Download Math 20 Module 4 Review - Westwind Alternate School

Math 20 Module 4 Review
Review the following about polynomial functions:
1) Note:
-The degree of the function is the exponent of the highest power on x in the equation.
-The leading coefficient is the coefficient on the term with the highest exponent.
2) The definition of a polynomial:
For a function to be a polynomial the terms in the equation must
contain powers of x with whole-number exponents.
-Whole numbers are all positive integers (no fractions or decimals) from 0 to
infinity
Are the following polynomials?
1) Y= X4+2X3+6X2+X+10
YES ALL THE EXPONENTS ARE WHOLE NUMBERS
2) Y=X4+10
YES ALL THE EXPONENTS ARE WHOLE NUMBERS (IT DOESN’T
MATTER IF ONLY ONE TERM HAS AN EXPONENT IT IS STILL
CONSIDERED AN EXPONENT)
3) Y= X ½ +X3
NO (1/2) IS NOT A WHOLE NUMBER IT IS A FRACTION
4) Y=X-2+10
NO -2 IS NOT A WHOLE NUMBER
5) Y=Xπ
NO π IS NOT A WHOLE NUMBER
3) Graphing Polynomials
-First off be able to identify generalized graph shaped
Linear: y=mx+b Parabola/Quadratic y=ax2+bx+c Polynomial y=x(3,4,6,6….)
Inverse function y=1/(x-1)
When Graphing Polynomials remember:
Rule 1
The number of roots of a polynomial graph is equal to the degree of the
polynomial (ie. A quartic polynomial (of degree 4) has AT MOST 4 roots)
Rule 2
The number of 'changes of direction' of a polynomial graph is one less than
the degree of the polynomial (ie. A quartic polynomial (of degree 4) has AT
MOST 3 changes in direction - it could have 0, 1, 2 or 3 changes of direction.)
Rule 3
For polynomials with odd degrees (3, 5, 7, etc.):
If the leading coefficient is positive the graph falls to the left and rises to the
right.
If the leading coefficient is negative the graph rises to the left and falls to the
right.
Rule 4
For polynomials with even degrees (2, 4, 6, etc):
If the leading coefficient is negative the graph falls to the left and right.
If the leading coefficient is positive the graph rises to the left and right.
4) Example:
Match the equation to the graphs:
A) Y=2(x+6)(x+3)(x+1)
B) Y=-2(X+6)(X+3)(X+1)
C) Y=2(X+4)2(X-4)
D) Y=-2(X+4)2(X-4)
- This has 3 different x-intercepts so it
- This has 2 different x-intercepts so it
can’t be C/D
can’t be A/B
- Since the graph rises and falls it has an - Since the graph rises and falls it has an
odd degree
odd degree
- Since it falls to the left and rises in the - Since it rises to the left and falls in the
right it has to have an odd degree and
right it has to have an odd degree and
a positive leading coefficient.
a negative leading coefficient.
-Thus it has to be equation B
-Thus it has to be equation D
Polynomial questions:
5) What is the Equation of the quadratic function that passes through the point
(1,4) and has zeros -1 and 2
-if it has 2 zeros then it is a parabola
thus y=a(x+1)(x-2)
-then plug in the point (1,4)
4=a(1+1)(1-2)
4=a(2)(-1)
a=-2
y=-2(x+1)(x-2)
6) What is the solution to the equation x3+5x2-24x=0
-this means you need to find the roots
-you could graph it or if they give you a selection of answers you can plug those
numbers in and see if they are zeros.
Review the following
Inverse Functions:
f -1(x) means inverse (exchange domain/range) of the function f(x).
For f(x): If f(x) = 2x + 3, then y = 2x + 3
But for f -1(x): x = 2y + 3 y = (x - 3)/2
Note: The graphs of a an original function and it's inverse are always reflections in
the line y = x.
Example:
7) Graph the inverse of the following function:
Remember back to grade 10 math and the vertical line test. A function is a
function if a vertical line can intersect the graph of a function in at most one
point.
Thus the inverse of some functions may not be a real function
For example this graph doesn’t represent a
function as it fails the vertical line test
In this case the inverse graph of the function
f(x) is not a function as again it fails the
vertical line test.
However we can put the restriction x  0
for our original function and then both the
original function and the inverse will be a
function.
Ex when x  0
Rational Functions
Equation: y= 1/(x-8), y= 2/(x2-1), y= 2x/(x2-1)
Graph:
-an asymptote is a straight line to which a graph comes closer and closer but never
touches.
-the vertical asymptote occurs at the value of x in which the denominator is zero
EX
y=
2
1-x
This graph has a vertical asymptotes x=1
- the horizontal asymptote occurs as x greats bigger and bigger or smaller and
smaller
EX y= 2
1-x
This graph has a horizontal asymptote y=0
8) Graph the following rational function
y= 3
3-x
Step 1) determine the vertical asymptote 3-x=0 when x=3
Step 2) Determine the horizontal asymptote
-Look what happens when x increase or decreases….for this graph y approaches 0
so the horizontal asymptote is y=0
Step 3) Pick a couple of test points
Step 4) Draw the graph through the test points and approaching the asymptotes
Answer
9) Maurice drove 400 km from Edmonton to Battleford in one hour less time than it
took Martin to drive the same route from Battleford to Edmonton. If Maurice drove
20 km/h faster than Martin, at what speed did each of them drive? Show a complete
algebraic solution.
10. a Given f ( x ) = x2 - 4 and g( x ) = 2x - 1 , determine the value of each of the
following:
( f + g) (3)
10. b Given f ( x ) = x + 2 and g( x ) = x , find the value(s) of x for which (fg)( x) = 8.
10. c Given f ( x ) = x2 + 4 and g( x ) = 2x + 1 , determine the following: ( f x(g))
Answer
b
c
Review polynomial long division
11) What are the roots of the following function?
F(x)= X3+2x2-5x-6
Step 1) look for an obvious root
In this case one is x=1
x2 +x-6
x+1 X3+2x2-5x-6
-(x3+x2 )
x2 -5x-6
-(x2 +x)
-6x-6
-(-6x -6)
0
y=(x-1)( x2 +x-6)
y=(x-1)(x+3)(x-2)
Review the remainder theorem
This leads us to the Remainder Theorem which states:
If a polynomial f(x) is divided by (x − r) and a remainder R is obtained, then f(r) = R.
12) When P(x) = x3 - 3x2 + kx + 2 is divided by x - 2 the remainder is 4.
a. Determine the value of k.
b. What is the remainder when P(x) is divided by x + 2?
Inequalities
Solve the following inequality
13) 2x-3  5
Method 1)
Graph y=2x+3 and y=5 and find the point where y=2x-3 is less then the line y=5
Method 2
2x-3  5
2x-8  0
2(x-4)  0
-Test values of x above 4 and below 4 to see which make the statement true.
14) I2x-3I  5
Method 1)
Graph y=I2x+3I and y=5 and find the point where y=I2x-3I is less then the line y=5