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:

- Graphing polynomial functions.
- The Intermediate Value Theorem.


Graphing a polynomial function, P(x) with degree n.

Make sure you use the leading term test to find the zeros.

And there are at most n zeros and x-intercepts.

There are at most n-1 turning points.

Relative maxima or minima indicate a turning point, which is where the graph changes direction, from increasing to decreasing or decreasing to increasing.



Graph the polynomial function.

`P(x) = -x^4-5x^3`.

Let's start with the leading term test. Our leading term is `-x^4`. Our degree is 4, and it's an even number, and our coefficient is -1, a negative value. This means that both ends of our graph point down.

Now let's determine the zeros of our function. We have to set our polynomial function equal to zero, and then factor. We can pull a `-x^3` out of both of our terms, so that we're left with `x+5`. And then setting `-x^3 = 0`, gives us our first zero, `x = 0`, and it has a multiplicity of 3, because of the power 3. And then for our second zero, we have to solve `x+5 = 0`, and we get `x = -5`, which has a multiplicity of 1, because our power is 1 here.

Since the multiplicity is odd on our first zero, that means that we have an x-intercept at the point (0, 0). And since the multiplicity is again odd at our second zero, that means we have another x-intercept at the point (-5, 0).

We can plot these two points on our graph.

Now we need to determine if the graph is above or below our x-axis between each of our intervals. We have three intervals, `(-oo, -5), (-5, 0)," and "(0, oo)`. So choose a test value in each interval. We'll choose -6, -1, and positive 1.

P(x) is either positive in an entire interval, or it's negative in the interval.

So if we plug in and find P(-6), P(-1), and P(1), if we get a negative or positive output, that'll determine whether P(x) is either positive or negative in each of our intervals.

So `P(-6) = -(-6)^4-5(-6)^3`; that's equal to -216.

And then `P(-1) = -(-1)^4-5(-1)^3`, which is equal to positive 4.

And then `P(1) = -(1)^4-5(1)^3`, which is equal to -6.

So since we got a negative value at -6, that means that for the interval, `(-oo, -5)`, our graph is below the x-axis. And then since we got a positive value at -1, that means that between -5 and 0, our graph is above the x-axis. And then since we got a negative value at 1, that means that from `(0, oo)`, our graph is below the x-axis, again.

We can plot the points (-6, -216), (-1, 4), and (1, -6), and then complete the graph.

Before graphing your polynomial function, you may need to solve for additional points or the y-intercept. In this case, we have the y-intercept at (0, 0), which is the same as our x-intercept. Also, since our polynomial has a degree 4, the graph of P can have at most 4 x-intercepts, and at most 3 turning points. It actually has 2 x-intercepts and 1 turning point. And remember that when we have multiplicities that are odd numbers, that means that we can have x-intercepts that are zeros.



The Intermediate Value Theorem.

For any polynomial function, P(x), given two inputs a and b, where `a != b`, if the outputs P(a) and P(b) have opposite signs, then there is at least one real zero between a and b.

The Intermediate Value Theorem cannot be used if P(a) and P(b) have the same sign.



Use the Intermediate Value Theorem to determine if there is at least one real zero between a and b.

`f(x) = x^3+x^2-3x-10`, where `a = 0`, and `b = 3`.

So let's start by finding f(a) or f(0). We get `0^3+0^2-3*0-10`, which is equal to -10, a negative value. Then let's find f(b), `f(3) = 3^3+3^2-3*3-10`, this is equal to positive 17, a positive value.

So by the Intermediate Value Theorem, f(0) and f(3) have opposite signs, therefore f(x) has at least one zero between 0 and 3.