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5 Questions on Intermediate Algebra with Solution - Assignment | MTH 95, Assignments of Algebra

Material Type: Assignment; Class: Intermediate Algebra; Subject: Math; University: Portland Community College; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

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Mr. Simonds MTH 95 โ€“ Week 1 Extension Solutions
Page 1 of 2
1. The line passes through the points
(
)
8,2 and
(
)
2,7โˆ’, so its slope is 72 1
28 2
mโˆ’
=
=โˆ’
โˆ’โˆ’ .
We now know half of the formula:
()
1
2
gx x b
=
โˆ’+
.
Using the fact that
()
82g= gives us
()
182 6
2bb
โˆ’
+= โ‡’ =.
So the function is
()
16
2
gx x=โˆ’ + .
Please note that the other point checks:
() ()
1
226
2
7
g
โˆ’
=โˆ’ โˆ’ +
=
2. a.
()()
44 412
41612
44
fx x
x
x
+= +โˆ’
=+โˆ’
=+
b.
(
)
44 124
48
fx x
x
+
=โˆ’+
=โˆ’
c.
()()( )
(
)
444412
44412
16
fx fx x x
xx
+โˆ’ = +โˆ’ โˆ’
=+โˆ’+
=
d. 4120
412
3
t
t
t
โˆ’=
=
=
The only zero of
f
is 3.
e.
() ()
10 4 10 12
28
f=โˆ’
=
f. 41248
460
15
t
t
t
โˆ’=
=
=
The function value equals 48 at 15.
g.
The t-intercept is
()
3,0 (See part d.)
Since
()
012f=โˆ’ , the y-intercept is
(
)
0,12
h.
()
40
412 40
428
7
ft
t
t
t
>โˆ’
โˆ’>โˆ’
>โˆ’
>โˆ’
()
40ft>โˆ’ at every value of t on
(
)
7,
โˆž.
pf2

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Mr. Simonds MTH 95 โ€“ Week 1 Extension Solutions

Page 1 of 2

1. The line passes through the points ( 8, 2) and ( โˆ’2, 7 ), so its slope is 7 2 1

m = โˆ’ = โˆ’ โˆ’ โˆ’ .

We now know half of the formula: ( ) 1

g x = โˆ’ x + b.

Using the fact that g ( ) 8 = 2 gives us 1 ( ) 8 2 6

โˆ’ + b = โ‡’ b =.

So the function is ( ) 1 6

g x = โˆ’ x +.

Please note that the other point checks: ( ) ( )

g โˆ’ = โˆ’ โˆ’ +

2. a. ( 4 ) 4 ( 4 ) 12

f x x x x

b. ( ) 4 4 12 4

f x x x

c. ( 4 ) ( ) ( 4 4 ) ( 4 12 )

f x f x x x x x

d. 4 12 0 4 12 3

t t t

The only zero of f is 3.

e. ( 10 ) 4 10( ) 12

f = โˆ’

f. 4 12 48 4 60 15

t t t

The function value equals 48 at 15.

g. The t- intercept is ( 3, 0 ) (See part d.)

Since f ( 0 )= โˆ’ 12 , the y- intercept is ( 0,12)

h. ( ) 40

f t t t t

f ( ) t > โˆ’ 40 at every value of t on ( โˆ’7, โˆž ).

Mr. Simonds MTH 95 โ€“ Week 1 Extension Solutions

Page 2 of 2

  1. We could get the answer to the first question from a graph but we need a formula to answer the second question, so we might as well go directly to the formula route.

The line passes through the points ( 3, โˆ’ 2 )and ( 7, โˆ’ 14 ), so its slope is (^ )

m

.

We now know half of the formula: f ( x ) = โˆ’ 3 x + b.

Using the fact that f ( ) 3 = โˆ’ 2 gives us โˆ’3 3 ( ) + b = โˆ’ 2 โ‡’ b = 7.

So the function is f ( x ) = โˆ’ 3 x + 7.

Please note that the other point checks:

( 7 ) 3 7( )^7

f = โˆ’ + = โˆ’

a. ( 9 ) 3 9( ) 7

f = โˆ’ + = โˆ’ b. 3 7 93 3 86 86 3

x x x

The function value equals 93 at 86 3

4. The domain of f is [10, 70 ), the range is [ 0, 40] , and the only zero is 10.

The domain of g is [ 0,3 ) โˆช ( 3,โˆž ), the range is ( โˆ’โˆž, 3 ), and the zeros are 2 and 6.

5. g is the constant function g ( x ) = 12.

a. Since the value of g is always 12 , the value of g ainโ€™t never 0 ; that is, g has no zeros. b. We can put any old number we want inside those parentheses, so the domain of g is

( โˆ’โˆž โˆž,^ ).^ Since the value of^ g^ is always^^12 , the range of^ g^ is^ {^12 }^.

The function g