Algebra Homework Solutions: Detailed Answers and Explanations

This document presents a complete set of solutions to an algebra homework assignment. The solutions cover a range of algebraic concepts, including solving equations, finding intercepts, determining domains, and working with functions. The assignment includes questions on linear and quadratic equations, calculating costs and profits, and analyzing graphical representations of functions. Detailed step-by-step solutions are provided for each question, offering clear explanations and methodologies. The solutions cover a wide range of topics, providing a comprehensive review of core algebra principles. The document is a valuable resource for students seeking to understand and master algebra concepts, offering a clear pathway to solving complex problems and improving their overall mathematical abilities.

Question 1
B and C
Question 2
a) (-3, 2)
b) The ˚Equation is givenby ( x−h ) 2 + ( y−k ) 2=r2
Where ( h , k ) is thecenter of the circe , r isthe radius of the ˚¿ ( x , y ) is a point onthe ˚.
Therefore ,
( −1+3 ) 2 + ( 10−2 ) 2=r2
r2=4+ 64=68
r = √68= √4∗17
¿ 2 √17
Question 3
a) The x-intercepts are -1,1, and 5
b) The y-intercepts is -3
c) [-1,6]
d) [-3,2]
Question 4
a) f ( −2 ) = −2+5
( −2−8 ) 2 = 3
100
b) (−∞, 8 ) U ( 8 , ∞ )
c) f ( a+7 ) = ( a+ 7 ) +5
( ( a+7 )−8 )2 = a+ 12
( a−1 )2 = a+12
a2−2a+ 1
Question 5
The domain of the quotient function is ( −∞,−7 ) ∪ (−7 , ∞ ). The domain of the quotient function is only
influenced by the nature of the denominator function. Given that we take absolute values of the
denominator, we can have both negative values as long as the function is defined. However, for negative
infinity, infinity and -7, the quotient function is undefined. Thus we exclude them from the domain.
Question 6
a) $6300 lies in the second bracket. Therefore,
T ( 6300 )=70+0.035 ( 6300−2500 )=$ 203
b) $63000 lies in the third bracket, so
T ( 63000 )=245+0.05 ( x−7500 )
¿ 245+0.05∗55500
¿ $ 3020
Question 7

a) At x−intercepts , y=0 ,
Therefore 2−2 x2=0
2 x2 =2 x2=1 ,
Thus x=1∨−1
Therefore x−intercepts are1∧−1
b) At y intercept , x=0.
Therefore y=2−2∗0=2
Therefore y−intercept is 2.
c)
x y (x, y)
-3 -16 (-3,-16)
-2 -6 (-2,-6)
-1 0 (-1,0)
0 2 (0,2)
1 0 (-1,0)
2 -6 (2,-6)

d) Yes
e) No. Taking (0,2), the appropriate reflecting across x-axis gives (0,-2). But we do not have (0,-2)
on the graph.
f) By definition , the graph is symmetirc about the origin if f ( x )=−f ( x ) .
¿ the equation∧graph , the graph is NOT symmetric about the origin .
For example, point (0,2), when reflected across the origin, it will become (0,-2). But (0,-
2) is not on the graph.
Question 8
a) g−f =1−2 x−4 x2−2 x +8=−4 x−4 x2 +9=¿
at x =−3 ,
g−f =−15
b) f∗g= ( 4 x2 +2 x−8 ) ( 1−2 x )
¿ 22∗7
¿ 154
c) f −g ( x )= ( 4 x2+2 x−8 )−(1−2 x)
¿ 4 x2 +2 x+2 x−8−1
¿ 4 x2 −4 x−9
Question 9
a) C(300) = 5250 + 7.00 x = 5250 + 7.00 ( 300) = 7350 dollars
Thus $7350 is the total cost for producing 300 widgets by the company.
b) The average Cost Function=C ( x )
x ,Thus when x=300 ,
C ( x )
x = 7350
300 =24.5 dollars .
Thus $ 24.5is the cost of producing a single widget .
c) R ( x )=xP ( x )
¿ x ( 45 x−0.02 x )
¿ 45 x−0.02 x2
d) P ( x )=R ( x ) −C ( x )
¿ 45 x−0.02 x2−5250−7 x
¿−0.02 x2 +38 x−5250

e) P ( x ) =−0.02 x2 +38 x−5250
P ( 300 ) =−0.02 ( 300 ) 2 +38 ( 300 ) −5250=4350 dollars .
Thus the profit of the company ¿ 300 widgets is $ 4350