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## MATH121, Lesson 5.1 Notes

by: Mallory McClurg

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# MATH121, Lesson 5.1 Notes Math 121

Marketplace > University of Mississippi > Math > Math 121 > MATH121 Lesson 5 1 Notes
Mallory McClurg
OleMiss
GPA 3.37

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## About this Document

These notes cover all of Lesson 5.1 - Introduction to Polynomial Equations and Graphs! I give some pretty good tips in these notes that should help you be able to work these pretty quickly!
COURSE
College Algebra
PROF.
Dr. KHAZHAKANUSH NAVOYAN
TYPE
Class Notes
PAGES
8
WORDS
KARMA
25 ?

## Popular in Math

This 8 page Class Notes was uploaded by Mallory McClurg on Tuesday October 11, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dr. KHAZHAKANUSH NAVOYAN in Fall 2016. Since its upload, it has received 16 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 10/11/16
MATH121 Chhaappteer 55 Noottees Lesson 5.1 – Introduction to Polynomial Equations and Graphs BE ABLE TO VISUALIZE GRAPHS. Knowing this information will help you a lot with this section: v EVEN-DEGREE POLYNOMIALS: (where the highest exponent a variable is an EVEN number – x , x , x , etc.) • Points up on both ends with positive leading coefficient • Points down on both ends with negative leading coefficient Negative leading coefficient with x , x , x , etc. Positive leading coefficient with x , x , x , etc. v ODD-DEGREE POLYNOMIALS: (where the highest exponent a variable is an ODD number – x , x , x , etc.) • Ends head off in different directions o If the end enters the graph through the bottom, and leaves through the top of the graph (left to right), it is positive! o If the end enters the graph through the top, and leaves through the bottom (left to right), it’s negative! Negative leading coefficient with 3 6 x , x , etc. Positive leading c3ef6icient with x , x , etc. JUST REMEMBER: EVEN : SAME DIRECTION BOTH é = + LEADING COEFFICIENT BOTH ê = - LEADING COEFFICIENT ODD : DIFFERENT DIRECTION (WHEN READING GRAPH FROM LEFT SIDE TO RIGHT SIDE) ENTERS FROM THE BOTTOM = + LEADING COEFFICIENT ENTERS FROM TOP = – LEADING COEFFICIENT MATH121 Chhapptter 55 Noottes EXAMPLE 1. 4x > 7x + 2 (First, we need to get this inequality with a zero on one side. Subtract 7x and 2 from both sides.) 2 4x – 7x – 2 > 0 (Since the left side of the equation is not easily factorable, use the quadratic equation. Note: since it’s asking for what’s greater than zero, our final interval is going to be what’s above the x-axis on the graph of our function.) x = (7 +/- √((-7) – (4)(4)(-2))) / 2(4) = (7 +/- √(81)) /8 = (7 + 9)/8 = (7 – 9)/8 x = 2 x = (-1/4) (This means that on the graph, the function will cross the x-axis at (2, 0) and (-1/4, 0). This helps us visualize our graph. Now, use the rules. Note: Always apply these rules when the inequality has zero on a side by itself. Using our first step, we should recognize that it’s an even-degree polynomial, so either both sides of the graph will be pointing up or both will be pointing down. It would also help to know that “x ” polynomials look like a U or an upside-down U. So, picture a U (because it has a positive leading coefficient of 4x ) that intersects the horizontal x-axis at (2, 0) and (-1/4, 0). The question asks us to find what’s above the x-axis, so our answer is….) (- ∞, -1/4) U (2, ∞) MATH121 Chhapptter 55 Noottes EXAMPLE 2. (x – 5)(x + 3)(2 – x) ≤ 0 (The first thing we should do is determine the points where the function crosses the x-axis by finding the “zeroes”. They’re pretty easy to find. Set each of the three expressions on the left side of the equation equal to zero to solve for x.) x = 5 x = -3 x = 2 (The function will cross the x-axis at (-3, 0), (2, 0), and (5, 0). Now, we need to work out the whole function. FOIL two sets of expressions on the left, then multiply in the remaining expression like this.) ((x – 5)(x + 3))(2 – x) 2 (x2– 5x + 3x – 15)(2 – x) (x – 2x – 15)(2 – x) (2x – 4x – 30) – x + 2x + 15x -x + 4x + 11x – 30 (So the largest exponent is 3, which is odd, so go back to the rules. Also, the leading coefficient is negative so the graph will look a little something like a curved backwards letter N with the x-axis drawn horizontally through the center. The function touches the x-axis at -3, 2, and 5, which we found earlier. The original question wants us to find everything that is below or equal to the x-axis on the graph. So use this information to write your interval of solutions!) [-3, 2] U [5, ∞) MATH121 Chhaaptterr 5 NNoottess EXAMPLE 3. x (x + 1)(x – 4) > 0 (First, find the zeroes.) 2 x = 0 x + 1 = 0 x – 4 = 0 x = 0, 0 x = -1 x = 4 (Now, work it out!) x ((x + 1)(x – 4)) 2 2 x (x + x – 4x – 4) x (x – 3x – 4) x – 3x – 4x 2 (Now, looking at the highest exponent and the coefficient in front of it, use the rules to visualize your graph. Both leading ends of the function will point up and will at least touch the x-axis at -1, 0, and 4. But notice how x touches zero twice, here because of x in the original problem. This means it doesn’t pass through the axis, but only touches it and turns back around. This function looks like a curved W with the x-axis through the center, with the top of the middle hump at (0, 0). Since the problem asks us to find everything above the x- axis, this is our answer….) (- ∞, -3) U (4, ∞) EXAMPLE 4. (x + 3)(x – 4)(4 + x) < 0 (Find the zeroes.) x + 3 = 0 x – 4 = 0 4 + x = 0 x = -3 x = 4 x = -4 (Now, work it out!) ((x + 3)(x – 4))(4 + x) 2 (x2+ 3x – 4x – 12)(4 + x) (x – x – 12)(4 + x) 4x – 4x – 60 + x – x – 12x 3 2 x + 3x – 16x – 60 (Use the rules to visualize your graph. The highest exponent is odd, so the leading ends of the function on the graph will be going different ways. Also, the leading coefficient is positive, so the graph will enter from the bottom left. Since it’s positive x , it will look like the letter N with the x- axis through the center of it. The function will touch the x-axis at -4, -3, and 4. The question asks us to find everything less than zero, or everything below the x-axis.) (- ∞, -4) U (-3, 4) MATH121 Chhapptterr 5 NNootees EXAMPLE 5. x(x + 3)(x – 1) ≥ 0 (Find the zeroes.) x = 0 x + 3 = 0 x – 1 = 0 x = 0 x = -3 x = 1 (Now work it out!) x((x + 3)(x – 1)) x(x + 3x – x – 3) 2 x(x + 2x – 3) x + 2x – 3x (So the highest exponent is odd, and the leading coefficient is positive. This means the graphed function will look like a curved letter N with the x- axis running through it. The question asks us to find everything greater than and equal to zero, so use brackets in your interval!) [-3, 0] U [1, ∞) EXAMPLE 6. x(x + 2) (x – 5) < 0 (Find the zeroes.) x = 0 x + 2 = 0 x – 5 = 0 x = 0 x = -2, -2 x = 5 (Now, work it out!) x((x + 2)(x + 2))(x – 5) (x + 2x + 2x + 4x)(x – 5) 3 2 (x + 4x + 4x)(x – 5) (x + 4x + 4x ) – 5x – 20x – 20x x – x – 16x – 20x (So the highest exponent is even and the leading coefficient is positive, the graphed function will have both leading ends pointing up. Notice that x = -2 twice, so that means it must touch the x-axis and then immediately turn back around. So this function looks like a lop-sided letter W with the bottom of the first hump at (-2, 0). The question is asking us for everywhere on the graph that the function falls below the x-axis (< 0).) (0, 5) MATH121 Chaappteer 5 NNootees EXAMPLE 7. x(x + 2) (x – 5) > 0 (Now, using the same function as the last problem, find the solutions when the function is greater than zero. First, find the zeroes.) x = 0 x + 2 = 0 x – 5 = 0 x = 0 x = -2, -2 x = 5(Now, work it out! Or just look back to EXAMPLE 6.) x((x + 2)(x + 2))(x – 5) 3 2 2 (x3+ 2x 2 2x + 4x)(x – 5) (x + 4x + 4x)(x – 5) (x + 4x + 4x ) – 5x – 20x – 20x x – x – 16x – 20x (So, the highest exponent is even, and the leading coefficient is positive, so the graph will have both leading ends pointing up and will touch the x-axis at -2 twice, meaning it only touches it – not passes through it. But the question doesn’t ask us to find what’s touching the graph, only what positive numbers not including zero solve the function. So, what parts of the graph are above the x-axis (> 0)?) (-∞, -2) U (-2, 0) U (5, ∞) MATH121 Chhaaptterr 5 NNoottess EXAMPLE 8. 2 x(x + 5) (x – 8) ≤ 0 (First, as usual, find the zeroes.) x = 0 x + 5 = 0 x – 8 = 0 x = 0 x = -5 x = 8 (Now, work it out!) 2 x(x + 10x + 25)(x – 8) x + 10x + 25x (x – 8) x + 10x + 25x – 8x – 80x 2 (So, the highest exponent is even and the leading coefficient is positive, so the graphed function will have both ends pointing up. There’s also a critical point at (0, -5) where the function will touch the x- axis but not pass through it. So we’re asked to find every solution that is less than or equal to zero. This graph also looks like a lop-sided letter W with the bottom of the first hump touching the point (0, -5). Only the bottom of the second hump on the W is below the axis. Since it is only touching the axis, but not passing through it, we must still include this in our final interval because it is equal to y = 0.) { - 5 } U [0, 8] (The brackets around the -5 do not indicate a range from one number to another’ the bracket cuts it off to only include the point within it.) EXAMPLE 9. 2 x(x + 4) (x – 7) ≥ 0 (Find the zeroes!) x = 0 x + 4 = 0 x – 7 = 0 x = 0 x = -4, -4 x = 7 (Now, work it out!) 2 x(x + 8x + 16)(x – 7) x + 8x + 16x(x – 7) x +8x + 16x – 7x – 56x – 112x 4 3 2 x + x – 40x – 112x (The highest exponent is even and the leading coefficient is positive. Notice that we have a repeating value when we found our zeroes in the first step. That means at (-4, 0) the function will only touch the x-axis, then turn back around. This will look like a lop-sided letter W with the bottom of the first hump on the x-axis at (-4, 0) and the second hump underneath the x-axis. The “≥ 0” means that we’re looking for everything above the x-axis.) (- ∞, 0] U [7, ∞) MATH121 Chaaptter 5 NNootess EXAMPLE210. r(x) = x + 6x – 7 (The question asks us to choose the correct graph, and to write out the x-intercepts and y-intercept. Since the highest power is even and the leading coefficient is positive, this will be shaped like the 2 letter U. To find the x-intercepts, find the zeroes.) r(x) = x + 6x – 7 = (x – 1)(x + 7) x – 1 = 0 x + 7 = 0 x = 1 x = -7 (The function will cross the x-axis at (1, 0) and (-7, 0). Now, to find the y-intercept, treat r(x) like y, substitute zero in for x in the original function and solve for y, this will be the y-value in the y- intercept point.) (0) + 6(0) – 7 = -7 y-intercept: (0, -7)

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