College Algebra

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Previous Lessons
Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Determine if a correspondence or relation is a function
- Find function values, or outputs, given graph or formula
- Graph functions
- Determine if a graph is that of a function
- Find domain and range of a function
- Solve an applied problem using a function


A function is a correspondence between two sets, the first set a domain and the second set is called the range. Each element of the first set corresponds to only one element of the second set.



Is the following correspondence a function?

In the first set, we have A, B, C, and D in our domain. And in our second set, we have 1, 2, 3, and 4 in our range. This is not a function, because B corresponds to 1 and 2.



This is another example of a correspondence. We are given a list of paintings in our first set, or our domain, and a list of artists in our second set, or our range.

Now, in order to tell whether this correspondence is a function or not, we need to check whether any of the values in our first set corresponds to more than one value in our second set. We have The Storm on the Sea of Galilee going to Rembrandt and The Return of the Prodigal Son going to Rembrandt. And then The Last Supper going to Leonardo da Vinci and Mona Lisa going to Leonardo da Vinci, and "The Weeping Woman" goes to Pablo Picasso.

Since none of the values in our domain corresponds to more than one value in our range, this is a function.



A relation is a correspondence between two sets; the first set called the domain and the second set called the range, where each element of the first set corresponds to one or more elements of the second set. And a relation is sometimes written as an ordered pair.



Let's determine whether the following relation is a funcion and then state the domain and range. We have the set of ordered pairs, {(1, 2), (3, 4), (5, 6)}, and this is a function, since none of the ordered pairs contain the same domain, but different range.

So the domain includes the set of all of the first values, 1, 3, and 5. And then the range contains the set of all our second values, 2, 4, and 6.



Determine whether the following relation is a function and then state the domain and range. Now we have the set of ordered pairs, {(-1, -2), (-3, -4), (-3, -5), (0, 1)}. This is not a function because we have two ordered pairs here, (-3, -4) and (-3, -5), which have the same domain, but different range.

So the domain will include the set of all of our first values, -1, -3, and 0. And the range will include the set of all of our second values, -2, -4, -5, and 1.



Functions are given by equations, where the y is replaced with function notation, such as f(x). Members of the domain, or the x-values, are called the inputs, and the outputs are members of the range, or the resulting values of f(x).



Now we're given the following function: `H(x) = 2x^2-3x+1`. And we need to find each of the following outputs.

The output, `H(0) = 2(0)^2-3(0)+1`. The first two terms cancel, and we're left with 1 as the output.

And then, `H(-1) = 2(-1)^2-3(-1)+1` which is equal to `2+3+1`, which is equal to 6.

And then, `H(2) = 2(2)^2-3(2)+1` which is equal to `8-6+1`, which is equal to 3.

And then we have `H(-x) = 2(-x)^2-3(-x)+1` which is equal to `2x^2+3x+1`.



Graph the function: `f(x) = -x^2+1`.

So we're going to start by drawing a table with our independent value, x, our dependent value, f(x), and our ordered pair, (x, f(x)). So let's go ahead and plug in values of x. Let's go ahead and try -2, -1, 0, 1, and 2. So if we go ahead and plug in -2, we'll get `-(-2)^2+1` which gives us `-4+1`, or -3. And then -1 will give us `-(-1)^2+1` which is equal to `-1+1`, which is 0. And then `-(0)^2+1` is equal to 1. And `-(1)^2+1` is equal to 0. And `-(2)^2+1` is equal to `-4+1`, which is -3.

So, our ordered pairs include: (-2, -3), (-1, 0), (0, 1), (1, 0), and (2, -3). Let's go ahead and plot those points and complete the graph. (-2, -3), (-1, 0), (0, 1), (1, 0), and (2, -3). So here we have an upside down parabola, which is our graph of f(x).



Given the following inputs, find the outputs for g(x).

We're given the inputs on the graph of g(x), and we're asked to find g(0), g(-1), and g(1). So let's go ahead and find out what an input of 0 will give us. So, an input of zero is here, and it gives the output of 5. And then, an input of -1 can be found with this ordered pair, (-1, 6), where 6 is the output. And then g(1); we want to find the input 1, which is with the ordered pair (1, 6) where the output is 6.



The vertical line test is a way that we can visually determine whether a graph is a function. If a vertical line drawn anywhere on the graph can only strike the graph once, then the graph is a function.



Use the vertical line test to determine whether the graph is a function. So if we draw a vertical line anywhere on this graph, whether it's here to the left of the y-axis or here to the right of the y-axis, or anywhere else along our curve, it will only strike the graph one time. And therefore, this is a function.



Use the vertical line test to determine whether the graph is a function. Now our graph is a horizontal parabola. So if we go ahead and draw a vertical line anywhere along this graph, it will strike twice, here and here, or here and here. So it is NOT a function.



The domain of a function includes the set of all inputs where the function is defined as a real number.



Find the domain of the function: `h(x) = (x+1)/(5-x)`.

The only input that results in an undefined output is `x=5`, since this results in a denominator of zero. Therefore the domain is `{x|x!=5}`, and this is called set notation. Interval notation, you can write the same thing as `(-oo, 5) cup (5, oo)` and this is called interval notation. So this is the domain, two ways of writing the domain of this function.



Determine the domain and range given the following graph.

Our domain includes all of the x-values from -4 to 4. And then the range includes all of our y-values from -2 to 2. So we'll write that in interval notation as [-4, 4], all in brackets, because they include those points, and range is [-2, 2].



Graph the given function and then visually estimate the domain and the range.

`g(x) = 1/(x-1)`. The graph of g(x) looks like this, where it doesn't touch the line x=1 or y=0. And so the domain includes all of the x-values greater than 1 and less than 1 and the range includes all of the y-values greater than zero or less than zero.

So the domain in interval notation is `(-oo, 1) cup (1, oo)` and the range is `(-oo, 0) cup (0, oo)`. And notice the parentheses, because the graph does not touch those points, where x=1 or y=0.



The daily cost of renting a Toyota Corolla is $25/a day plus 30 cents per mile and $20/a day for car insurance. So if we let x represent the number of miles we drive in a day and function C(x) represent the daily cost of renting our car, then we'll have the function `0.30x+45`, where the 45 is the $25/a day plus the $20 for insurance.

Now if a customer drives 300 miles in one day, how much is his rental charge that day? We need to plug 300 in for x, so we have `C(300) = 0.30(300) + 45`. This will give us the value 135, which means that it costs $135 that day for the rental charge.

If the customer has a budget of $200 to spend on his daily rental charge, what is the maximum number of miles that he can drive each day? So if he's spending the maximum amount of $200, then C(x) = 200, and we need to find out what x equals. So that means `200 = 0.30x + 45`. And if we subtract 45 from both sides, we'll get `155 = 0.30x`, and then divide by 0.30 on both sides, gives us the value of x. That means that the maximum number of miles that he can drive each day is about 516.7 miles.