Exam

• • • • May 15 th 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/spring 2009/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

x-intercept (-2,0)

Intercepts

y 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

x x f f(x) * f(x)g(x) g g(x) g Is not the same as f g(x) f(g(x))

COMPARISON

f

(

x

) = 2

x

1,

g

(

x

) = (

x

3) 2 (f ∘ g)(3)=f(g(3))

g

(3) = (3 3) 2

g

(3) = ƒ(

g

(3)) 0 = ƒ(0) = ƒ(

g

(3)) = 1 2 * 0 1 = 1 1≠0 (fg)(3)=f(3)g(3)

f

(3)

g

(3) = = 5 0

f

(3)

g

(3) = (

fg

)(3) = 0 = 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: x 2 • A function ƒ is ODD if ƒ(-x)=-ƒ(x). Example: x 3 • A function ƒ is NEITHER if ƒ(-x)=something else. Example: x 3 +1

Function inverse

Cubing to cube root

y=x 3 x= ∛ y

Cubing to cube root

The relationship between x and y stays the same Only my point of view changes y=x 3 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

) = 7

x

numbers, find

f

+ 1 on the domain of all real -1 (

x

). Be sure to write your answer as a function of

x

.

a) b) c)

f

-1 (

x

) = (1/7)

x

− 1/7

f

-1 (

x

) =

x

− 1/7

f

-1 (

x

) = 1/(7x+1)

d) Both (a) and (c) e) None of the above

Given

f

(

x

) = 7

x

numbers, find

f

+ 1 on the domain of all real -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 (a 1 ,b 1 ) and (a 2 ,b 2 ), then the slope m is just the slope formula for those points.

m

=

b a

2 2 -

b a

1 1

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 4

x

= 4 + 2

x

6 = 6 6 = 6 2

x

= 2

x

2

x

= 4 /2 = /2 3 = 3 /2 = /2

x

= 2

WARNING

• Each side is a number . When multiplying (or dividing) multiply (or divide) the whole number .

BAD 2+4 = 6 2 + 4 / 2 2 + 2 4 = 3 = 3 = 6 / 2 GOOD 2+4 = 6 (2 + 4) / 2 = 6 / 2 3 = 3 = 3 6 / 2 GOOD 2+4x (2 + 4 = 6

x

) / 2 2 / 2 + 4

x

/ 2 1 + 2

x

= 3

x

= 1 = 6 / 2 = 3

Rearranging an equation to solve 4 (

x

 1 )  8  2

x

4

x

 4  8  2

x

 4   4 4

x

 4  2

x

 2

x

  2

x

2

x

 4

x

 2

Solving Inequalitites

• When I divide by a negative, I can have the same Moved x effect as “moving to the other side” by “flipping the sign” 8 < 4 2

x

8 < 4 2

x

8 + 2

x

< 4 8 4 < 2

x

2

x

< -

x

< 2 4 Answers Match 4 < 2

x

2 >

x

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] = ë é ê ê ê 4 8 2 3 5 8 4 1 4 8 4 7 û ú ú ú ù So the answer is x=-3.5

y=3 z=0.5

Then I used the command rref([A]) The calculator output was é ê ê ê ë 1 0 0 0 1 0 0 0 1 3.5

3 0.5

û ù ú ú ú

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 i 2 =-1 and

√

(-1)=i – This can be simplified

2 + 3

i

1 + 2

i

= (2 + 3

i

) (1 + 2

i

) (1 (1 2

i

) 2

i

) = (2 + 3

i

)(1 2

i

) 1 2 (2

i

) 2

Examples

You are not done until you have the real and imaginary parts completely separate = (2 + 3

i

)(1 2

i

) 1 ( 4) = (2 + 3

i

)(1 2

i

) 5 = 8 -

i

5 = 8 5 1 5

i

Two Formulas

Quadratic Formula

x

= -

b

2

a

±

b

2 4

ac

When

ax

2 + 2

a bx

+

c

= 0 Vertex Formula

h

= 2

b a

,

k

=

f

(

h

) When

f

(

x

)=

ax

2 +

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 (x 1 ,y 1 ).

– Plug in h,k, x 1 ,y 1 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

3x 2 +2x -2x 4 -3 Constant term = y-intercept Leading Term 3x 2 +2x -2 x 4 -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

4 x 3 -3x 2 +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 / 4 , -1 / 1 / 4 , 1 , 1 / 5 / 4 1 , 5 / 1 , -5 / 2 , -1 / 2 , 1 / 2 , 5 / 2 , -5 / 4 , -

Review: Synthetic Division

• x 3 +x 2 -4x-4. Root at x=2 2 1 3 2 0 | 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 1 x 3 + 1 x 2 -4 x -4 =(x 2 )( 1 x 2 + 3 x+ 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 = .25x

2 + 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 4/4 2 =a a=1/4 y=0.25(x+8) 2 +3 y=0.25(x+8)(x+8)+3 y=0.25x

2 +4x+19

D

Solving Polynomial Inequalities

• • • • • x 2 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 -12 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