## Introduction

Have you ever wondered how a business knows how to maximize their profit or how a human cannon successfully works? Have you ever seen roller coasters, bridges or water fountains? These are examples of quadratic relations. In grade 9, we learned about linear systems. We learned the equation y=mx+b, and that it's graph was a straight line. A quadratic relation's graph is a U-shape. There are many different forms of equations of a quadratic relation.

The Different Equations of A Quadratic Relation Are:

• Vertex Form: a(x-h)^2+k
• Factored Form: a(x-r)(x-s)
• Standard Form: ax^2+bx+c

Chapter 4

• Characteristics of a Parabola
• Finite Differences
• Vertex Form Equation
• Making Equations
• Mapping Notation and the Mapping Notation Formula
• Finding the Intercepts From Vertex Form
• Quadratic Word Problems and Examples
• Assessment 1: Chapter 4 Unit Test

Chapter 5

• Multiplying Polynomials and Review
• Special Products
• Common Factoring
• Simple Trinomial Factoring
• Complex Trinomial Factoring
• Factoring Special Products
• Factoring Summary
• Solving Quadratic Equations in Factored Form
• Graphing Quadratic Relations in Factored Form
• Assessment 2: Factoring Quiz

Chapter 6

• Completing The Square
• Discriminants
• Graphing Quadratic Relations in Standard Form
• Solving Word Problems Using The Quadratic Formula
• Assessment 3: Chapter 6 Unit Test

Reflection

## The Parabola

The graph of a quadratic relation is U-shaped and it is called a parabola.

Characteristics of a Parabola

• The zeros of the parabola is where the graph crosses the x-axis. Zeros can be called x-intercepts or roots.
• The axis of symmetry divides the parabola into two equal halves. The axis of symmetry is the x-coordinate of the vertex. It is labeled x=
• The optimal value is the value of the maximum or minimum point of the parabola. It is the y-coordinate of the vertex. It is labeled y=
• The vertex of the parabola is the point where the axis of symmetry and the optimal value meet. It is the point where the parabola is at its maximum or minimum value.
• The y-intercept of the parabola is where the graph crosses the y-axis.
• Parabolas can open up or down
• Parabolas are symmetrical

• The graph of a quadratic relation is U-shaped and it requires a larger number of points to draw its graph than a linear relation.
• An easy method for graphing parabolas involves preparing a chart. (This method is not always accurate since it will not always give you some of the key features of a parabola such as the vertex and the x-intercepts)

## Finite Differences

To tell if a relation is quadratic, linear or neither we use finite differences (first and second differences).

How do we find first differences?

To find first differences, we subtract the second y-value with the first, then the third y-value with the second and so on. If the first differences are the same, that means the relation is linear.

Therefore this relation is linear.

How do we find second differences?

To find second differences, we subtract the first differences. We subtract the second first-difference with the first first-difference and we subtract the third first-difference with the second first-difference, and so on. If the second differences are the same that means the relation is quadratic. If the second differences are not the same, that means the relation is neither quadratic nor linear.

## Vertex Form Equation and Transformations of Quadratics

• The most basic form of the parabola is y=x^2 (where no transformations have been made).
• The equation of a graphed parabola in vertex form is:

Where:

• y and x do not change
• a makes the parabola narrower or wider. It is the vertical stretch or compression factor. If the a value is higher than 1 (or less than than -1), the graph will be vertically stretched. This means that the parabola will be narrower. If the a value is less than 1 (or more than -1), the graph will be vertically compressed. This means the parabola will be wider. If the a value is negative, the parabola will open downwards. If the a value is positive, the parabola will open upwards.
• h is the x-coordinate of the vertex. It is the horizontal translation. (Side to side)
• k the y-coordinate of the vertex. It is the vertical translation. (Up or down)
• The vertex=(h,k)
• The values a,h, and k change the graph of y=x^2 by making it narrower or wider or by moving it up, down, or to the side

For example:

## Making Equations

Making equations when given your vertex and translations

• The vertex: The x-value of the vertex will be your h in the equation. In the equation, the h-value is always the opposite sign of the x-value of the vertex. (For example, if the vertex is (-4,-5), the h value will be 4).
• The y value of the vertex is your k value in the equation.
• The vertical stretch or compression factor and the direction of the opening is your a value. If the parabola is opening up, the a value will be positive. If the parabola is opening down, the a value will be negative.

For example:

Making equations when given a vertex and a point.

• The vertex: The x-value of the vertex (the axis of symmetry) will be your h in the equation. In the equation, the h-value is always the opposite sign of the x-value in the vertex.
• The y value of the vertex (the optimal value) is your k value in the equation.
• Sub in the given y and x values into your equation and isolate a to find the a value. Once you have found the a value, sub it into your equation with the h and k values.

For example:

To check if your answer is correct, sub in the x-value from the given point into your equation and see if it gives you the y-value of the given point.

## Mapping Notation

Mapping notation is a strategy used to graph any quadratic relation. Before we can use mapping notation, we need to know the “key” points of the basic quadratic relation (y=x^2) which is the simplest parabolic curve with no transformations applied.

The key points are:

## Mapping Notation Formula

Mapping formula shows exactly how the graph of y=x^2 changes into the equation

y=a(x-h)^2+k.

Mapping Notation Formula:

x+h, ay+k

For example, the equation y=3(x-2)^2-4 would become x+2 because the parabola is being horizontally translated 2 units to the right, changing only the x values (it is x+2 instead of x-2 because the signs always become the opposite for the h value), and 3y-4 because the parabola is being vertically stretched by a factor of 3 and it is being vertically translated down by 4 units.

The Chart Method

When you have to graph a parabola and you were only given the equation, it is helpful to make a table. Use the key points for your x-values (-2, -1, 0, 1, 2) and plug them into the equation to find your y-values. Then plot your points!

• This method is not always accurate since it will not always give you some of the key features of a parabola such as the vertex and the x-intercepts

Mapping Notion Formula

You can also use the mapping notion formula. You can use the mapping notation formula when you are given an equation or transformation by creating the formula and subbing some of the key points for the graph of y=x^2 into the formula.
• This method is more accurate and it will give you the vertex and other key points.

Mapping Notation

When you have to graph a parabola and you were only given a transformation or an equation, you can use mapping notation. Look at the points of the graph y=x^2 and see how your y-values and x-values have changed. Be sure to also look at the vertical stretch and compression factor and see if the parabola is opening up or down.

The Step Pattern

The step pattern is helpful for graphing parabolas when given the vertex form equation of a quadratic relation.

For example:

## Finding The Intercepts From Vertex Form

The vertex from is y=a(x-h)^2+k

To find the y-intercept, sub in x=0

To find the x-intercept, sub in y=0

For example:

When you are square rooting to get rid of the squared sign on one side, the answer of the square root must be positive and negative. This is because a parabola can have two x-intercepts. If you are square rooting a negative number, that means the quadratic relation does not have any x-intercepts.

1. What is the maximum height of the ball?

Ans. Find the vertex, the y-value is the maximum height y=k

2. At what time did the ball reach the maximum height?

Ans. Find the vertex, the x-value is the time when the max height occurred x=h

3. What was the initial height of the ball (firework, rocket, etc)?

Ans. Set x=0 and solve for y (the height of the ball when time=0) or find the y-intercept

4. How long was the ball in the air?

Ans. Set y=0 and solve for x (the time when the ball hit the ground, when height=0). Find the x-intercept

5. When did the ball hit the ground?

Ans. See above………….. The time when the ball hit the ground, when height=0

6. At what time did the ball reach the maximum height?

Ans. Find the vertex, the x-value is the time when the max height occurred x=h (the axis of symmetry)

7. What is the height of the ball at 3 seconds?

Ans. Sub the x-value into the equation to solve for y (sub t into the equation to solve for h)

Profits (revenue) can be maximized and this relation can be modeled by a quadratic relation.

1. The relationship between selling price and profit can be modeled by a quadratic relation.

2. The relationship between selling price and the number of units sold can be modeled by a quadratic relation

EXAMPLE:

A concert promoter models the profit from her next concert, P dollars, by the equation P=-1(t-55)^2+10571, where t dollars is the cost per ticket.

A) What price should she charge to maximize her profit? What is the maximum profit?

What is the x value of the vertex? That is the price she should charge to maximize the profit. What is the y value of the vertex? That is the the maximum profit.

How do we solve it?

Look at the equation. The h value is the x-value of the vertex. That would be the price she would charge to maximize her profit. (Remember: the h value is always the opposite sign!) Look at the k value. The k value is the y-value of the vertex. That would me the maximum profit.

Solution:

She should charge \$55 to maximize the profit. The maximum profit is \$10571.

B) What will be her profit if the tickets were free?

How much money will she make if the price of the tickets is \$0?

How do we solve it?

Find the y-intercept. Sub in 0 for t and solve your equation.

Solution:

P= -11(t-55)^2+10571

P= -11(0-55)^2+10571

P= -11(-55)^2+10571

P= -11(3025)+10571

P= -33275+10571

P=-22704

Therefore if her tickets were free, her profit would be \$-22704.

C) How much should she charge per ticket to break-even?

How much would the tickets cost so that her profit would be \$0?

How do we solve it?

Find the x-intercepts. Sub in 0 for P and solve the equation.

Solution:

EXAMPLE:

A flare is released into the air following the path, h=-5(t-6)2+182, where h is the height in meters and t is the time in seconds.

a) What was the flare's maximum height?

The maximum height of the flare was 182m.

b) What was the flare's initial height?

h=-5(t-6)^2+182

h=-5(0-6)^2+182

h=-5(-6)^2+182

h=-5(36)+182

h=-180+182

h=2

Therefore, the flare's initial height was 2m.

c) How long was the flare in the air?

h=-5(t-6)^2+182

0=-5(0-6)^2+182

0-182=-5(0-6)^2

-182=-5(0-6)^2

-182/-5=-5(0-6)^2/-5

/36.4=(t-6)^2

6.03=t-6

For the next step, make one of your answers to the square root positive and the other one negative.

6.03+6=t

12.03=t

-6.03+6=t

-0.3=t

Therefore the flare was in the air for 12.03 seconds

## Polynomials Review and Multiplying Polynomials

You can classify polynomials by two ways. You can classify them by the number of terms or the highest exponent.

1. When we classify by the number of terms, the suffix 'nominal' is used behind the prefix accorded each number.

1-MONOMIAL

2-BINOMIAL

3-TRINOMIAL

4/MORE- POLYNOMIAL

2. When we classify by the highest exponent, we use:

• Linear for any polynomial with highest exponent of 1
• Quadratic for any polynomial with the highest exponent for 2
• Cubic for any polynomial with the highest exponent of 3
• Quartic for any polynomial with the highest exponent of 4

Simplifying polynomials includes both collecting like terms and use of distributive property. It is important to be sure to use BEDMAS at all times. Exponent rules must be used for both multiplication and division when simplifying.

Exponent Rules

• All numbers and variables without exponents have an exponent of 1
• When multiplying, you add the exponents and multiply the coefficients
• When dividing, you subtract the exponents and divide coefficient

To multiply polynomials, use distributive property, then collect like terms

For Example:

## Special Products

Perfect Square

You can use the pattern for squaring a polynomial.

1. Square the first term
2. Twice the product of the terms/add two middle terms after expansion
3. Square the last term
4. e.g. (a+b)^2=a^2+2ab+b^2 or (a-b)^2=a^2-2ab+b^2

For example:

Product of Sum and Differences

When you multiply the sum and the difference of two terms, the two middle terms are opposite, so they add to zero.

(a-b)(a+b)=a^2-b^2

For example:

## Common Factoring

Factoring is the opposite of expanding. Expanding involves multiplying, while factoring involves looking for the exprssions to multiply.
One method of factoring a polynomial is to look for the greatest common factor of its terms.

Monomial Common Factor

1. Find the GCF of coefficients and variables
2. Divide each term by the GCF

Binomial Common Factor

If there are two binomials that are exactly the same, consider that as a binomial common factor.

x(x-2)+2(x-2)

(x-2)(x+2)

5x(3x+2)+4(3x+2)

(3x+2)(5x+4)

Factor By Grouping

Factoring groups of two terms with a common factor to produce a binomial common factor.

ax+ay+2x+2y

(ax+ay)+(2x+2y)

a(x+y)+2(x+y)

(x+y)(a+2)

## Simple Trinomial Factoring

Simple Trinomial: ax^2+bx+c

Where x is a variable and a,b,c are constants where a does not equal 0. A simple trinomial is quadratic where a=1. Given a quadratic in standard form, you can factor to get factored form.

ax^2+bx+c=(x+s)(x+r)

Step 1: Find the Product and Sum

• Find two numbers whose product is c
• Find two numbers whose sum is b

For example: x^2+12x+27

Find two numbers whose product is 27.

(1)(27)

(3)(9)

Find two numbers whose sum is 12.

3+9=12

x^2+bx+c=(x+r)(x+s)

x^2+12x+27=(x+3)(x+9)

Step 2: Look at the signs of b and c in the given expression x^2+bx+c

• If b and c are positive, both r and s are positive.
• If b is negative and c is positive, both r and s are negative.
• If c is negative one of r or s is negative.
• If both b and c are negative, one of r or s is negative.

Simple Trinomials With A GCF

Factor completely by first removing the GCF

Divide the equation by the GCF

Put the GCF outside of the bracket

For example:

3x^2+12x+9

3(x^2+4x+3)

(1)(3)

3(x+1)(x+3)

Word Problems

Dimensions of a Water Garden

The area of a water garden is x^2+5x+6.

Determine binomials that represent the dimensions of the rectangular water garden.

• Change the expression x^2+5x+6 into factored form by factoring

x^2+5x+6

(3)(2)

(x+3)(x+2)

Therefore the dimensions of the rectangular water garden are (x+3) and (x+2).

Determine the dimensions if x represents 1m.

• Sub in x=1 for each bracket and solve.

(x+3)

(1+3)

4

(x+2)

(1+2)

3

Therefore the dimensions of the water garden if x represents 1m are 4m and 3m.

## Complex Trinomial Factoring

1. Always look at the common factor first when factoring trinomials
2. To factor ax^2+bx+c, find two integers whose product is ac and whose sum is b.
3. Break up the middle term and factor by grouping.

Example:

3x^2+8x+4

(3)(4)=12

(2)(6)= 8

3x^2+2x+6x+4

(3x^2+2x)(6x+4)

x(3x+2)+2(3x+2)

(x+2)(3x+2)

Word Problem

The area of the rectangle is 6x^2+17x+12.

Factor to find the algebraic expression for the length and width of the rectangle.

6x^2+17x+12

(6)(12)=72

9+8=17

6x+9x+8x+12

(6x+9x)(8x+12)

(3x+4)(2x+3)

Therefore the length of the rectangle is (3x+4) and the width of the rectangle is (2x+3)

If x=5cm, determine the perimeter and area of the rectangle.

• Sub in the dimensions into the perimeter and area formulas. Then sub x=5 and solve.

P=2l+2w

P=2(3x+4)+2(2x+3)

P=2(3(5)+4)+2(2(5)+3)

P=2(15+4)+2(10+3)

P=2(19)+2(13)

P=38+26

P=64cm

A=(l)(w)

A=(3x+4)(2x+3)

A=(3(5)+4)(2(5)+3)

A=(15+4)(10+3)

A=(19)(13)

A=247cm^2

Therefore the perimeter of the rectangle is 64cm and the area of the rectangle is 247cm^2.

## Factoring Special Products

Factoring Perfect Square Trinomials

(a+b)^2=a^2+2ab+b^2

(a-b)2=a^2-2ab+b^2

To factor perfect squares, look at the fist term and the last term and find their square root. Make sure double the product of the square roots equals the middle term. Look at the middle term to see if it is positive or negative. If it is positive, the equation will be (a+b)^2. If the middle value is negative, the equation will be (a-b)^2.

For example:

Factoring Differences of Squares

a^2-b^2=(a+b)(a-b)

To factor differences of squares, look for GCF and divide all the terms by it. Then find the square root of both the values and put the values into factored form.

For example:

## Solving Quadratic Equations In Factored Form

To "solve" a quadratic equation means to find the x-intercepts which are also called the roots, or zeros.

Steps
1. Make one side equal zero 0=(x-r)(x-s)
2. Set each bracket equal zero (x-r)=0 (x-s)=0
3. Solve for x: x=r x=s (the zeros are r and s, so (r,0) and (s,0)

Use factoring to graph the quadratic equation
1. Find the x-intercepts
2. Find the vertex
3. Graph

How to Find the Vertex
1. Axis of Symmetry (AOS)= Add zeros/2
2. Sub in the axis of symmetry into the equation to find your optimal value

For example:

## Graphing Quadratic Relations in Factored Form

1. To graph equations in factored form, first find the x-intercepts by making each bracket equal zero.
2. Then, find the axis of symmetry. To find the axis of symmetry, add the two x-intercepts and divide them by two.
3. Once you have found the axis of symmetry, plug it into the equation and solve for y. To find the y-intercept, set x to 0 in your equation and solve for y. Now you have 4 points to plot on a graph.

## Word Problems

Area Word Problems

A rectangle has dimensions 2x^2+2x-4 and x^2+2. Its area is 11m^2, determine the value of x.

• Make an expression by multiplying the length and the width of the rectangle. Then expand and simplify the equation. Next, sub in the area for y. Make one side equal zero, and move the area to the other side. Then collect the like terms.

A= (l)(w)

A= (2x^2+2x-4)(x^2+2x)

A= 2x^2+2x-4+x^2+2x

A= 2x^2+x^2+2x+2x-4

A= 3x^2+4x-4

11=3x^2+4x-4

0= 3x^2+4x-4-11

0= 3x^2+4x-15

0= 3x^2+9x-5x-15

Therefore the value of x is 3x^2+9x-5x-15.

Find The Area of The Shaded Region

a) Write an algebraic expression for the area of the shaded region.

• Subtract the area of the white rectangle by the area of the square. Then expand and simplify the expression.

b) Substitute x=7 into the expression to find the area.
Flight Word Problems

The path of a toy rocket is defined by the relation y= -3x^2+11x+4, where x is the horizontal distance traveled in meters and y is the height, in meters, above the ground.

a) Determine the zeros in the relation.

• Change the equation into factored form. Set the equation to 0. Find the x-intercepts by setting the brackets to 0.
b) How far has the rocket traveled horizontally when it lands on the ground?
• Find the x-intercepts. If one of the x-intercepts is negative, it is invalid and you use the positive zero.

For this relation, the x-intercepts are (4,0) and (-1/3,0) (as shown above). Since -1/3 is negative, it is invalid in this relation.

Therefore the rocket has traveled 4m horizontally when it lands on the ground.

c) What is the maximum height of the rocket above the ground, to the nearest hundredth of a meter?

• Find the x-intercepts by factoring and setting the brackets to zero. Next, find the axis of symmetry by adding the two x-intercepts and dividing them by two. Once you have found the axis of symmetry, plug it into the equation for x and then solve for y. The y-value will be the maximum height of the rocket.

## Completing The Square

Not all quadratic equations are written in Vertex Form, y=a(x-h)+k. Functions can be written in standard form, y=ax^2+bx+c, and can be converted to vertex form using a process called completing the square. It is useful for us to convert functions into vertex form so we can easily see the vertex or graph it using transformations.

Completing the square steps:

1. Group the x-terms
2. Common factor the 'a' from the x-terms. If the a value is -1, factor out the negative. Remember to just factor out the number in front, you do not want to factor out an x! If there is no 'a' value, skip this step.
3. Divide the coefficient of the middle term by 2, square it, then add and subtract that number inside the brackets.
4. Remove the subtracted term from brackets. On the way out of the brackets, it is multiplied by the 'a' value that you factored out.
5. Factor the brackets (trinomial) as a perfect square trinomial

For example:

Graphing

• There is always a solution
• Not as accurate

Factoring

• There is sometimes a solution

Completing The Squares

• Always a solution
• Best method when the b value is even

• There is always a solution
• It is the only formula that gives you the exact answer

The quadratic formula is mostly used when a certain quadratic cannot be factored. This formula was developed by completing the square and solving the quadratic ax^2+bx+c=0. The formula gives us the solutions or roots of the quadratic equation.

TIP: If your square root has decimals round to three decimal places to get the most accurate answer. Round your final answer to the number of decimal places specified in the question.
For example:

## Discriminants

• All quadratic equations of the form ax^2+bx+c=0 can be solved using the quadratic formula.
• The value of the discriminant (under the square root) tells you how many real solutions a quadratic equation has.
For example:

## Graphing Quadratic Relations In Standard Form

Completing the Square

To graph a quadratic relation in standard form, you can use the method of completing the squares. First convert your standard form equation into vertex form by completing the squares. You now know your vertex. Next find the x-intercepts by subbing y=0 into your vertex form equation. The c value in the standard form equation is your y-intercept. Now plot these points on your graph. Since a parabola is symmetrical, plot a point that is opposite to your y-intercept (a point that has the same y-value as the y-intercept and it is the same distance away from the vertex, however it is on the opposite side of the vertex).

To graph a quadratic relation in standard form, you can also use the quadratic formula. Find the x-intercepts using the quadratic formula. Then find your axis of symmetry and optimal value. Next, use the c value in the standard form equation (it is your y-intercept). Once you have plotted your points on the graph, plot the point that is opposite to your y-intercept.

1) A fireball is fired and follows a path modeled by y=-0.1x^2+1.6x+5.7, where x is the horizontal distance from the start, and y is the vertical height of the ball.

a) How high does the fireball start?

• Look for the y-intercept. It is the c value in the equation.

The fireball starts at a vertical height of 5.7m.

b) How high must a protective wall be to protect us from the fireball?

• Find the vertex. To find the vertex, find the x-intercepts using the quadratic formula. Next, add the two x-intercepts and divide by 2. This will give you the axis of symmetry. Sub this value into x in the equation to find the y-value of the vertex.

c) Where will the fireball land?

• Find the x-intercepts using the quadratic formula. A negative number is not valid in this relation.

2) Jagdeep the pirate has a cannon that's underwater. The height of the cannonball above water level t second after it was fired is given by h=-4.9t^2+20.6t-8

a) How far underwater is the cannon?

• Look at the y-intercept. It is the c value in the equation. If the c value is negative, write your answer as a positive number.

The cannon is 8m underwater.

b) How long is the cannon ball above water?

• Find the x-intercepts using the quadratic formula. If both the x-intercepts are negative, subtract the x-intercepts. If only one of the intercepts is positive, use the positive value since the negative value is invalid.

c) How long is the cannon ball above a height of 2m?

• Sub 2 into the equation for the value of y. Set the equation to zero (move the 2 to the other side) and use the quadratic formula.

3) A baseball is thrown from the top of a building and falls to the ground. It's path is approximated by the relation: h=-4t^2+32+4, where h is the height above the ground in meters and t is the elapsed time in seconds.

Area Word Problems

1) The length of a rectangle is 6 inches more than it's width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.

• Create a diagram. Define your variables. Since area equals length times width, multiple the length by the width. Expand and simply the equation. Plug the area into the equation and set one side to zero. Then use the quadratic formula.

2) The length and width of a rectangle are 4cm and 3cm. A frame is placed around the rectangle. The frame increases the length and the width by the same amount. The area of the new rectangle is 30cm^2. Find the dimensions of the new rectangle to the nearest tenth of a meter.

• Create a diagram. Define your variables. Since area equals length times width, multiply the length by the width. Expand and simply the equation. Plug the area into the equation and set one side to zero. Then use the quadratic formula.
Triangle Word Problems

The length of one leg of a right triangle is 17cm more than that of other leg. The length of the hypotenuse is 4cm more than triple that of the shorter leg. Find the lengths of each of the three sides.

• Make a diagram and define you variables. Label each side of the triangle. Put your dimensions into the equation a^2+b^2=c^2. Remember that c is the length of the hypotenuse (the longest side of a right triangle). Expand and simplify your equation. Make one side equals zero. Put your equation into the quadratic formula and solve. Sub in the x-intercepts into your dimensions.

Consecutive Number Word Problems

1) The product of two consecutive even integers is 63. What are the numbers?

• Define your variables. Multiply your two variables and make it equal 63. Expand and simplify the equation. Move 63 to the other side to make the equation equal zero. Use the quadratic formula.

2) The sum of the squares of two consecutive integers in 365. Find the integers.

• Define your variables. Square your two variables and add them. Make the equation equal 365. Expand and simplify the equation. Move 365 to the other side to make the equation equal zero. Use the quadratic formula.

Revenue Word Problems

Calculators are sold to students for 20 dollars each. Three hundred students are willing to buy them at that price. For every 5 dollar increase in price, there are 30 fewer students willing to buy the calculator. What selling price will produce the maximum revenue and what will the maximum revenue be?

1. Make an equation in factored form. One bracket will represent price and the other bracket will represent the number of calculators sold.
2. Find the x-intercepts
3. Find the axis of symmetry. Remember this will not represent the price.
4. Find the optimal value by plugging the axis of symmetry into the equation.

## Reflection

This unit on quadratics has been very interesting. It is one of the most used functions in math and we see quadratic relations in everyday life. For example, we see parabolas in nature such as rainbows and valleys; in structures such as bridges, roller coasters, buildings, and water fountains; and in logos. Quadratics also helps us to maximize or minimize profits and costs and measure the maximum or minimum heights of an object.

In this unit I learned a lot. I learned about the different equations for quadratic relations and how to solve them. I also learned about how to apply quadratics to real life. I think that once you understand quadratics and the different equations for the relations, it is not hard to solve the questions; however, it is important to look for small mistakes such as looking for a GCF in equations and making sure you have added or multiplied correctly.