphs can be used to find solutions of equations containing radical expressions.
Problem
What is the minimum braking distance of a bicycle with a speed of 22 mph?
You can find the minimum braking distance d, in feet, of a bicycle travelling s
miles per hour using the equation s 5 5.5 !d 1 0.002 .
We want to find the value of d when s 5 22. In other words, solve the equation
5.5 !d 1 0.002 5 22. Graph Y1=5.5√(X+0.002) and Y2=22. Try different values
until you find an appropriate window. Then use the intersect feature to find
the coordinates of the point of intersection.
Intersection
x ⫽ 15.998 Y ⫽ 22
The minimum braking distance will be about 16 ft.
Exercises
Solve the equation by graphing. Round the answer to the nearest hundredth, if
necessary. If there is no solution, explain why.
11. !3x 1 1 5 5 8
12. !4x 1 1 5 9 20
13. !2 2 5x 5 4 22.8
14. !3x 1 5 5 7 14.67
15. !7x 1 2 5 11 17
16. !2x 2 1 5 !1 2 2x 0.5
17. !x 2 2 5 !2 2 3x no solution; x 5 1
is extraneous
18. 7!x 2 3 5 2!2x 1 1 3.68
19. !2x 2 5 5 !4 2 x 3
20. !2x 1 7 5 3!5x 1 2 20.26
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