Transcript Math 143 Final Review Spring 2007

Math 143
Final Review
Spring 2007
1.
Given line
Line in question
perpendicular to the given line
4x – 5y = 6
m = -5
-5y = -4x + 6
4
y = 4x– 6
5
m = 4
5
goes through (-2, 3)
5
Equation:
-5 =
4
y–3
x+2
4y – 12 = -5x – 10
4y = -5x + 2
y = -5 x + 1
4
5x + 4y = 2
2
Slope-intercept
form
Standard form
2.
Given f(x). Draw g(x) = -f(-x + 1) + 3
1. Hor shifts
2. Flips
3. Vert shifts
3
-3
3
-3
•Shift one unit left
•Flip over the y-axis
•Flip over the x-axis
•Shift three units up
f(x) = x2 – 2x
g(x) = 3x + 1
3. (f o g)(x) =
(3x + 1)2 – 2(3x + 1)
h(x) = x – 2
x+3
= 9x2 + 6x + 1 – 6x – 2
= 9x2 – 1
4.
g
2
x–1
=
3 x2– 1 + 1 =
=
6
+
x–1
=
x+5
x–1
x–1
x–1
6
x–1
+1
f(x) = x2 – 2x
5. g(x)2– 1 =
6.
f(x + h) – f(x)
h
g(x) = 3x + 1
2
(3x + 1) – 1
=
h(x) = x – 2
x+3
2
3x
=
[(x + h)2 – 2(x + h)] – [x2 – 2x]
h
=
x2 + 2xh + h2 – 2x – 2h – x2 + 2x
h
=
2xh + h2 – 2h
=
h
=
2x + h – 2
h(2x + h – 2)
h
f(x) = x2 – 2x
7.
h-1(x) = ?
Original function
h(x) =
x–2
x+3
y=
x–2
x+3
h(x) = x – 2
x+3
g(x) = 3x + 1
Inverse
x=
y–2
y+3
xy + 3x = y – 2
xy – y = -3x – 2
y(x – 1) = -3x – 2
y = -3x - 2
x–1
h-1(x)
-3x - 2
= x–1
f(x) = x2 – 2x
8.
g(x) = 3x + 1
(4) – 2
h(4) =
(4) + 3
=
2
7
h(x) = x – 2
x+3
9. Is f(x) a function?
f(x)
3
-3
Yes
3
-3
Why or why not?
It passes the
vertical line test
Does f(x) have an inverse function?
No
Why or why not?
It does not pass the horizontal line test
f(x)
10. a. What is the
domain of f(x) ?
3
(-4, 5]
-3
3
-3
b.
What is the
range of f(x) ?
[0, 3]
c.
What is f(5) ?
f(5) = 3
f(x)
11. a. For what value(s)
of x does f(x) = 2 ?
3
-3
When x = 3
or
x = -3
3
-3
b.
Identify any
x-intercepts
(0, 0)
c.
Identify any y-intercepts
(0, 0)
f(x)
12. a. For what values of x
is the graph of f(x)
increasing ?
3
-3
From x = - 4 to x = -3
From x = 0 to x = 5
3
-3
b.
For what values of x is the
graph of f(x) decreasing ?
From x = -3 to x = 0
c.
For what values of x is the graph of f(x) constant ?
None
f(x)
3
13.
-3
3
-3
-9
Is the degree of f(x)
even or odd? How do
you know?
The degree is odd
The endpoint
behavior is different
on the right and left.
14.
What can be said about the
leading coefficient of f(x)?
Why?
The leading coefficient is positive.
The right side of the graph rises.
f(x)
3
15.
-3
3
-3
-9
What is the minimum
degree of f(x)? How
do you know?
Minimum degree = 5
The graph makes
four turns.
16.
Counting multiplicities,
what is the minimum
number of real zeros of f(x)
The minimum number of real zeros is 5.
At least 2 zeros at x = -5, at least two zeros at x = -1,
and at least 1 zero at x = 3.
f(x)
3
17.
-3
3
-3
-9
State each zero of f(x)
and whether its
multiplicity is even or
odd.
Xeros at x =
-5
even multiplicity
-1
even multiplicity
3
odd multiplicity
18.
Write one possible
equation for f(x)
f(x) = (x + 5)2(x+ 1)2(x – 3)
2x2 + 4 , x > 3
19.
f(x) =
7 – x , x 3
a. f(4)
b. f(3)
c. f(- 9)
=
2(4)2 + 4
= 7 – 3
= 7 + 9
= 36
= 2
= 4
Evaluate each of
the following
20. List all the possible rational zeros of:
f(x) = 3x4 – 13x3 + 22x2 – 18x + 4
factors of p: ±4, ± 2, ± 1
factors of q: ±3, ± 1
possible rational zeros: ±4, ± 2, ± 4/3, ±1, ± 2/3, ± 1/3
21.
Find all the zeros of:
f(x) = 3x4 – 13x3 + 22x2 – 18x + 4
By seeing the graph of the function or by plugging the values from
problem 20 into the function you can determine that there are zeros at 2
and at 1/3
1/3
2
3
3
3
-13
22
-18
4
1 -4
-12 18
6 -1
-6
6
6
-12
12
0
-4
0
The other two zeros
must be the solutions of
3x2 – 6x + 6 = 0
x2 – 2x + 2 = 0
x=
=
Zeros: 1/3, 2, 1+ i, 1 - i
2 ± 4 – 4(1)(2)
2
2±-4
=
2
1±i
22.Graph the system: x2 + y2  9 and
y>x+1
Border: x2 + y2 = 9
Test (0, 0)
0+09
True
Border: y = x + 1
Test (0, 0)
0>0+1
False
23. Match each equation to its type
E
A
D
H
C
(x – 3)2 + (y + 1)2 = 16
y = 2x – 3
2 + 3x + 2
x
y =
x2 – 9
A. Linear
B. Quadratic
C. Higher order
polynomial
D. Rational
E. Circle
(x – 3)2 – 2(y + 1)2 = 9
y = 4x4 – 3x2 + 5
F. Ellipse
H. Hyperbola
J. Parabola
23. Complete the square and draw the graph
a.
9x2 + 16y2 – 18x + 64y – 71 = 0
9(x2 – 2x + __)
1 + 16(y2 + 4y + __)
4 = 71 + __
9 + __
64
9(x – 1)2 + 16(y + 2)2 = 144
(x – 1)2 + (y + 2)2 = 1
16
9
23. Complete the square and draw the graph
b.
16x2 –y2 + 64x – 2y + 67 = 0
16(x2 + 4x + __)
4 – 1 (y2 + 2y + __)
1 = -67 + __
64 + __
-1
16(x + 2)2 – 1(y + 1)2 = -4
(y + 1)2 – (x + 2)2 = 1
4
9
25. Evaluate each of the following.
0! =
1
1! =
1
5! =
120
26. Evaluate:
8
3
=
8!
3! 5!
=
56
=
(8)(7)(6) 5!
3! 5!
On the calculator:
8
3
=
8 nCr 3 =
56
27. Find the third term of (x – 3)9
3rd
term =
9 (x)7 (-3)2
2
= 36(x7)(9)
= 324x7
1.5
Solve 2x + 3 – x – 2 = 2
2x + 3
2
= 2 – x – 2
2
2x + 3 = 4 – 4x – 2 + (x – 2)
x+1
x2 + 2x + 1
2
=
-4x – 2
2
= 16(x – 2)
x2 – 14x + 33 = 0
(x – 3)(x – 11) = 0
x = 3 or x = 11
Both answers
work in the
original problem
1.6
Solve:
4
x – 3 x – 10 = 0
x1/2 – 3x1/4 – 10 = 0
a2
The form of this equation
looks like a quadratic
equation.
Let: a = x1/4
– 3a – 10 = 0
(a – 5)(a + 2) = 0
a = 5 or a = -2
so
x1/4 = 5
(x1/4)4 = (5)4
x = 625
x = 625
or
x1/4 = -2
(x1/4)4 = (-2)4
x = 16
16 does not work in the
original equation, but
625 does work.
1.8a.
Solve:
|5x – 2| > 13
Find the key values of x by solving |5x – 2| = 13
5x – 2 = 13
or
5x = 15
5x – 2 = -13
5x = -11
x=3
x = -11/5
Now test numbers from the intervals created by these key values.
Use the original problem to test.
T
F
T
3
-11/5
Solution: x < -11/5 or
x>3
1.8b.
Solve:
4x2 + 7x < -3
Find the key values of x by solving 4x2 + 7x = -3
4x2 + 7x + 3 = 0
(4x + 3)(x + 1) = 0
x = -3/4 or x = -1
Now test numbers from the intervals created by these key values.
Use the original problem to test.
F
T
-1
Solution: -1 < x < -3/4
F
-3/4
2.7 a.
f(x) = (1/2)x3 – 4
Original function
f(x) =
(1/2)x3
y=
(1/2)x3
Inverse
-4
x = (1/2)y3 – 4
-4
x + 4 = (1/2)y3
y3 = 2x + 8
3
y = 2x + 8
3
f-1(x) = 2x + 8
3.4. f(x) = 6x3 + 25x2 – 24x + 5
factors of p: ±5, ± 1
factors of q: ±6, ± 3, ±2, ± 1
possible rational zeros: ±5, ± 5/2, ± 5/3, ±1, ± 5/6, ± 1/2, ± 1/3, ± 1/6
one of the zeros is -5
-5
6
25
-30
6 -5
-24 5
25 - 5
1
0
The other two zeros are
the solutions to
6x2 – 5x + 1 = 0
(3x – 1)(2x – 1) = 0
x = 1/3
zeros:
- 5, 1/3, 1/2
x = 1/2
3.5
Find all the roots of x4 – 4x3 + 16x2 – 24x + 20 = 0
given that 1 – 3i is a root.
• Since 1 – 3i is a root, 1 + 3i is also a root
so (x – 1 + 3i) and (x – 1 – 3i) are factors of the equation
(x – 1 – 3i)(x – 1 + 3i) =
x2 – 2x + 10
–2
See problem 21
x4 – 4x3 + 16x2 – 24x + 20
The solutions for
x2 – 2x + 2 = 0
are
x= 1±i
x2 – 2x
x2 – 2x + 10
- x4 +– 2x3 +- 10x2
– 2x3 + 6x2 – 24x + 20
+ – 2x3 +- 4x2 –+ 20x
2x2 – 4x + 20
2x2 – 4x + 20
Roots: 1 – 3i, 1 + 3i, 1 – i, 1 + i
3.6
Graph the function: f(x) =
Vertical asymptote: x = -2
Horizontal asymptote: y = -3
x-intercept: (0,0)
y-intercept: (0,0)
x
y
-4
-3
-1
1
-6
-9
3
-1
-3x
x+2
4.2 f(x) = logb x is shown on the graph
a. When x = 5,
f-1(x)
logb 5 = 1
so b = 5
5
-5
5
c. For f-1(x)
Domain: all real numbers
Range: y > 0
-5
4.3
4.4 a.
log3 140.3 =
Solve:
log 140.3
log 3
22x – 1 + 3 = 35
22x – 1 = 32
ln 22x – 1 = ln 32
(2x – 1) ln 2 = ln 32
2x – 1 = 5
2x = 6
x = 3
 4.5
4.4 b.
Solve: log3 (x – 5) + log3 (x + 3) = 2
log3 (x2 – 2x – 15) = 2
x2 – 2x – 15 = 32
x2 – 2x – 24 = 0
(x – 6)(x + 4) = 0
x = 6 or x = -4
x = -4 does not work in the
original equation.
Solution: x = 6
4.5
R = 6e12.77x
If R = 25% accident risk
25 = 6e12.77x
25/6 = e12.77x
ln (25/6) = ln e12.77x
ln (25/6) = 12.77x
x = 0.112
A blood alcohol level of 0.112
corresponds to an accident
risk of 25%
7.1 Write the standard form of the equation of an ellipse given
foci: (0,-3) and (0,3)
vertices: (0,-4) and (0,4)
Center: (0, 0)
V
The equation must be in
the form:
F
x2
y2
+ 2 = 1
a2
b
b = distance from the center to a vertex = 4
c = distance from the center to a focus point
c=3
c 2 = b 2 – a2
9 = 16 – a2
a2 = 7
Equation:
F
V
x2
y2
+
= 1
7
16
7.3 Find the equation of a parabola with vertex at (2, -3) and
focus at (2, -5)
The parabola is vertical with an
equation in the form:
(x – h)2 = 4p(y – k)
(h, k ) = (2, -3)
p = -2
Equation:
V
F
(x – 2)2 = 8(y + 3)