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:

- Graph a function by using transformations.


Memorize the graphs of these basic functions, which can be considered building blocks for many other functions.

The identity function, `y = x`. This is a diagonal line passing through the origin.

The squaring function, `y = x^2`. This is a parabola, opening up, and it has the points, (0, 0), (1, 1), and (-1, 1).

The square root function, `y = sqrt(x)`. This function goes through the points (0, 0) and (1, 1).

The cubing function, `y = x^3`. This goes through the points (0, 0), (1, 1), and (-1, -1).

The absolute value function, `y = abs(x)`. This function has the points (0, 0), (1, 1), and (-1, 1) and looks like a 'v'.

The cube root function, `y = root(3)(x)`. This function has the points (0, 0), (1, 1), and (1, -1).

And the reciprocal function, `y = 1/x`. This function doesn't touch the x or y-axis, and has the points (1, 1) and (-1, -1).

We can create graphs of new functions by shifting, stretching, shrinking and reflecting these graphs.



For any function, `y = f(x)` and any positive constant, b:

If we add b to our function, then we shift up b units.

If we subtract b from our function, then we shift down b units.

If we add b inside of the parentheses to our input, then we're going to shift left b units.

And if we subtract b from the input, then we shift to the right b units.



Describe how the graph of the function can be obtained from one of the basic graphs and then graph the function.

`g(x) = x-5`

So think of the graph of `f(x) = x`. This is a straight line through the origin. So since `g(x)` is the same as `f(x)-5`, then we need to shift the graph of `f(x)` down 5 units. So (1, 1) becomes (1, 1-5) and (0, 0) becomes (0, 0-5) and (-1, -1) becomes (-1, -1-5). Or (1, -4), (0, -5), and (-1, -6).

So we get the following graph.


Describe how the graph of the function can be obtained from one of the basic graphs and then graph the function.

`g(x) = (x+3)^2-1`

Think of the graph of `f(x) = x^2`. It's a parabola opening up. Now `g(x)` is equivalent to `f(x+3)-1`. This means that we need to shift left 3 units and down 1 unit. Or in other words, subtract 3 from the x-coordinates and subtract 1 from the y-coordinates.

So (-1, 1) becomes (-4, 0) and (0, 0) becomes (-3, -1) and (1, 1) becomes (-2, 0). So here's the graph of g(x).



For any function, `y = f(x)`, the graph of `y = -f(x)` is the reflection of the graph of `y = f(x)` across the x-axis.

And the graph of `y = f(-x)` is the reflection of the graph of `y = f(x)` across the y-axis.



For any function, `y = f(x)`, and any constant b:

If we multiply b on the outside of the function, then we either stretch vertically, if `abs(b) > 1` or shrink vertically if `abs(b)` is between 0 and 1. And we also reflect across the x-axis if `b < 0`.

Now if b is multiplied by our input, then we stretch horizontally if `abs(b)` is between 0 and 1, and shrink horizontally `abs(b) > 1` and we also reflect across the y-axis if `b < 0`.



A graph of `y = f(x)` is as follows.

Graph the given equation, `g(x) = -3f(x)`.

So we're multiplying a negative number on the outside of f(x). So since `-3 < 0`, we reflect across the x-axis. And since the `abs(-3) > 1`, we also stretch vertically.

The new coordinates for g(x) can be obtained by multiplying all of our y-values by -3. So (-2, -3) becomes (-2, 9); and (1, 2) becomes (1, -6); and (0, 0) becomes (0, 0); and (3, 2) becomes (3, -6); and (6, 0) becomes (6, 0).

So here's the graph of g(x).



The graph of `y = f(x)` is as follows.

Graph the given equation, `g(x) = f(-x/2)`.

This is the same as `g(x) = f(-1/2*x)`. Since we're multiplying our input by a negative value, that means that we're going to reflect the graph of f(x) across the y-axis. And since the `abs(-1/2)` is between 0 and 1, we're going to stretch horizontally.

The new coordinates for g(x) can be obtained from the coordinates of f(x) by dividing each of our x-values by `-1/2` or multiplying each of our x-values by -2. So (-2, -3) becomes (4, -3); (1, 2) becomes (-2, 2); (0, 0) stays the same; (3, 2) becomes (-6, 2); and (6, 0) becomes (-12, 0).

So this is the graph of g(x).



A graph of `y = f(x)` is as follows.

Graph the given equation, `g(x) = (-f(x-1))/2+4` So we can rewrite this as `g(x) = -1/2*f(x-1)+4`. Since we're multiplying the outside of our function by `-1/2`, that means that because of the negative value, we're reflecting the graph of f(x) about the x-axis. And then since 1/2 is between 0 and 1, that means that we're shrinking vertically.

So let's start by multiplying all of our y-values by `-1/2`. So (-2, -3) becomes (-2, 3/2); (1, 2) becomes (1, -1); (0, 0) stays the same; (3, 2) becomes (3, -1); and (6, 0) stays the same.

Now since we're subtracting 1 on the inside of our parentheses, that means that we're shifting the graph of f(x) to the right one unit, or adding one to each of our x-values. So (-2, 3/2) becomes (-1, 3/2); (1, -1) becomes (2, -1); (0, 0) becomes (1, 0); (3, -1) becomes (4, -1); (6, 0) becomes (7, 0).

Then adding 4 to everything means that we're shifting up 4 units, or adding 4 to each of our y-coordinates. So (-1, 3/2) becomes (-1, 11/2); (2, -1) becomes (2, 3); (1, 0) becomes (1, 4); (4, -1) becomes (4, 3); and (7, 0) becomes (7,4). So this is the graph of g(x).