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 the equation of a line.
- Determine whether two lines are parallel or perpendicular.
- Model a set of data with a linear function.


Write the equation of the line in slope-intercept form with the given characteristics. The slope, m, is equal to 1/3 and the y-intercept is equal to (0, 5).

`y = mx+b` is the slope-intercept form of a line.

So with the given information, let's just plug in 1/3 for m, and 5 for b. So `y = 1/3x + 5`.



Another formula that can be used to determine an equation of a line is called the point-slope equation. For two points, (x, y) and `(x_1, y_1)`, the slope is `(y-y_1)/(x-x_1)`. We can get the point-slope equation of a line if we multiply both sides by `(x-x_1)`. So, `m(x-x_1) = y-y_1`. Then if we switch around what's on either side of the equals sign, we'll get `y-y_1 = m(x-x_1)`.

This is the point-slope equation of a line with the slope m that passes through the point `(x_1, y_1)`.



Write the point-slope equation for a line if it passes through the points (-2, 5) and (1, -3).

Our first step is to write the slope. The slope is going to be the difference between y's, `5-(-3)`, over the difference in x's, `-2-1`, which gives us -8/3.

The point-slope equation for a line is `y-y_1 = m(x-x_1)`. So for (x_1, y_1), we could pick either one of our points. So let's pick the first one, (-2, 5). That'll give us `y-5 = (-8/3)(x-(-2))`. You'll get `y-5 = -8/3x-16/3`. Then add 5 to both sides. This gives us `y = -8/3x-1/3`. Which is the point-slope equation for a line that passes through (-2, 5) and (1, -3).



Parallel lines and perpendicular lines.

Two vertical lines are parallel. Non-vertical lines are parallel if they have the same slope but different y-intercepts.

Unlike parallel lines, perpendicular lines cross in such a way that they form a 90 degree angle. If they have the slopes `m_1` and `m_2`, then their slopes multiplied together will equal -1. Also, their slopes are opposite reciprocals of each other, which means `m_2 = -1/m_1` or `m_1 = -1/m_2`. A horizontal line is always perpendicular line to a vertical line.



Given `y_1 = 2/5x-13` and `y_2 = -5/2x-13`, determine whether the lines are parallel, perpendicular, or neither.

So let's identify the slopes. `m_1`, the slope of the first line is 2/5. And then `m_2`, the slope of the second line is -5/2. Since the slopes are not identical, we do not have parallel lines.

The slopes of two perpendicular lines have a product of -1. 2/5 and -5/2 are opposite reciprocals of each other. So their product is -1.

Therefore, the lines `y_1` and y_2` are perpendicular.



Determine whether the pair of lines are parallel, perpendicular, or neither.

We're given `x+4y = 10` and `4x+16y = 11`. Let's go ahead and solve each equation for y.

So if we subtract x from both sides in our first equation, we get `4y = -x+10`. And then divide everything by 4. We get `y = -1/4x+5/2`.

Now let's solve the second equation. Subtract 4x from both sides. `16y = -4x+11`. Divide by 16 and you get `y = -1/4x + 11/16`.

So for our first equation, our slope is -1/4 and our y-intercept is equal to (0, 5/2). And then for our second equation, we have our slope is equal to -1/4 and our y-intercept is equal to (0, 11/16).

So therefore since we have identical slopes, but different y-intercepts, we have two lines that are parallel to each other.



Find the equation of a line parallel to `y = 3/8x+1`, if it contains the point (1, 5).

So since our line is parallel, it's going to have the same slope of 3/8. And our point, `(x_1, y_1)` is (1, 5). Now let's plug our slope and our `(x_1, y_1)` into our point-slope equation. This gives us `y - 5 = 3/8(x - 1)`. Therefore the equation of the line parallel to our given line is `y = 3/8x + 37/8`.

Now let's find the equation of the line perpendicular to our given line, if it contains the point (1, 5).

Our new slope will be the opposite reciprocal of 3/8, which is -8/3. And our point `(x_1, y_1)` is (1, 5). Now let's plug in our point and our slope into our point-slope formula. We get `y - 5 = -8/3(x - 1)`. Therefore the equation of the line perpendicular to our given line is `y = -8/3x + 23/3`.



Linear functions are one type of function that can be used to model data. Data can also be modeled using a quadratic function, a cubic function, or an exponential function.

Notice that these are all non-linear.



Now let's model the following data with a linear function. The following table illustrates the growth in world-wide cell phone usage. Here we're given the year and the number of cell phone users in billions.

Let the independent variable represent the number of years after 2004. Our answer will vary depending on the data points used. We're going to choose two data points, find the slope and use the point-slope equation to make our linear model.

So for our two points, I'm going to use 2006 and 2008. Therefore our two ordered pairs and (2, 1.584) and (4, 1.981). Using these two points, we get a slope of 0.1985. And now we can choose either one of these points for our point-slope equation. Let's go ahead and pick (2, 1.584). So if we plug in `y_1`, `x_1`, and m, `y = 0.1985x + 1.187`.

Now using the function you found in part (a), estimate the number of world cell phone users in 2018. Since we're letting our independent value represent the number of years after 2004, we have to subtract 2018-2004 to get our independent variable x. Therefore, x = 14. So, plugging this into our function in part (a), we get `f(14) = 0.1985(14) + 1.187`. The estimated number of world cell phone users in 2018 is 3.966 billion.