College Algebra

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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:

- Plotting points on a Cartesian plane
- Verifying solutions to an equation
- Graphing the equation of a line
- Finding distance and midpoint for two points
- Equation of a circle


Let's graph and label the following points. A cartesian plane consists of a y-axis and an x-axis. The first value in your ordered pair is your x-value, and your second value is your y-value.

So (5,0), then (-2, -2), is two units to the left and two units down. Then (-1, 3), then (0, 7), then (-7, 7).



Determine whether (1, 2) and (3, 10) is a solution of the equation `7x-3y = 1`.

Now, since the first value is our x-value and the second value is our y-value, we're just going to plug each of these ordered pairs into our equation and see if we get a true statement.

So, for the first one, we have `7(1)-3(2) = 1`, or `7-6 = 1`, which is `1 = 1`. Since this is a true statement, (1, 2) is a solution.

Next, (3, 10). `7(3)-3(10) = 1`, gives us `21-30 = 1`, or `-9 = 1`. Since this is a false statement, that means (3, 10) is not a solution.



Now, graph the following equation by solving for the x-intercept and y-intercept first.

This is one way of graphing an equation. The x-intercept occurs where the y-value is zero. It's the point at which the graph crosses the x-axis. The y-axis occurs where the x-value is zero, and happens where the graph crosses the y-axis.

So, x-intercept: `5x-2(0) = -10`. `5x = -10`, `x = -2`. So, the x-intercept is (-2, 0).

The y-intercept happens where the x-value is zero. So, we have `5(0)-2y = -10`, or `-2y = -10`, or `y = 5`. So we have (0, 5).

Now, let's plot the points and complete the graph.



Graph the following equation by solving for y first: `5x-2y = 6`. So, this is another way to graph a linear equation, by solving for y first.

So, we'll subtract 5x from both sides, and we'll get `-2y = -5x + 6`. And then divide everything by -2. That gives us `y = 5/2x - 3`.

Now we need to set up a chart of our independent values, x, our dependent values, y, and our ordered pairs, (x, y). So, if we plug in multiples of two, we can cancel out the 5/2, and then if we also plug in zero for x, that will cancel this term out altogether.

So, let's do -2, 2, and 0. So then we have `5/2(-2) - 3` or `-8`. So our ordered pair is (-2, -8). And then, if we plug 2 in, we have `5/2(2) - 3`, or `5-3`, which is 2. So the ordered pair is (2, 2). And then for zero, `5/2(0)` is `0-3`, so we're left with -3. And our ordered pair is (0, -3).

Now all that's left to do is plot these points and complete the graph. So we have (-2, -8), (2, 2), and (0, -3), and we get the following line for `5x-2y = 6`.



Distance formula is derived from the Pythagorean Theorem. The Pythagorean Theorem says that for any right triangle, the first leg squared plus the second leg squared is equal to the hypotenuse squared. `a^2 + b^2 = c^2`.

So, for distance formula, let's consider two points, `(x_1, y_1)` and `(x_2, y_2)`. And then, let's construct the line between them, which we'll call d, the distance between these points. And then, if we make a right triangle here, where the first leg is the difference between the x-values, `x_2-x_1`, and then the other leg is the distance between the y-values, y_2-y_1. We can go ahead and plug into the pythagorean theorem here, and we'll get the distance formula. So that we have `(x_2-x_1)^2 + (y_2-y_1)^2 = d^2`. So that the distance formula, `d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)`



What is the radius of a circle whose diameter has endpoints at (-2, -1) and (10, 4)?

So now we have a circle and they've given us two endpoints on the diameter, and they want us to know what the radius is. So let's go ahead and solve for diameter first. That's going to be to the distance between these two endpoints.

The difference between the x-values is `-2-10`, and then let's square that and add the difference between the y-values, `-1-4`. So that's going to give us the `sqrt(12^2+5^2) = sqrt(169)`, which is the same as 13.

Therefore, the radius of our circle is half of that; 13/2 units.



Given two endpoints of the diameter of a circle, `(sqrt(3), -5)` and `(sqrt(7), 5)`, where is the center of the circle?

So, if we have a circle, we're given two endpoints, the center of the circle will be that point exactly between those two endpoints. We just need to use the midpoint formula, which is just an average of the x-values and an average of the y-values.

So, the midpoint, or the center of the circle, which a lot of times is called (h, k), is the `(sqrt(7) + sqrt(3))/2`. `(5+ -5)/2`.

So, `5+ -5 = 0`, so since we have zero at the top of the fraction, it's the same as zero. And `(sqrt(7) + sqrt(3))/2` does not simplify any further, and it's the center of our circle.



The standard equation of a circle can be derived from the distance formula. The definition of a circle tells us, that if we have the center, (h, k), then every single point that lies on the circumference of the circle, say, (x, y), is the distance, r, from the center. And that distance, r, is taken from the Pythagorean Theorem or the distance formula, and you get `r^2 = (x-h)^2 + (y-k)^2`, which is the standard form for the equation of a circle.



So let's find the center and radius of this circle and graph it. We're given `(x-2)^2 + (y-7)^2 = 64`.

So we need to start by putting this in standard form. `(x-2)^2 + (y-7)^2 = 8^2`

So that we know that the center is at (2, 7), and then the radius is at 8, or the radius is eight units. And then we graph this by finding the center at (2, 7), then counting eight units from that, which is at (10, 7) and then sketching the rest of the circle.