chapter 5 a detailed summary of all of the information within chapter 5

Chapter 5

A detailed summary of all of the information within chapter 5

Table of ContentsReview- Exponents…………………………………………….…….1 slidePractice- Exponents…………………………….…..........................1 slideLesson 5.1: Graphing Quadratic Equations ……………….……..8 slidesLesson 5.2: Solving Quadratic Equations with factoring…………5

slidesLesson 5.3: Solving Quadratic Equations with square roots……2 slidesLesson 5.4: Complex Numbers……………………………….……8 slidesLesson 5.5: Completing the Square……………………….………4 slidesLesson 5.6: The Quadratic Formula and the Discriminant………6 slidesLesson 5.7: Graphing and Solving Quadratic Inequalities………2 slidesLesson 5.8: Modeling with Quadratic Functions………………….5 slidesReview………………………………………………………………...5 slides

Review- Exponents

An Exponent is a multiple determining how many times the base is multiplied.

X2

Base

Exponent

Practice- Exponents

What is the value of 32 ?

32 = 3∙3 = 9

Remember that a Negative Exponent refers to “1/” the exponent.

What is the value of 3-4?

3-4 = 1/34 = 1/3∙3∙3∙3 = 1/81

Lesson 5.1: Graphing Quadratic Functions

A standard quadratic function is written in the form

Y=ax2+bx+cAnd appears in the shape of a parabola, or U-Shape

Curve. Parabolas are symmetrical over the Y-axis,

because exponents always create positive numbers.

Graph of Y=X2


Most Quadratic Equations (Such as x2+4x+3) have a zero, root, or solution.

x2+4x+3=0 (Which is equal to (x+1)(x+3)=0)

Roots

X=-1,-3 Which are solutions.

They are also Zeros, where (x,0) occurs.


The highest or lowest point on the U-Shaped Curve is known as the Vertex.

When an equation is written as Y=ax2+bx+c, the vertex’s X-value is located at –b/2a.


What is the vertex of Y=2x2-3x+4?

-b/2a = -(-3)/2(2) = ¾

Y=2(3/4)2 -3(3/4) +4 = 2(9/16) – 9/4 +4;

9/8 – 9/4 + 4 = 9/8 – 18/8 +4 = -9/8 + 32/8 = 23/8

So, the Vertex of this Equation is (3/4, 23/8).

Lesson 5.1: Graphing Quadratic Equations

After plotting the Vertex (3/4, 23/8), plot nearby points and connect them to form a parabola.

Y=2X2-3X+4Points based on Equation:

(0,4)

(-1,9)

(1,3)

(2,6)


Besides Standard Form, there are two other common Quadratic Expressions. Both are graphed in the same way, but provide information that can help

you.

Vertex Form Intercept Form

Y=a(x-h)2 +k Y=a(x-m)(x-n)

Where the vertex can easily be identified as

(h,k).

Where the intercepts, or zeros, (Y=0) can

easily be identified as (m, 0) and (n, 0), and the vertex X-value is the number between

the two points.


Graph Y=2(x-1)2 +4.

1. Find Vertex (h,k): (1,4)

2. Graph nearby points; (0,6) (2,5) (-1, 12) (3,12)

3. Connect the points.


Graph Y=3(x-3)(x-7).

1. Find the Y-Intercepts: (3,0) (7,0)

2. Find the X-value in-between, which is the vertex: (7+3)= 10; 10/2=5

3. Plot Vertex, (5,-12) and other points (4,-9), (6,-9)

4. Connect the points to draw the graph.

Lesson 5.2: Solving Quadratic Equations with Factoring

Solving a quadratic equation involves changing the changing the standard

quadratic equation

Y=ax2+bx+c

To a binomial term (intercept form) which will have two roots.

Y=(x-m)(x-n)


To do this, first set up two sets of parentheses.

Factor 2x2+x-3.

( )( )Then, examine a and c. Are either of them prime? If they are prime, there only factors are 1 and itself. Fill in the parentheses as necessary.

Somewhere in the parentheses, there is a 2x, an x, a 1 or -1, and a 3 or -3, Because 3 and 2 are prime.Now that you know the factors (2x, x, 3, 1), you need to organize them so that the two numbers multiply to c, and the x values add to b.


Try out some combinations using F.O.I.L.

FIRST: 2x∙x=2x2

INSIDE: -3∙x=-3x

OUTSIDE: 2x∙-1=-2x

LAST: -3∙-1=3


Combine like terms to get 2x2-5x+3; not the solution. But if 3 were positive, would it work?

Use F.O.I.L.

You’ll get 2x2+x-3; you’ve found the solution.


Here’s another method.Factor 6x2+x-2.

First, find factors of 6: 2/3, 3/2, -2/-3, -3/-2

Then, find factors of -2: -2/1, 1/-2, 2/-1, -1/2

Which pair would make sense? You want a positive x; therefore, you should use a pair of 2 and -1. 2∙3=6; so use the pair of positive 3.

You’ll get (2x+2)(3x-1). Using FOIL, change the digits to match the equation. The correct combination is (2x-1)(3x+2).

Lesson 5.3: Solving Quadratic Equations with Square Roots

Solve 2x2+5=41.

This isn’t too hard; start out by solving it normally:

2x2=36

x2=18

x=√18, OR

x=-√18.

However, remember that exponents always create positive numbers. That means there are two solutions here:

Lesson 5.3: Solving Quadratic Equations with Square Roots

Solve (2x-5)2=81.

Again, it will solve normally; but don’t forget that there are two solutions when it comes to exponents.

(2x-5)2=y+81

2x-5=9 –OR- 2x-5=-9

2x=14 –OR- 2x=-4

x=7, -2

Both solutions work perfectly well; it’s almost exactly like absolute value equations.

Lesson 5.4: Complex Numbers

Solve x2+15=-21.

This problem doesn’t make sense; you would need a negative square root. How can a square root be negative? It can’t. But, the imaginary number i is used to represent negative square roots.

Remember that i=√-1.

With the number i, unreal solutions can be simplified.

x2+15=-21

x2=-36

x=√-1∙36

x=i√36

x=6i; the solution


Complex Numbers are graphed on a real-imaginary plane, where the Y-axis represents the number of i, and the X-Axis represents the number of real numbers.


Where does our solution, 6i, go on this plane?

On the Y-axis, because 6i has no real numbers, only i’s.


How would you add (4-i) + (3+2i)?

Simply combine like terms.

(4+3) + (-i+2i)

7+i

Remember that i is always goes behind the real number, in this case 7 goes before the i.


How do you multiply Complex numbers?

To multiply Complex Numbers, remember that i2=-1. Other than that fact, the multiplication is standard.

What is 4i(6-i)?

4∙6(i) – 4(i2)

24i-4(-1)

24i+4

Solution: 4+24i


How do you divide Complex Numbers?Dividing complex numbers is more difficult than other operations. To divide Complex Numbers, it is similar to normal division except you must multiply by a complex conjugate (similar to an inverse).

A complex conjugate of a+bi is a-bi.

To practice this, we will solve -7+6i ∕9-4i.


-7+6i∕9-4i

Find C.C.: 9+4i

(9-4i)(9+4i)=97

-7+6i/97

Solution: -7/97 + 6i/97


Finally, how do you find the absolute value of a complex number?

To find the absolute value of a complex number, ignore the i and square all the terms, then find the square root. (This is similar to a distance formula; your finding the distance from 0). Remember absolute value is always positive.

What is the absolute value of 4-8i?

(4)2 + (-8)2

16+64=80; Find √80

Simplifies to 4√5

The absolute value of 4-8i is 4√5.

Remember, absolute values are always positive.

Lesson 5.5: Completing the Square

The Ultimate Goal of completing a square is to change a trinomial, such as this:

To a binomial squared, such as this:

x2+20x+100

(x+10)2


Find the value of c to make x2+15x+c a perfect square trinomial.

The easiest way to do this is to use the formula x2+bx+(b/2)2 = (x + b/2)2.

x2+15x+(15/2)2=(x+15/2)2

x2+15x+(225/4) = (x+15/2)2

Perfect Square Trinomial Resulting square of a binomial


It is very important not to forget the formula, x2 + bx +(b/2) = (x+(b/2)2, because it can be used for many different problems.


Solve x2+3x-1=0.

Remember (b/2)2

(3/2)2=9/4

Add one to both sides, x2 +3x = 1

Add 9/4 to both sides, x2 +3x +9/4 = 13/4

Regroup to a squ. Binomial, (x+3/2)2=13/4

Square Root both sides, x+3/2=√13/2 OR x+3/2=- √13/2

x= √13/2 -3/2, - √13/2 -3/2

Always, Always… Two Solutions!

Lesson 5.6: The Quadratic Formula and the Discriminant

There is a formula that can be used to find the exact value of a quadratic equation Y-intercept (When X=0).

Normally when graphing Quadratic Equations, you’re Y-intercept is difficult to determine, in terms of exact value.

Graph of Y=x2+3x-10


X= (-b +-√(b2-4ac) / 2a

Using ax2+bx+c, the value of the X-intercepts are:


So, on the equation we previously graphed, Y= x2+3x-10, the X-intercepts are:

X= (-(3) +√((3)^2-4(1)(-10)) / 2(1)

X= -3 +√9+40 / 2(1)

X= -3 +√49 / 2

X=4/2

X=2

X= (-(3) -√((3)^2-4(1)(-10)) / 2(1)

X= -3 -√9+40 / 2(1)

X= -3 -√49 / 2

X= -10/2

X=-5Final Solution: X=2,-5

Like completing the square, it’s all about remembering the basic equation.


A part of the Quadratic Formula can be used to tell how many solutions a Quadratic Equation has. It could have two real solutions, one real solution, or two imaginary solutions (i.e. no real solution). It is called the Discriminant.

Blue: Two Solutions

Green: One Solution

Red: Two Imaginaryhttp://www.algebra.com/


The Discriminant formula is b2-4ac.

If:

>0, then the equation has two real solutions.

=0, then the equation has one real solution.

<0, then the equation has two imaginary solutions.


How many solutions are there for:

(x-3)2

x2-6x+8

x2-6x+10

Change to standard form: x2-6x+9

(-6)2 -4(1)(9) = 36-36 = 0; One Real Solution

(-6)2 -4(1)(8) = 36-32 = 4 = 4>0; Two Real Solutions

(-6)2 -4(1)(10) = 36-40 = -4 = -4<0; Two Imaginary Solutions

Lesson 5.7: Graphing and Solving Quadratic Inequalities

To graph a Quadratic Inequality:

1. Graph the Equation, as if Y=x2…

2. Remember to dash the line for < or >, shade for ≤ or ≥.

3. Use a Test Point to determine whether to shade inside or outside the parabola.

Lesson 5.7: Graphing and Solving Quadratic Inequalities

Graph Y ≥ -2x2-x+3.

1. Find the Vertex (-b/2a): 1/-4 = -(1/4); (-1/4, 25/8)

2. Plot the graph as normal: (0,3) (-1,2) (1,0) (-2,-3)

3. ≥ Means you should have a solid line (Draw that)

4. Test (0,0). (0,0) is NOT a solution (0≥3?), so shade OUTSIDE the parabola.

5. The Graph is Complete.

Systems of Inequalities work in the exact same way.

Lesson 5.8: Modeling with Quadratic Functions

Sometimes with Quadratics you will already have a graph, and instead need to find out the equation of the graph.

A good memory of the different forms of equations (Intercept, Standard, Vertex) can help you find out the equation quickly.


What is the equation for this graph?

(1,0)(-2, 0)

(-1,-6)

You know the two intercepts, so try graphing it in intercept form.

Y=a(x-m)(x-n)

Y=a(x+2)(x-1)

Test (-1,-6); -6=a(1)(-2)

-6=-2a

a=3Therefore, the equation of this graph is

Y=3(x+2)(x-1)


What is the equation for this graph?

Vertex: (1,0)

(-1,-3)

You know the Vertex, so try using Vertex Form this time.

Y=a(x-h)2 +k

Y=a(x-1)2+0

Test (-1,-3); -3=a(-1-1)2

-3=a(4)

a=-3/4

So, the equation of this graph is Y=-3/4(x-1)2


What is the equation of this graph?

(0,1)(-4,0)

(-5,-4)

You can’t use Intercept- or Vertex-Form here… but because there are three points, you can make three equations with standard form.

Y=ax2+bx+c

(0,1): 1=a(0)2+b(0)+c; 1=c

(-4,0): 0=a(-4)2+b(-4)+c; 0=16a-4b+c

(-5,-4): -4=a(-5)2+b(-5)+c; -4=25a-5b+c


With these three equations, you can find the value of a, b, and c.

c=1

16a-4b+c=0

25a-5b+c=-4

a= -3/4

b= -11/4

c=1

Y=(-3/4)x2-(11/4)x+1So, the final equation is

Chapter 5 Review

Factor 8y2-28y-60.

Start off by removing like terms (In this case, 4):

2y2-7y-15

Find the factors of the numbers:

2: 1,2 or -1,-2

-7: -7,1 or -1,7

-15: 3,-5 5,-3 or 15,-1 1,-15

By trial-and-error, find the correct combination:

(2y+3)(y-5) = 2y2-7y-15 (2y-15)(y+1) = 2y2-13y-15

(2y-5)(y+3) = 2y2+y-15

Chapter 5 Review

Find y if the area of the trapezoid is 71units2. y+4

9y-10

y

Recall the formula for area of a trapezoid:

A=((b1+b2)/2) *h

Add in the information given:

((y+9y-10)/2) *(y+4)

(5y-5) *(y+4)

5y2+15y-14 = 71

Solve with square roots:

5y2+15y-14 = 71 5y2+15y=85

y2+5y=17 y2+5y+6.25=23.25

(y+2.5)2=23.25

y+2.5=√23.25

y= √23.25 +2.5

y=~7.32

Chapter 5 Review

How many natural numbers between 1 and 50 have a value of 3 when the number’s square is added to eight times the number and subtracted by 2?This is a word problem. It is important in word problems to convert the words into an algebraic expression.

the number’s square is added to eight times the number and subtracted by 2

have a value of 3

x2+8x-2

= 3

Knowing the solution needs to be between 1 and 50 is not useful here until the solution is found!

Chapter 5 Review

Simplify: x2+8x-2=3

Simplify: x2+8x-5=0

Recall the quadratic formula:

X= (-b +-√(b2-4ac) / 2a

x= (-8+-√(82-4(1)(-5) /2(1)

x= (-8+-√(84) /2

x= ~.58258 or ~-8.58258

Chapter 5 Review

Recall the list of possible answers.

.58258<1; not a solution, because 1<x<50

-8.58258<1; not a solution, because 1<x<50

So, the answer is that there are no whole numbers that fit the requirement.

chapter 5 a detailed summary of all of the information within chapter 5

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