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:

- Find the slope, given two points on a line.
- Solve an applied problem using the slope formula.
- Find the slope and y-intercept of a linear equation.
- Graph a linear equation by using the slope and y-intercept.
- Solve an applied problem involving a linear equation.


A function is linear if it could be written in the form `y = mx + b`. m is the slope, and the point (0, b) is the y-intercept.

A function that can be written in the form `f(x) = b` is called a constant function. Its slope, m, is equal to zero, which generates a horizontal line.

An identity function is one that can be written `f(x) = x`. This time the slope, m, is equal to 1, and it produces a diagonal line through the origin.

Vertical lines are not functions, their slopes are undefined. `x = a` or x is equal to some constant.



The slope given two points, `(x_1, y_1)` and `(x_2, y_2)` is equal to the `(rise)/(run)` or the change in y, `y_2 - y_1`, over the change in x, `x_2 - x_1`, which is the same as `(y_1 - y_2)/(x_1 - x_2)`.



Find the slope of the line containing the given points, (0, 4) and (7, 0).

So slope is equal to `("the change in y")/("the change in x")`, which is equal to `(4-0)/(0-7)`, and that is equal to `-4/7`.



Now let's cover positive slopes and negative slopes. A line has a positive slope if it slants up from left-to-right. A line has a negative slope if it slants down from left-to-right.

The greater the slope, the steeper the line. A horizontal line has a slope of zero, and a vertical line has an undefined slope, or a slope of infinity.



Find the slope of the line containing the given points: (1, 5) and (7, 5).

m is going to equal the change in y, `5-5`, over the change in x, `1-7`. That gives us `0/-6` or 0. A slope of zero, a slope of zero, this means this is a horizontal line.

Now let's find the slope of (5, 1) and (5, 7).

That means the slope is equal to the change in y, `1-7`, over the change in x, `5-5`. That gives us `-6/0`, which is equal to `-oo`.

A slope of `-oo` or a slope that's undefined indicates that we have a vertical line.



Slope means average rate of change.

In 1992, the U.S. sold about 300,000 motorcycles. By 2005, the total number of motorcycles sold reached 1,100,000. What's the average rate of change from 1992 to 2005?

Let x represent the year and y represent the total number of motorcycles sold. Then we could setup two ordered pairs: in 1992, they sold 300,000 motorcycles, and in 2005, they sold 1,100,000 motorcycles. So then the slope is the average rate of change, which is the change in y, `1,100,000 - 300,000`, all over `2005-1992`. This gives us 61,538.5 motorcycles. The average rate of change in sales of motorcycles from 1992 to 2005 was about 61,538.5 motorcycles.



Find the slope and y-intercept of the line. `y = (2/5)x-3` is already in slope-intercept form. So the slope is `2/5` and the y-intercept is at the point (0, -3).

Now find the slope and y-intercept of the line, `3y-x+5 = 0`. We're going to need to put this into slope-intercept form. We're going to need to solve for y. If we add x to both sides, we get `3y+5 = x` and then subtract from 5 from both sides gives us `3y = x-5` and then if we divide everything by 3, this gives us `y = 1/3x-5/3`. Therefore, the slope is `1/3` and the y-intercept is at (0, -5/3).



We can also graph a linear equation by using its slope and y-intercept. Graph `y = 1/2x+4`. The slope is `1/2` and the y-intercept is (0, 4). Now we'll just need to plot the y-intercept and then use the slope to find another point. Rise one unit, run 2 units. So the next point should be at the point (2, 5).



A boy on a bicycle is pedaling on a slick road and needs to stop suddenly when he sees a car is headed in his direction. He travels a distance, d meters, before his brain is able to send the signal to his foot to hit the brakes. His reaction distance is given by the function, d(s), where the s is his speed in m/s.

What is the slope of this line and what does it represent? So notice d(s) is a linear formula and its in the slope-intercept form already. So the slope is 5/4. The slope represents `("the change in distance in meters")/("the change in speed of his bike in meters per second")` and it means that for each meter per second faster that he's travelling, it's going to take another 5/4 meters to stop.

Now find d(4) and d(16). The distance that he's going to travel before stopping when he's travelling 4 meters per second and 16 meters per second. `d(4) = 5/4(4) + 1/2`, which is equal to 5.5 meters. And `d(16) = 5/4(16) + 1/2`, which is equal to 20.5 meters.