This section consists of Algebra II hints, based on common errors that I see on student tests. I also provide a helpful hints for Algebra I, which would also be helpful.
Algebra I
These are the topics in the Algebra I page.
- Order of Operation
- Distribution
- Rates
- Percentage Problems
- Solutions to Solution Problems
- Trains Approaching
- Representation of Inequalities
- Set Builder Notation
- Interval Notation
Algebra II
As common problems come up in Algebra II, I will add to this section as needed. The topics are as follows:
Combining Function
Functions come in the general format of f(x) = function of x. I.E. f(x) = x2 -3x + 2. Functions can be combined or act on one another. There are specific rules for combining functions. Following are a few basic examples.
Function | How to Use |
---|---|
Functions | f(x)=2x2+4x-8, g(x)= 5x-3, h(x)=1/(x-2) |
(f+g)(x) | f(x)+g(x) = (2x2+4x-8) + (5x-3) = 2x2+4x-8+5x-3 = 2x2+9x-11 |
(f+g)(-2) | f(-2)+g(-2) = (2(-2)2+4(-2)-8)+(5(-2)-3) = 2(-2)2+4(-2)-8+5(-2)-3 = 2(4)+9(-2)-11=8-18-11=-21 |
(f-g)(x) | f(x)-g(x) = (2x2+4x-8) – (5x-3) = 2x2+4x-8-5x+3 = 2x2-x-5 |
(f-g)(-2) | f(-2)-g(-2) = (2(-2)2+4(-2)-8)-(5(-2)-3) = 2(-2)2+4(-2)-8-5(-2)+3 = 2(4)-(-2)-5=8+2-5=5 |
(f.g)(x) | f(x).g(x) = (2x2+4x-8).(5x-3) = 10x3+20x2-40x-6x2-12x+24 = 10x3+14x2-52x+24 |
(f.g)(-2) | f(-2).g(-2) = |
(f/g)(x) | f(x)/g(x) = (2x2+4x-8)/(5x-3) |
(f/g)(-2) | f(-2)/g(-2) = (2(-2)2+4(-2)-8)/(5(-2)-3) = (8+4)/(-10-3) = 12/(-13) = -12/13 |
(goh)(x) | (g(h(x)) = g(1/(x-2)) = 5(1/(x-2)) – 3 = 5/(x-2) – 3(x-2)/(x-2) = 5/(x – 2) – (3x – 6)/(x-2) = (5 – 3x + 6)/(x -2) = (-3x + 11)/(x-2) |
(goh)(-2) | (g(h(x)) = g(1/(-2-2)) = 5(1/(-4)) – 3 = 5/(-4) – 3(-4)/(-4) = 5/(-4) – (-12)/(-4) = (5 + 12)/(-4) = (17)/(-4) = -4 1/4 |
(hog)(x) | (h(g(x)) |
FOIL
In multiplying expressions, every element of the first expression must be multiplied by every element of the second expression. Normally, this is a two by two multiplication and the way to remember how this is done is through FOIL – First, Outer, Inner, and Last. For example:
(ax + b)(cx + d) = acx2+ adx + bcx + bd = acx2 + (ad + bc)x + bd In most cases a and c are 1, so the middle expression becomes (b + d)x. In this case, the number in front of the x is the sum of the two number that when multiplied gives the final element of the resulting equation. For example:
(x + 3)(x + 5) = x2 + 5x+3x + 15 = x2 + (5+3)x + 15 = x2 + 8x + 15 (x + 3)(x - 5) = x2 - 5x+3x - 15 = x2 + (-5+3)x - 15 = x2 - 2x - 15 (x - 3)(x + 5) = x2 + 5x-3x - 15 = x2 + (5-3)x - 15 = x2 + 2x - 15 (x - 3)(x - 5) = x2 - 5x-3x + 15 = x2 - (5+3)x + 15 = x2 - 8x + 15
Factoring
Use FOIL in reverse to calculate the factors from a polynomial.
Rational Expressions
A Rational Expression is a ratio, or fraction, that includes variables in the elements of the ratio. For example:
x - 3 = x - 3 = 1 x2-9 (x-3)(x+3) (x+3)
Simplifying
Simplifying is finding common factors, like in the example above, and cancelling them out.
Restricted Values
Restricted Values are values not possible for the value of the variable. In the example above, you would set the denominator to 0 and then solve for X. The two possible solutions are given for before and after simplification:
Before: (x-3)(x+3)=0->(x-3)=0 or x=+3 and (x+3) = 0 or x=-3 After: (x+3) = 0 or x = -3
It is better to simplify first and then find the domain from -∞ through +∞, with the exception for the Restricted Values.
Before: (-∞, -3) ∪ (-3, 3) ∪ (3, ∞) After: (-∞, -3) ∪ (-3, ∞)
Two Variable set of equations
As a rule of thumb, you need as many equations as you do variables in order to solve a problem. In the examples here, there will be two unknowns so we need to create two equations and solve for them.
Two person bike ride
If Josh and Janice go for a bike ride. They decide on a 24-mile route. Janice rides 2 miles/hour faster than Josh and finishes the course 1 hour sooner. What are their speeds and times. The variables are:
Time = Distance/Speed
Rider | Speed | Time | Time2 = Distance/Speed |
---|---|---|---|
Josh | x mph | y hours | 24 miles/x mph |
Janice | x+2 mph | y-1 hours | 24 miles/((x+2) mph) |
Distance = speed * time. Therefore, plugging in for y from the Josh line into the Janice equation, I have:
((24 hours)/(x mph) -1 hour)((x + 2) mph) = 24 miles Multiplying both sides by x produces: (24 - 1x)(x+2) =24x Using FOIL: 24x + 48 - x2 - 2x = 24x Combining: -x2 + 22x + 48 = 24x +x2 - 22x - 48 = +x2 - 22x - 48 0 = x2 +2x -48 Factoring: (x + 8)(x - 6) = 0 Since speed cannot be a negative number, the only answer is: x = 6 mph y = 24 miles/6 mph = 4 hours
Then the final table is:
Rider | Speed | Time | Time2 = Distance/Speed |
---|---|---|---|
Josh | 6 mph | 4 hours | 24 miles/6 mph |
Janice | 8 mph | 3 hours | 24 miles/8 mph |