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:

- Use the leading-term test to determine the end behavior of the graph of a polynomial function.
- Find the zeros and multiplicities of polynomial functions.
- Solve an applied problem using a polynomial equation.


Polynomial functions include constant functions, linear functions, and quadratic functions, to name a few.

An example of a constant function would be `y = 5x^0` or `y = 5`. Notice that the degree of x is zero.

A linear function would be `y = 5x^1` or `y = 5x`. Notice that the degree of x is one, here.

And a quadratic function would be `y = 5x^2`, where the degree of x is two.

A polynomial function is any function of the form `P(x) = a_nx^n+a_(n-1)x^(n-1)+a_(n-2)x^(n-2)+...+a_1x+a_0`. `a_n` is called the leading coefficient, and `a_nx^n` is the leading term, where n is the degree of our polynomial. Now, all of our exponents must be whole numbers, and all of the coefficients must be real numbers.

So, for example, `f(x) = 1/x+5` is the same as `x^(-1)+5`. This is not a polynomial, because we have a negative exponent. The exponent is not a whole number.

Similarly, `f(x) = sqrt(x)+5` is the same as `x^(1/2)+5`. This is not a polynomial, because we have a fraction, not a whole number as our exponent.

So again, these are not polynomials.



The domain of a polynomial function is `(-oo, oo)`, or all real numbers.

The Leading-Term Test uses the degree of our polynomial and its leading coefficient to determine the end behavior of the graph of our polynomial function.

So, if n is even and `a_n > 0`, both ends of our graph will point up.

And if n is even and `a_n < 0`, then both ends of our graph will point down.

And if n is odd and `a_n > 0`, then the left side of our graph will point down and the right side of our graph will point up.

And if n is odd and `a_n < 0`, then the left side of our graph points up and the right side of our graph points down.

The leading term test cannot determine the middle portion of our graph.



Use the leading term test to match each polynomial function to its graph.

So, given `f(x) = 1/2x^6+5`, the leading coefficient, `1/2 > 0`, and the degree, 6 is even. So therefore this is the graph of our polynomial function, because both ends point up.

Now for `f(x) = -x^4-3x^3+x-1`, the leading coefficient is -1, it's less than zero, and the degree of our polynomial is 4, an even number. So therefore both ends point down, and this is the graph of our polynomial.

`f(x) = -x^5+x^3-2x^2+4` has -1 for our leading coefficient. It's less than zero, and the degree of our polynomial is 5, an odd number. So this is the graph of our polynomial. The left end points up, and the right end points down.

`f(x) = 1/2x^9-20x+3`. The leading coefficient, `1/2`, is greater than zero, and the degree, 9, is odd. So this is the graph of our polynomial, because the left end points down and the right end points up.



Even and Odd Multiplicity:

Begin by factoring your polynomial. The number of times, k, that a factor occurs, determines the multiplicity of the zero obtained from that factor.

So, for `(x-c)^k`, where `k >= 1`, as long as `(x-c)^k+1` is not a factor, then if k is an odd number, the point (c, 0) is an x-intercept of the graph.

But if k is even, then the graph is tangent to the x-axis at the point (c, 0).



Find the zeros of the polynomial function and state the multiplicity.

So we're given `f(x) = x^3-6x^2-x+6`. Let's start by factoring our first two terms. We'll get `x^2(x-6)`, and then we can factor an `x-6` out of our last two terms as well, and we'll get `-(x-6)`. So we have `(x^2-1)(x-6)`. And then let's factor `(x^2-1)`, and we get `(x-1)(x+1)(x-6)`.

Then by the principle of zero products, `f(x) = 0`, when `x = 1, -1, 6`. So the zeros of our polynomial 1, -1, and 6. And they each have a multiplicity of one.

For any polynomial where `n >= 1`, it has at least one zero, and at most n zeros.



Three workmen, A, B, and C, dig a ditch. 'A' can dig it alone in 2 days more time, 'B' in one more day, than the time it takes the three to dig the ditch together. If 'C' can dig the ditch in 3 times the time the three dig it in, then how long does it take the three, working together, to dig the ditch?

So if we let `a` represent the time it takes 'A' to dig alone, and little `b` represent the time it takes 'B' to dig alone, and let little `c` equal the time it takes 'C' to dig alone, then `1/a+1/b+1/c = 1/x`, where `x` is the time it takes all of them working together to complete the ditch. So we're given that `a` is equal to 2 more days than `x`. So, `1/(2+x)`, plus where `b` is one more day than `x`, `1+x`, and `c` is 3 times `x`, `1/(3x)`. All of it is equal to `1/x`.

And they want to know how long it takes when they're working together to dig the ditch, so we need to solve for x. And we're going to do that by multiplying both sides of our equation by our LCD, which is `(3x)(2+x)(1+x)`. So now we can go through and cancel. On the right, the x cancels; on the left, the 3x cancels, the `1+x` cancels, and the `2+x` cancels, so that we don't have any more fractions. We have `3x(1+x)+3x(2+x)+(2+x)(1+x) = 3(2+x)(1+x)`.

So if we distribute, we get `3x+3x^2+6x+3x^2+2+2x+x+x^2 = (6+3x)(1+x)`, or `6+6x+3x+3x^2`.

And then we can combine like terms, so we get `7x^2`, and then combine the 6x, the 3x, the 2x, and the x, then we get `12x+2` on the left.

Now let's combine like terms on the right. We can combine the 6x and the 3x to 9x, and we have `3x^2+9x+6` on the right.

And now we can bring `7x^2+12x+2` on the left, and move everything to the left. So we'll subtract 6 on both sides, subtract 9x on both sides, and subtract `3x^2` on both sides. So we get `4x^2+3x-4 = 0`. So that we have `a = 4`, `b = 3`, and `c = -4`.

And using the quadratic formula, `x = (-b+-sqrt(b^2-4ac))/(2a)`, we have `x = (-3+-sqrt(3^2-4(4)(-4)))/(2(4))`. We get `(-3+-sqrt(73))/8`, or 0.693 or -1.443. And since a negative answer is not valid here, we can go ahead and cross out the -1.443. And we're left with 0.693 days, which we multiply by 24 to get 16.632 hours, and we multiply this .632 times 60, to get 37.92, or 37 minutes, and `.92*60" seconds"`, which is 55 seconds.

So when all three workmen work together, it takes 16 hours, 37 minutes and 55 seconds to dig the ditch.