# College Algebra Exam 4 Material.

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College Algebra Exam 4 Material

Quadratic Function A function that can be written in the form:
f(x) = ax2 + bx + c (standard form) where “a”, “b”, and “c” are real numbers with a ≠ 0 A second degree polynomial function Examples: f(x) = x2 – 8x + 15 f(x) = -4x2 – 16x + 33

Every quadratic function has a graph that is a “U-shaped” curve called a “parabola” that opens upward when “a” is positive and downward when “a” is negative The highest point on a parabola that opens downward and the lowest point on a parabola that opens upward is called the “vertex” The vertical line through the vertex is called the “axis” The parabola is always “symmetric” with respect to its axis (if the graph were folded along the axis it would lay on top of the other half

“Vertex Form” of a Quadratic Function
Every quadratic function can be written in “vertex form”: f(x) = a(x – h)2 + k where “a”, “h”, and “k” are real numbers with a ≠ 0, and where (h, k) is the vertex of the parabola As before, parabola opens upward when “a” is positive and downward when “a” is negative Examples: f(x) = 2(x – 3)2 – 4 parabola that opens upward with vertex (3, -4) f(x) = -3(x + 5)2 + 1 parabola that opens downward with vertex (-5, 1)

Converting a Quadratic Function to Vertex Form
Given f(x) = ax2 + bx + c, factor “a” from variable terms leaving variables in parentheses Complete the square on the expression inside the parentheses and balance on the same side of the equation Factor the expression inside parentheses and combine like terms outside parentheses to get vertex form:

Example One: Converting to Vertex Form
f(x) = x2 – 8x + 15 Factor “a” from variable terms: f(x) = 1(x2 – 8x ) + 15 Complete the square on the expression inside parentheses and balance on the same side of the equation: f(x) = 1(x2 – 8x + 16) + 15 – 16 Factor the expression inside parentheses and combine like terms on outside: f(x) = 1(x – 4)2 – 1 Note: parabola opens upward with vertex (4, -1)

Example Two: Converting to Vertex Form
f(x) = -4x2 – 16x + 33 f(x) = -4(x2 + 4x ) + 33 f(x) = -4(x2 + 4x + 4) f(x) = -4(x + 2)2 + 49 Note: parabola opens downward with vertex (-2, 49)

Convert to “vertex form” to find vertex, (h, k) Note the “x” value of vertex, h, and pick two other x values that are bigger than h, or two values that are smaller than h, and calculate corresponding y values Plot these three points on a graph and use symmetric properties of parabola to plot two more points on the other side of the axis Connect points with smooth curve with arrows at both ends

f(x) = x2 – 8x + 15 f(x) = 1(x – 4)2 – 1 Parabola opens upward with vertex (4, -1) Note: h = 4 (x coordinate of vertex) If we pick x = 3, we get y = 32 – 8(3) + 15 (3, 0) is a point on parabola If we pick x = 2, we get y = 22 – 8(2) + 15 (2, 3) is a point on parabola

Graph f(x) = 1(x – 4)2 – 1

f(x) = -2x2 + 4x + 3 f(x) = -2(x - 1)2 + 5 Parabola opens downward with vertex: (1, 5) If x = 0, then y = -2(0)2 + 4(0) + 3 = 3 (0, 3) is a point on the graph If x = -1, then y = -2(-1)2 + 4(-1) + 3 = -3 (-1, -3) is a point on the graph

Graph f(x) = -2(x - 1)2 + 5

Equation of Axis, Domain and Range of Quadratic Functions
Given vertex (h, k), the equation of the axis is: x = h Unless otherwise specified, the domain of a quadratic function is “all real numbers” The range depends on whether the parabola opens upward or downward: If upward the range is from “k” to positive infinity If downward, the range is from negative infinity to “k” Example: If vertex is (-2, 3) and “a” is positive: Equation of axis is: Domain is: Range is:

Homework Problems Section: 3.1 Page: 303
Problems: All: 1 – 4, Odd: 13 – 25 MyMathLab Assignment 3.1 for practice MyMathLab Quiz 3.1 is due for a grade on the date of our next class meeting

Polynomial Functions A polynomial function in “x” is defined as f(x) = a finite sum of terms, each of which has the form anxn where “n” represents a whole number and an is the coefficient of the variable xn The largest exponent in the finite sum of terms is called the degree of the function (the only exception being where the polynomial function is of the form: f(x) = 0. This is called the “zero polynomial” and has no degree.)

Examples of Polynomial Functions

Dividing Polynomial Function by “x – k” Using Long Division
This method has been discussed in previous classes: First write each polynomial in descending powers If a term of some power is missing, write that term with a zero coefficient Complete the problem exactly like a long division problem in basic math

Example

Dividing Polynomial Function by “x – k” Using Synthetic Division
First write each polynomial in descending powers If a term of some power is missing, write that term with a zero coefficient Set the problem up as a long division problem, but write only the coefficients of the dividend inside division symbol and “k” outside Put a blank line under dividend coefficients with an underline Drop the first coefficient below this line Multiply this coefficient by “k” and write the answer above the line below the second coefficient Add the two numbers and put answer beneath the line Repeat this process until you reach the end The numbers beneath the line represent coefficients of the answer and the remainder where the first term of the answer is of degree one less than the dividend and the last number is the remainder

Last Example by Synthetic Division

Homework Problems Section: 3.2 Page: 319 Problems: Odd: 1 – 17
MyMathLab Assignment 3.2(a) for practice

Remainder Theorem When a polynomial, ,is divided by the remainder is
Consider the last example where was divided by using synthetic division: The Theorem says that And it is:

Evaluating a Polynomial
It is usually easiest to evaluate a polynomial for a specific value of “x” by using synthetic division and the remainder theorem Example: Given Find:

Homework Problems Section: 3.2 Page: 319 Problems: Odd: 27 – 37
MyMathLab Assignment 3.2(b) for practice

Zeros of Polynomial Functions
Zeros of polynomial functions are the values of “x” that make “y = 0” “k” is a zero of f(x) if f(k) = 0 If a zero, k, of a polynomial function is a real number, then k is also an x-intercept of the graph of f(x) Polynomial functions may also have zeros that are non-real complex numbers (these are not x-intercepts of the graph)

Relationship Between Zeros and Factors
If k is a zero of f(x) and f(x) has degree “n”, then x – k is one factor of f(x), and the other factor is the polynomial of degree “n – 1” whose coefficients are shown in the bottom row of synthetic division Example: Given: Find:

Example Continued Synthetic Division shows that is a zero of , so is one factor, and the other factor is:

Homework Problems Section: 3.2 Page: 320 Problems: Odd: 39 – 55
MyMathLab Assignment 3.2(c) for practice MyMathLab Quiz 3.2 is due for a grade on the date of our next class meeting Section: 3.3 Page: 329 Problems: Odd: 5 – 21 MyMathLab Assignment 3.3(a) for practice

Rational Zeros Theorem
Given a polynomial function , with integer coefficients, where is the coefficient of the highest degree term and is the coefficient of the constant term, rational zeros must be the ratio of some factor of to some factor of Example: Find all possible rational zeros:

Example Continued

Using Rational Zeros Theorem on Polynomials with Non-Integer Coefficients
“k” is a zero of f(x), if and only if “k” is also a zero of the polynomial formed by multiplying a constant by f(x) To find zeros of f(x) when f(x) does not have integer coefficients, find zeros of af(x) where “a” is a factor that forms a polynomial with integer coefficients

Relationship Between Degree of a Polynomial Function and the Number of Distinct Zeros
A polynomial function of degree “n” has at most “n” distinct zeros Examples: A polynomial of: degree 2, has at most ___ distinct zeros degree 5, has at most ___ distinct zeros

Other Examples Given We can find zeros by solving

Non-Real Complex Zeros of Polynomial Functions
If a non-real complex number is a zero of f(x), then its conjugate is also a zero of f(x) Examples: If 3 - 2i is a zero, then _____ is also a zero If 1+ 5i is a zero, then _____ is also a zero

Exact Number of Zeros in a Polynomial of Degree “n”
A polynomial function, f(x), of degree “n” has exactly “n” zeros, but they may not all be distinct A number “k” is said to be a zero of multiplicity “m” if “m” is greater than 1 and is the largest exponent for which (x – k)m is a factor of f(x) In this case “k” counts as “m” of the “n” zeros of f(x)

Example Given: Degree: Exact Number of Zeros:
Number of Distinct Zeros: Distinct Zeros: What is the multiplicity of each zero?

Finding Zeros of a Polynomial Function
Use Rational Zeros Theorem to find all possible rational zeros Use Synthetic Division and Remainder Theorem to try to find one rational zero (remainder will be zero) If “n” is a rational zero, factor original polynomial as (x – n)q(x) Test remaining possible rational zeros in q(x). If one is found, factor again as in previous step Continue in this way until all rational zeros have been found See if additional irrational or non-real complex zeros can be found by solving a quadratic equation

Example Find all zeros of: Find all solutions to:
Rational Zeros Theorem says the only possible rational zeros are: See if -1 is a zero: Conclusion:

Example Continued This new factor has the same possible rational zeros: Check to see if -1 is also a zero of this: Conclusion:

Example Continued This new factor has as possible rational zeros:
Check to see if -1 is also a zero of this: Conclusion:

Example Continued Check to see if 1 is a zero: Conclusion:

Example Continued Check to see if 2 is a zero: Conclusion:

Example Continued Summary of work done:

Homework Problems Section: 3.3 Page: 330 Problems: Odd: 29 – 47
MyMathLab Assignment 3.3(b) for practice MyMathLab Quiz 3.3 (Shortened Version) is due for a grade on the date of our next class meeting

Further Hints on Finding All Zeros of a Polynomial Function
In trying to find all zeros of a polynomial function, it would be useful to know the number of positive and negative real zeros Descartes’ Rule of Signs - If f(x) is a polynomial function with real coefficients, written in descending powers, with a non-zero constant term The number of positive real zeros is equal to the number of sign changes in the terms of f(x), or is less than the number of sign changes by an even positive integer The number of negative real zeros is equal to the number of sign changes in the terms of f(-x), or is less than the number of sign changes by an even positive integer In both considerations, missing terms do not count as a sign change

Application of Descartes’ Rule of Signs
Find the number of positive and negative real zeros of: How many sign changes? Number of positive real zeros? Find f(-x): Number of negative real zeros?

Further Consideration of Previous Example
Consider all possible zeros: Total Zeros: Positive Negative Non-Real* * Since non-real zeros occur only as conjugate pairs there must always be an even number

Homework Problems Section: 3.3 Page: 331 Problems: All: 73 – 78
MyMathLab Assignment 3.3(c) for practice

Example of Finding Polynomial Functions with Specific Zeros
Given the fact that k is a zero of f(x) if and only if (x – k ) is a factor: Find a polynomial with zeros: 3, 1, and -2 Note: Other polynomial functions with the same zeros will be non-zero multiples of this one Other examples:

Additional Considerations Relative to Last Example
For the previous example, if we are also told that f(2) = 8, find the polynomial function with the specified zeros

Homework Problems Section: 3.3 Page: 330 Problems: Odd: 49 – 71
MyMathLab Assignment 3.3(d) for practice MyMathLab Quiz 3.3 is due for a grade on the date of our next class meeting

Turning Points of Polynomial Functions
The point at which the graph of a polynomial function changes from increasing to decreasing or vice versa is called a turning point of the function

Turning Points of Polynomial Functions
A polynomial function of degree n has at most n – 1 turning points with at least one turning point between successive zeros Given that the following function has zeros at 1 and -1, what do you know about its turning points? It has at most how many? Where is at least one of those located?

Graphs of Polynomial Functions
The domain of every polynomial function is: Polynomial functions are continuous over their domain (entire graph can be drawn without lifting pencil) Even polynomial functions, , are symmetric with respect to y-axis Odd polynomial functions, , are symmetric with respect to origin Note: Many polynomial functions are neither even nor odd

Graphs of Polynomial Functions
End behavior of graphs of polynomial functions is determined by characteristics of highest degree term: If degree of polynomial is odd and coefficient of highest degree is positive, ends look like: If degree of polynomial is odd and coefficient of highest degree is negative, ends look like: If degree of polynomial is even and coefficient of highest degree is positive, ends look like: If degree of polynomial is even and coefficient of highest degree is negative, ends look like:

Graphs of Polynomial Functions
x-intercepts of polynomial functions are real zeros of the function The y-intercept of a polynomial function, f(x), is f(0)

Sketching Graphs of Polynomial Functions
Find the real zeros of the function and plot them as x-intercepts Find and plot the y-intercept Use synthetic division to find values of the polynomial function between zeros Show end behavior of graph based on characteristics of highest degree term

Sketch Graph: Descartes’ Rule of Signs: Possible rational roots:
Test for rational roots:

Example Continued Solving equation by zero factor: Find y-intercept:
Find values of function between zeros:

Example Continued Find f(2) by synthetic division:

Sketch of Graph

Homework Problems Section: 3.4 Page: 343
Problems: Odd: 21 – 27, All: 29, 30, 37 – 40 MyMathLab Assignment 3.4(a) for practice

Intermediate Value Theorem
If “a” and “b” are in the domain of a polynomial function, f(x), and f(a) and f(b) have opposite signs, then f(x) has a least one real zero between “a” and “b” We can use this theorem to approximate the value of irrational zeros to any desired degree of accuracy as shown by the following example

Find a Zero to the Nearest Tenth:
Find f(1) Find f(2) There must be a zero between 1 and 2 Find f(1.5) Now we know there is a zero between:

Find a Zero to the Nearest Tenth:
Find f(1.8) There must be a zero between: Find f(1.7) Find f(1.75)

Find a Zero to the Nearest Tenth:
Since and We have now found that, to the nearest tenth, a zero is:

Homework Problems Section: 3.4 Page: 343 Problems: Odd: 43 – 51
Also, without using a graphing calculator, approximate the zero discussed in problems 45 and 47 to the nearest tenth MyMathLab Assignment 3.4(b) for practice

Boundedness Theorem Given a polynomial function, f(x), with real coefficients, of degree 1 or more, and a real number “c” in the domain of f(x), and finding f(c) by synthetic division: If c > 0 and all numbers on bottom row of synthetic division are non-negative, then f(x) has no zero greater than c If c < 0 and all numbers on bottom row of synthetic division alternate in sign (with 0 considered positive or negative as needed), then f(x) has no zero less than c

Example of Boundedness Theorem
In a previous example: we found f(2) by synthetic division: Notice bottom row. What does this tell us about zeros?

Example of Boundedness Theorem
Using the same function: find f(-3) by synthetic division: Notice bottom row. What does this tell us about zeros?

Homework Problems Section: 3.4 Page: 343 Problems: Odd: 53 – 59
MyMathLab Assignment 3.4(c) for practice MyMathLab Quiz 3.4 is due for a grade on the date of our next class meeting

Rational Functions A “rational function” is a function that can be written as: Any “real “ zeros of will make the rational function undefined and will establish “vertical asymptotes” for the graph of the function When the zeros of are non-real complex numbers, there will be no vertical asymptotes

Vertical Asymptote A “vertical asymptote” is a vertical line located at each real zero of the denominator polynomial, On either side a vertical asymptote the value of the rational function, ,approaches as as x gets closer to the real zero of The possible behaviors of the graph of a rational function near a vertical asymptote are illustrated on the next slide

Behavior of Rational Function Graphs Near Vertical Asymptotes

Finding Vertical Asymptotes of Rational Functions
Given a rational function: Find zeros of For each number, , that is a real zero of , is a vertical asymptote

Example of Finding Vertical Asymptotes
Find the vertical asymptotes: Find zeros of denominator:

Approx. Graph:

Homework Problems Section: 3.5 Page: 364
Problems: 37 – 45 (Vertical Asymptote) No MyMathLab Assignment

Linear Systems of Equations
When two or more linear equations are considered simultaneously they are referred to as a system of equations Example of System of Linear Equations in Two Variables: 2x – y = 7 3x + 7y = 2 The solution to this system is the set of all points, (x, y) pairs, that make both equations true at the same time If these represent different non-parallel lines, the solution is a single point. If they represent parallel lines, there is no solution. If they are different forms of the same line, then any point on the line is a solution.

Solving a System of Linear Equations in Two Variables by Substitution
Solve either equation for one variable (choose easiest) Substitute this value into the second equation Solve the second equation Substitute the solution to the second equation back into the first equation and solve it The solution to the system is shown as an (x, y) pair

Example 2x – y = 7 3x + 7y = 2 It is easiest to solve first equation for y: 2x – 7 = y Substitute into the second equation: 3x + 7(2x – 7) = 2 3x + 14x – 49 = 2 17x = 51 x = 3 Substitute into first equation: 2(3) – y = 7 - y = 1 y = -1 Solution: (3, -1)

Homework Problems Section: 5.1 Page: 484 Problems: Odd: 7 – 17
MyMathLab Assignment 5.1(a) for practice

Solving a System of Linear Equations in Two Variables by Elimination
Multiply one or both equations by a constant so that when the two equations are added, the result is a single equation in one variable Solve the resulting equation for that variable Substitute that value back into either original equation to find value of other variable Show solution as an ordered pair

Example of Solving a System of Linear Equations by Elimination
2x – y = 7 3x + 7y = 2 Multiply the first equation by 7 and add the result to the second equation to eliminate “y”: 14x – 7y = 49 3x + 7y = 2 17x = 51 x = 3 Substitute into either equation (first): 2(3) – y = 7 y = -1 Solution: (3, -1) (Same as before)

Homework Problems Section: 5.1 Page: 484 Problems: Odd: 19 – 29
MyMathLab Assignment 5.1(b) for practice MyMathLab Quiz 5.1(Shortened Version) is due for a grade on the date of our next class meeting This is the last assignment for a grade!!!!

Linear Equations in Three Variables
Equations of the form: Ax + By + Cz = D Solutions to these equations are “ordered triples” Example: 4x + 3y – z = -5 There are an infinite number of ordered triples that are solutions. One solution is (3, -5, 2)

Solving a System of Three Linear Equations by Elimination
Multiply one or two equations by a constant and add to eliminate one variable Use the same process on two different equations to eliminate the same variable Now find the solution to this system of two equations in two variables as previously learned Find the number solution for the third variable by substituting into any of the three original equations

Example 3x + 9y + 6z = 3 2x + y – z = 2 x + y + z = 2
Multiply second by 6 and add to first to eliminate “z”: 3x + 9y + 6z = 3 12x + 6y – 6z = 12 15x + 15y = 15 Also add second and third to eliminate “z”: 3x + 2y = 4

Example Continued 3x + 2y = 4 Multiply second by -5 and add to first:
Solve the system: 15x + 15y = 15 3x + 2y = 4 Multiply second by -5 and add to first: 15x + 15y = 15 -15x – 10y = -20 5y = -5 y = -1 Substitute into second equation at top to get: 3x + 2(-1) = 4 3x = 6 x = 2

Example Continued Substitute both x = 2 and y = -1 into any of the original equations (last): x y + z = 2 (2) + (-1) + z = 2 1 + z = 2 z = 1 Solution is: (2, -1, 1)

Homework Problems Section: 5.1 Page: 485 Problems: Odd: 47 – 57
MyMathLab Assignment 5.1(c) for practice MyMathLab Quiz 5.1 is due for a grade on the date of our next class meeting

Determinant Solution of Linear Systems
A “matrix” is a rectangular arrangement of numbers A matrix is designated as being an “m x n matrix” with “m” representing the number of rows, and “n” representing the number of columns Examples: A 2 x 2 matrix has two rows and two columns A 3 x 4 matrix has three rows and four columns

Determinant of n x n Matrix
Every matrix, A, that has an equal number of rows and columns has a real number associated with it called its “determinant” The determinant of a matrix, A, is indicated by the symbol: |A| The way in which the determinant is found depends on the number of rows and columns

Determinant of a 2 x 2 Matrix
The determinant of a 2 x 2 matrix: a b c d is defined as: ad – bc Example: If matrix B is: 4 -2 3 5 then |B| is: (4)(5) – (3)(-2) = = 26

Solving a System of Two Linear Equations in Two Variables using Cramer’s Rule
Write each equation in standard form Make a 2 x 2 “coefficient” matrix by taking the coefficients of “x” as the first column and the coefficients of “y” as the second column Call the determinant of this “coefficient matrix” by the name “D” Make an “X” 2 x 2 matrix by substituting in the “coefficient” matrix the constants on the right side of the equal sign for the “x” coefficients and find the determinant of this matrix, Dx Make a “Y” 2 x 2 matrix by substituting in the “coefficient” matrix the constants on the right side of the equal sign for the “y” coefficients and find the determinant of this matrix, Dy The solution to the equation will be: x = Dx/D and y = Dy/D except in the case where D = 0 when this method fails

Example of Solving a System by Cramer’s Rule
2x – y = 7 3x + 7y = 2 Coefficient Matrix: 2 -1 3 7 D = (2)(7) – (3)(-1) = 17 “X” Matrix: 7 -1 2 7 Dx = (7)(7) – (2)(-1) = = 51 “Y” Matrix: 2 7 3 2 Dy = (2)(2) – (3)(7) = 4 – 21 = -17

Example Continued From previous page: D = 17, Dx = 51, and Dy = -17
x = Dx / D x = 51 / 17 = 3 y = Dy / D y = -17 / 17 = -1 Solution for System: (3, -1)

Cramer’s Rule Applied to “Larger Systems”
Cramer’s Rule can be applied to systems containing “n” equations and “n” variables, but will not be discussed at this time.

Homework Problems Section: 5.1 Page: 484 Problems: Odd: 7 – 17
(Solve by Cramer’s Rule) No MyMathLab Assignment

Solving Systems of Equations by Gauss-Jordan Method
Must have same number of equations as variables Put each equation in standard form (variable terms in alphabetical order on left side of = sign and constants on right side) Make an “augmented matrix” consisting of a rectangular arrangement of variable coefficients, a vertical line, and constants as appear on the right side of the = sign

Gauss-Jordan Method Continued
With a goal of achieving an arrangement of “ones” down the diagonal and “zeros” in all other positions on the left side of the vertical line in the augmented matrix: Interchange any two rows Multiply numbers in any row by any nonzero number Replace any row by adding to its numbers the multiples of the numbers of another row

Gauss-Jordan Method Continued
When “ones” have been obtained down the diagonal from upper left to lower right, and “zeros” are in all other positions, the solutions to the system will be seen in alphabetical order on the right side of the vertical bar arranged from top to bottom

Example of Gauss-Jordan Method Applied to System with 2 Variables
2x + 3y = -1 5x – 2y = 26 Augmented Matrix:

Example Continued

Example Continued

Example of Gauss-Jordan Method Applied to System with 3 Variables
x + y – z = 0 2x + 3y + z = 5 -3x – y +2z = -4 Augmented Matrix:

Example Continued

Example Continued

Example Continued 1 0 0 3 0 1 0 -1 0 0 1 2 Solution to system:
Solution to system: x = 3, y = -1, z = 2 Correctly written as an ordered triple: (3, -1, 2)

Homework Problems Section: 5.1 Page: 484
Problems: Odd: 19 – 25, 47 – 53 (Solve by Gauss-Jordan) No MyMathLab Assignment

Direct Variation We say that y varies directly as x (or y is directly proportional to x) if there is a constant, k, such that: y = kx To solve a problem involving direct variation, the key is to find the value of k, from knowing one pair of (x, y) values Once the k value is determined, it establishes a formula, that can then find a value of y for any value of x

Direct Variation Example
Assuming that the pressure on an object underground varies directly as its depth, and the pressure is 5 psi when the depth is 10 feet, what would the pressure be at a depth of 30 feet?

Inverse Variation We say that y varies inversely as x (or y is inversely proportional to x) if there is a constant, k, such that: To solve a problem involving inverse variation, the key is to find the value of k, from knowing one pair of (x, y) values Once the k value is determined, it establishes a formula, that can then find a value of y for any value of x

Inverse Variation Example
Assuming that the illumination of light on an object varies inversely as the square of its distance from the object, and the illumination is 50 candela at 5 meters, what would the illumination be at 20 meters?

Joint Variation Joint variation occurs when y varies with multiple variables, directly with some and inversely with others In such cases, we write y as being equal to a constant, k, times the product of the other variables or inverses of the other variables, as appropriate To solve a problem involving joint variation, the key is to find the value of k, from knowing the value of one set of variables Once the k value is determined, it establishes a formula, that can then find a value of y given the values of the other variables

Joint Variation Example
Assuming that the gravitational attraction of an object varies directly as its mass and inversely as the square of the distance from the center of the mass, and the gravitational attraction is 33 newtons at 21 meters from the center when the mass is 2079 kilograms. What would the gravitational attraction be for the same mass at a distance of 3 meters from the center? .

Homework Problems Section: 3.6 Page: 373 Problems: Odd: 11 – 41
No MyMathLab Assignment