Exam
• May 15th 6:15pm. Be there early
• Exam rooms: On website
– http://www.math.ksu.edu/math100/
• Exam rooms are by recitation instructor (not
me)
• Bring your k-state student ID
How to study
• Focus on your previous exams.
– Review the problems and your own work
• Study Guide
– http://www.math.ksu.edu/math100/spring2009/FinalExamStudyGuideSpring09.pdf
• Take old finals (in realistic conditions)
– http://www.math.ksu.edu/course_info/oldtests/1
00tests/
Your test
• 75% questions from your previous tests this
semester with the numbers changed
• 25% new material
– Inequalities
– Composition
– Exponents & logs
– Systems
• STUDY YOUR OLD TESTS
Agenda
• Today
– Functions
– Linear problems (equalities, inequalities, systems)
– Polynomials (including quadratics)
• Thursday
– Radicals
– Rationals
– Exponents and logs
Functions
A function is a relationship between
two changing variables
• An “input” variable
• An “output” variable
– The result of “doing” the function to the output
variable
• Both variables change so that the “input”
variable always tells you exactly what the
“output” variable is.
– You never get two outputs for the same input.
Not a Function
output
input
Intercepts
y
x-intercept (-2,0)
x-intercept (2,0)
x
y-intercept (0,-4)
Combining Functions
•
•
•
•
•
(f+g)(x)=f(x)+g(x)
(f-g)(x)=f(x)-g(x)
(fg)(x)=f(x)*g(x)
(f/g)(x)=f(x)/g(x), g(x)≠0
(f∘g)(x)=f(g(x))
In picture form
f(x)
f
x
*
g
f(x)g(x)
g(x)
Is not the same as
g
x
f
g(x)
f(g(x))
COMPARISON
f (x) = 2x -1, g(x) = (x - 3)2
(f∘g)(3)=f(g(3))
(fg)(3)=f(3)g(3)
g(3) = (3- 3)2
g(3) = 0
ƒ(g(3)) = ƒ(0) = 2 * 0 -1 = -1
ƒ(g(3)) = -1
-1≠0
f (3) = 5
g(3) = 0
f (3)g(3) = ( 5) ( 0) = 0
( fg)(3) = 0
Graphing Transformations
• The graph for ƒ(x)+c is the graph of ƒ(x) shifted up by c.
• The graph for ƒ(x-a) is the graph of ƒ(x) shifted right by a.
– NOTE THE MINUS SIGN
• The graph for rƒ(x) is the graph of ƒ(x) stretched vertically by r.
– negative r causes the graph to flip vertically.
• The graph for ƒ(sx) is the graph of ƒ(x) squished horizontally by s.
– Negative s causes the graph to flip horizontally
• Note the difference!
Even and Odd
• A function ƒ is EVEN if ƒ(-x)=ƒ(x). Example: x2
• A function ƒ is ODD if ƒ(-x)=-ƒ(x). Example: x3
• A function ƒ is NEITHER if ƒ(-x)=something else. Example: x3+1
Function inverse
Cubing to cube root
y=x3
x=∛y
Cubing to cube root
The relationship between x and y stays the same
Only my point of view changes
y=x3
x=∛y
How to find a function inverse
•
•
•
•
•
ƒ(x)=………….x………….
Rewrite as y=……………x…………
Solve for x. x=~~~~y~~~~~~
Rewrite as an inverse ƒ-1(y)=~~~~y~~~~~~
OPTIONAL: change ys to xs.
• ƒ-1(x)=~~~~x~~~~~~
• WARNING: Always check that your inverse is
actually a function.
Given f (x) = 7x + 1 on the domain of all real
numbers, find f -1(x). Be sure to write your
answer as a function of x.
a)
b)
c)
d)
e)
f -1(x) = (1/7)x − 1/7
f -1(x) = x − 1/7
f -1(x) = 1/(7x+1)
Both (a) and (c)
None of the above
Given f (x) = 7x + 1 on the domain of all real
numbers, find f -1(x). Be sure to write your
answer as a function of x.
ƒ(x)=7x+1
y=7x+1
(y-1)/7=x
x=(1/7)y-(1/7)
ƒ-1(y)=(1/7)y-(1/7)
ƒ-1(x)=(1/7)x-(1/7)
A
Lines
Point slope form
• The equation of a line with slope m through
point (a,b) is
y - b = m(x - a)
• If you don’t know the slope, know two points
(a1,b1) and (a2,b2), then the slope m is just the
slope formula for those points.
b2 - b1
m=
a2 - a1
Slope intercept form
y = mx + b
• Slope intercept form is the “simplest” form of
a line
– “Simplify” means put in slope intercept form
Doing the same thing to both sides
• Adding, Subtracting, Multiplying, Dividing, a number
from both sides of the equation.
– Changes the value of both sides, but not the equality.
12 = 12
- 6 = -6
6=6
4x = 4 + 2x
- 2 x = -2 x
2x = 4
/2=/2
/2=/2
3=3
x=2
WARNING
• Each side is a number. When multiplying (or dividing) multiply
(or divide) the whole number.
GOOD
BAD
GOOD
2+4x = 6
2+4 = 6
2+4 = 6
2 + 4 / 2 = 6 / 2 (2 + 4) / 2 = 6 / 2 (2 + 4x) / 2 = 6 / 2
2 / 2 + 4x / 2 = 3
2+2 =3
6/2=3
1+ 2x = 3
4=3
3=3
x =1
Rearranging an equation to solve
4 ( x  1)  8  2 x
4x  4  8  2x
 4  4
4x  4  2x
 2 x  2 x
2x  4
x2
Solving Inequalitites
• When I divide by a negative, I can have the same
effect as “moving to the other side” by “flipping the
sign”
Moved x
8 < 4 - 2x
8 + 2x < 4
8 < 4 - 2x
8 - 4 < -2x
2x < -4
4 < -2x
x < -2
-2 > x
Answers Match
Flipped the sign
Solving a system of equations on your
calculator (and showing work)
• Solve
4x + 8y - 4z = 8
2x + 3y + 4z = 4
5x + 8y + 1z = 7
In my calculator, I set the matrix [A]
é 4 8 -4 8 ù
ê
ú
[ A] = ê 2 3 4 4 ú
ê 5 8 1 7 ú
ë
û
Then I used the command
rref([A])
The calculator output was
So the answer is
x=-3.5
y=3
z=0.5
é 1 0 0 -3.5 ù
ê
ú
ê 0 1 0 3 ú
ê 0 0 1 0.5 ú
ë
û
Polynomials
Arithmetic on complex numbers
• 1 and i cannot be combined. They are on
separate axes.
– 1+i can’t be simplified, just like x+y can’t be simplified.
• Treat i like a variable and you will be ok.
• Remember that i2=-1 and √(-1)=i
– This can be simplified
2 + 3i
1+ 2i
Examples
(2 + 3i)(1- 2i)
=
1- (-4)
(2 + 3i) (1- 2i)
=
(1+ 2i) (1- 2i)
(2 + 3i)(1- 2i)
=
5
(2 + 3i)(1- 2i)
=
12 - (2i)2
8-i
=
5
You are not done
until you have the
real and imaginary
parts completely
separate
8 1
= - i
5 5
Two Formulas
Quadratic Formula
-b
b 2 - 4ac
x=
±
2a
2a
2
When ax + bx + c = 0
Vertex Formula
-b
h = , k = f (h)
2a
2
When f (x)=ax + bx + c
Vertex is (h, k)
Vertex form
• y=a(x-h)2+k
• To find an equation of a parabola from vertex
(h,k) and point (x1,y1).
– Plug in h,k, x1,y1 and solve for a.
– Plug in h,k, and a.
– Answer should look something like: y=2(x-1)2-2
Standard Form of a Polynomial
3x2+2x-2x4-3
Constant term = y-intercept
Leading Term
3x2+2x-2x4-3
Leading Coefficient determines end behavior
Degree = number of roots = number of bends +1
Factored Form of a Polynomial
4(x-4)(x--1/3)(x-i)(x--i)
Leading Coefficient End Behavior
(+ means y is increasing for big x. – means y is
decreasing when x is big)
Roots
Solving Polynomials
1) Use the Rational Root Test to come up with
guesses for roots.
2) Use Synthetic Division to test roots and factor
the polynomial
3) When you have only a quadratic left, use the
quadratic formula
Rational Root Test
4x3-3x2+2x-5
Factors of -5: {-5,-1,1,5}
Factors of 4: {1,2,4}
The only possible rational roots of this
polynomial are
-5/1, -1/1, 1/1, 5/1, -5/2, -1/2, 1/2, 5/2, -5/4, 1/4, 1/4, 5/4
Review: Synthetic Division
• x3+x2-4x-4. Root at x=2
1 3 2 0
2 | 1 1 -4 -4
2 6 4
Add up
Multiply to the bottom
Add up
Multiply to the bottom
Add up
Multiply to the bottom
Add up
1x3+1x2-4x-4=(x-2)(1x2+3x+2)+0
You are given the coordinates of the vertex (-8,3) and of a
point (-4,7) on a parabola. Find the equation of the
parabola.
a) y = -.25(x+8)2 - 3
b) y = .25(x+8)2 - 3
c) y = .25(x-8)2 + 3
d) y = .25x2 + 4x + 19
e) Both (c) and (d)
You are given the coordinates of the vertex (-8,3) and of a
point (-4,7) on a parabola. Find the equation of the
parabola.
7=a(-4- -8)2+3
7=a(4)2+3
4/42=a
a=1/4
y=0.25(x+8)2+3
y=0.25(x+8)(x+8)+3
y=0.25x2+4x+19
D
Solving Polynomial Inequalities
•
•
•
•
x2<x3-3x
Get 0 on one side
[x2-x3+3x<0]
Graph the polynomial [y=x2-x3+3x]
Convert to an equality [x2-x3+3x=0]
Find the roots
x=0, x=0.5+0.5√(13), 0.5-0.5√(13)
• Use the roots and the graph to solve the
inequality
0.5+0.5√(13)<x<0 OR x>0.5-0.5√(13)
Test each interval
Is (3)2+2(3)-(3)3<0 ?
Is -12<0 ? YES.
<----------------|------------|--------------------|---|---->
-1
0
2 3
x<-1
-1<x<0
0<x<2
x>2
NO
YES
NO
YES
Solve the QUADRATIC INEQUALITY:
Hint: You might graph the parabola y=(x-3)(x+4) first
a) x > - 3
b) x < 4
c) x > 3 or x < - 4
d) - 4 < x < 3
e) None of the above
C) x > 3 or x < - 4