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MATH 101: College Algebra – Practice Exam I – Detailed Key; Chapters P.6 – 2
- Divide – leave in a + bi format : i
i
( )
( )
2
2
i
i
i
i i i
i i i
i
i
i
i
i 26
- Determine the value of the following – only use i , -1, - i , or 1 : i
484
484 ÷ 4 = 121 remainder 0 → i
484 = 1
- Solve the following: 4 x − 3 + 2 = 7
x
x
x
x
x
x
x
x
x
x
x
x
- Solve by completing the square : 2 x
2
2
2
2
2
2
2
2
2
x
x
x
x
x
x x
x x
x x
x x
x x
- Solve by using the quadratic formula : 3 6 1
2 x − x =−
3
3 6
6
23 6
6
6 2 6
6
6 4 6
6
6 24
6
6 36 12
23
6 6 431
3 6 1 0
3 6 1 1 1
2
2
2
±
±
±
± ⋅
±
± −
− − ± − −
− + =
− + =− +
x
x x
x x
- Solve by factoring over the complex numbers: 3 27 24 0
4 2 x + x + =
Let u = x
2
u
2 = x
4
2
2
2
u u
u u
u u
u u
2
2
x i
x
x
x
u
u
x i
x
x
u
u
2
2
{− 2 i 2 ,− i , i , 2 i 2 }
- Give the equation of the circle in standard form with center point (-1, -4) and passing
through the point (-2, -2)
Distance from the center to the point is the radius so:
2 2
2 2
2 2
d = r = − − − + − −−
Therefore, the equation in standard form is:
2 2 2 x − − 1 + y − − 4 = 5
2 2 x + + y + =
- Find the equation of a line in slope-intercept form that passes through the following
points: (2, -2) and (4, -14)
( )
m =
( ) ( )
y x
y x
y x
y = − 6 x + 10
- Determine the domain of f ( x )= 5 x − 8
Since x is defined only when x ≥ 0 , 5 x − 8 ≥ 0
x
x
x
x
x
- Determine the domain of ( ) 1
2 g x = x −
Since x is defined only when x ≥ 0 , 1 0
2 x − ≥
2 x − ≥ x + 1 – – – – | + + + | + + +
( x + 1 ) ( x − 1 ) ≥ 0 x – 1 – – – – |– – – – | + + +
x
x
x
x ( x + 1)( x – 1) + + + |– – – – | + + +
x ≤ − 1 or x ≥ 1 → ( − ∞,− 1 ] U[ 1 ,+∞)
- Graph y = | x | + 2 using a table of values – show the work
x = -3: y = |-3| + 2 = 3 + 2 = 5
x = -2: y = |-2| + 2 = 2 + 2 = 4
x = -1: y =|-1| + 2 = 1 + 2 = 3
x = 0: y = |0| + 2 = 0 + 2 = 2
x = 1: y = |1| + 2 = 1 + 2 = 3
x = 2: y = |2| + 2 = 2 + 2 = 4
x = 3: y = |3| + 2 = 3 + 2 = 5
In 16) – 20), given that f ( x ) = x – 4 and g ( x ) = 2 x
2
- x – 1, find the following:
- ( f + g )( x )
( )( ) ( ) ( )
( 4 ) ( 2 1 )
2 = − + − −
x x x
f g x f x gx
2 = x −
- ( f − g )( x )
( )( ) ( ) ( )
2
2
x x x
f g x x x x
2 = − x + x −
- ( f ⋅ g )( x )
( )( ) ( ) ( )
( ) ( )
( ) ( )
3 2 2
2 2
2
x x x x x
x x x x x
x x x
f g x f x gx
3 2 = x − x + x +
- ( g o f )( x )
( )( ) ( ( ))
( ) ( )
( )
2
2
2
x x x
x x x
x x
g o f x g f x
2 = x − x +
- ( f o g )( 1 )
( )( ) ( ( ))
( )
2
2
x x
x x
f o g x f gx ( )( ) ( ) ( )
f o g = − −
Use ( ) 2 8 4
2 f x = − x − x − for 21) – 26):
- Find the vertex of f ( x ) – use either method
Completing the square:
( )
( )
2 ( 2 ) 4
2
2
2
2
x
x x
x x
f x x x
Vertex Formula:
( )
( )
h = ( ) ( )
( )
2
k = f − =− − − − −
Using either method, the vertex is (-2, 4)
- Which direction does the parabola f ( x ) open?
Since a = -2 < 0, f ( x ) opens down
- Use the information gathered in 21) – 24) along with the axis of symmetry and some
additional points to graph f ( x ) – you MUST show the axis of symmetry on the graph
Axis of symmetry: x = -
When x = -1: 2 ( 1 ) 8 ( 1 ) 4 2 8 4 2 ( 1 , 2 ) lieson theparabola
2 − − − − − =− + − = → −
Using the axis of symmetry → (-3, 2) and (-4, -4) lie on the parabola
- State the domain AND range of f ( x )
Domain: ( − ∞, +∞) Range: ( −∞, 4 ]
