A Complete Visual Course

V Narayan

figures from animated-mathematics.net

No part of this book may be reproduced or used in any manner without the express written permission of the publisher except for the use of brief quotations in a book review

Preface

Chapter 1: Introduction

Chapter 3: Introduction to Quadratic Transformations

Chapter 4 Intersection With x and y Axis

Chapter 5 Factorization of Quadratic Equations with c=0.

Chapter 6 Factorization

Chapter 7: Designing a Quadratic Graph.

Chapter 8: A Perfect Square and Transformations.

Chapter 9: Completing the Square and Transformations

Chapter 10: The Quadratic Formula and Discriminant.

Chapter 11: Intersections with Lines and Quadratic Graphs

Chapter 12: Applications of Quadratic Equations

# Preface

The sequence of topics covered in a typical math program is shown below:

Quadratic equations is the final topic before calculus which is considered by many to be a complicated topic. The fundamentals of calculus are very elegant and simple, and should not be feared by students. A thorough graphical understanding of quadratic equations serves as good preparation for understanding visual concepts in calculus. The algebra of quadratic equations is a step above what students have previously encountered, and many may find it challenging. Students should take heart by the fact that the elegant way in which quadratic equations have. This book is aimed at students who wish to study advanced mathematics in their national. This textbook is based on my 7 years of experience of preparing students for higher level and standard level IB mathematics.

# Chapter 1: Introduction

In this section, quadratic equations are introduced visually by considering the area of a square. Before studying quadratic equations in detail, the reader should be familiar with the following: (i) simplifying quadratic equations, which often involves collecting like terms, (ii) multiply brackets, and (iii) graphing quadratic equations using a table of values. The first two concepts shall be explained through visual/geometric reasoning.

The simplest quadratic equation can be the area y of a square with sides x , here the area can be found by multiplying x by itself as shown in Fig 1:

Fig 1: A square with sides x and area y.

The x is a variable and different values can be substituted for x to create squares with different areas as shown in Fig 2:

Fig 2: Square with different dimensions; note the squares are not drawn to scale.

The square dimensions x and the area y can be plotted to create the quadratic curve shown in Figure 3:

Fig 3: The length x of a square and it’s area y.

The domain (which is the allowed values) of x is x>0 since the length of a square must be positive, and the range of y the area is y>0. The curve is an example of a parabolic curve or parabolic function. Mathematically, the x-values can be extended to negative values and since the product of two negative number is positive the value of y can only be positive (or zero) as shown in Fig 4:

Fig 4: The simplest parabolic curve.

The graph shown in Fig4 has a line of mirror symmetry x=0 (the y-axis); actually it will be shown later that all quadratic equations have a vertical line of mirror symmetry. The curve is only shown for integer values but there is no reason why x cannot have values in between integers to create a smooth continuous curve.

## The General Form of a Quadratic Equation

The general form of a quadratic equation is:

Where a,b, and c are constants; this means their values can be chosen and then kept fixed. The variables are x and y, and in practice the value of x is chosen and the value of y is then calculated. Below are examples of quadratic equations:

The values of the constants are a=2, b=-3, c=7 (top equation), a=-5, b=2, c=-6 (middle equation), and a=1, b=0, c=-16 (bottom equation). In all cases the value of a≠0 otherwise the equation will cease to be a quadratic equation.

The general form of a quadratic equation has 3 contributions, the first term can be thought as a square which is two-dimensional. The middle term with an x multiplied by a number can thought of as a one-dimensional line and is called linear term. The figure show the equation y=x for different values of x. The length of the line changes with x.

Fig 5: The equation y=x can represent a line.

The third term in the general form of a quadratic equation is just a single number c, and this is known as the constant. The graph of y=c is a horizontal line where the y coordinate is fixed to c.

A quadratic equation can be simplified by “collecting like terms, and this can be understood through a simple everyday example: you go shopping and buy 2 apples and 3 pears in one shop and 5 apples and 6 pears in another shop. Your brother/sister may ask you what did you buy? and most people will answer “7 apples and 9 pears”, since apples cannot be added to pears and vice-versa, it would be meaningless i.e. apples and pears are not like terms. The quadratic equation below contains many like terms:

The square terms cannot be added to the linear terms, and vice versa. The number terms must also be kept separated from the terms containing x. After collecting like terms the formula simplifies to:

## Expanding Brackets

The equation below on the right-hand side has two brackets next to each other

There is a multiplication sign between the brackets which is usually not shown:

The multiplication of brackets is often presented as an algorithm where terms are multiplied in order as shown by fig 6

Fig 6: Multiplication of brackets algorithm.

Bracket multiplication can be understood visually where each bracket represents a side of a rectangle as shown in Fig 7:

Fig 7: Representing brackets as the dimensions of a rectangle.

In fig 7 the horizontal width of the rectangle is (3x+5) and height is (x+6). The rectangle is subdivided into 4 smaller rectangles and the area of each rectangle is then calculated. The area of the large rectangle is the sum of the areas of the smaller rectangle. The final expression for the area y is then simplified by collecting like terms. All of this may seem like overkill and somewhat pointless, but it isn’t; in later chapters, visual arguments will be used to help simplify the understanding of difficult concepts in algebra. There is a lot of evidence that suggests that ancient mathematicians understood the algebra of quadratic equations through visual means. The separation between algebra and geometry maybe artificial and unhelpful.

## Exercise 1.2

For the following equations, expand the brackets by finding the area of the smaller rectangles, and then simplify the expression by collecting like terms.

## Checking Algebraic Steps

Algebraic manipulation be it expanding brackets or simplification can be checked by substituting an integer (apart from 0 and 19) for x, and then substituting the same integer after some algebraic step(s). The same value will be obtained provided the algebraic steps have been performed correctly.  All of this can be explained using an example; consider the equation:

After expanding and simplifying:

The expansion can be checked by substituting a value for x:

The reader is encouraged to use this substation method whenever there is a doubt about the correctness of an algebraic step. This substitution method helped me greatly to learn the fundamentals of algebra when I was student at school.

# Chapter 2: Graphing Quadratic Equations

The general features of the graph of quadratic equations can be understood by graphing a few quadratic equations, by hand. The process involves selecting an integer values of x and then calculating the y value and then plotting the point. The process is repeated for many x-values to trace the quadratic graph. This is how a computer plots a function by rapidly calculating points on the line and then fitting a smooth curve to those points, if possible.

## Example 2.1: Graph the Quadratic equation with a=1, b=-3, and c=-4

The quadratic equation that shall be graphed has the formula:

The y calculated y values for x=5, x=6, and x=7 are shown in Fig 2.1

Fig 2.1: Calculating points on the curve for three values. (5,6), (6,14), and (7,24) are points on the quadratic curve.

Three points are not enough to trace the shape of the quadratic graph, and more points are needed, and the calculated values are shown in the table below:

Table 2.1: Calculated points on the quadratic equation with with a=1, b=-3, and c=-4.

These points are plotted and a smooth curve is drawn through those points as show in Fig 2.2

Fig 2.2 The points in table 2.1 are plotted and a smooth curve is drawn through the points.

The graph is very symmetric with a vertical line of symmetry; this seems to be a very surprising result, could this it be a fluke? This can be investigated in the next exercise.

## Exercise 2.1

For the questions below complete the table and plot points on the quadratic curve, then draw a smooth line through the points.

(c) For the equation

calculate points on the line for -5≤x≤7 and plot these points, and then draw a smooth curve through the points.

(d) For the equation

calculate points on the line for 2≤x≤3 and plot these points, and then draw a smooth curve through the points.

## Evaluating y When the Squared Term has a Minus Sign

The minus sign in front of the x squared term must be interpreted correctly, for example:

The minus sign is outside the square of x, therefore the same y value is calculated for ±3.

## Features of A Quadratic Graph

Fig 2.3 show the quadratic graph for two more quadratic equations, and both graphs have a vertical line of mirror symmetry:

Fig 2.3 The graph of two quadratic equations.

The quadratic graphs from Exercise 2.1 and the above figure all have a vertical line of mirror symmetry. All quadratic graphs have an axis of symmetry, see Fig 2.4:

Fig 2.4: The graph of a concave up quadratic equation.

The axis of symmetry for the graph in figure 2.4 is the vertical line x=-3. The solutions of a quadratic graph are the x values for which y=0 i.e. the intercept of the quadratic graph with x-axis. The solution for the above equation are x=-5 and x=-1. The axis of symmetry is always half between the solutions (provided the quadratic equation intercepts the x-axis). A quadratic equation with a minimum point is said to be concave up. Fig 2.5 shows the graph of a concave down quadratic curve:

Fig 2.5: The graph of a concave down quadratic equation.

Fig 2.6: The sign of a determines the concavity.

The sign of a in the general form of the quadratic equation determines the concavity. The minimum or maximum point is often called the vertex of the quadratic graph. The y-intercept in all quadratic equation is determined by setting x=0 and then calculating the y value.

## Exercise 2.2

For the following quadratic graph, from the graph determine the solutions, the axis of symmetry, and the coordinate of the vertex. In addition, state the concavity of the quadratic function and determine the y-intercept.

## Discussion

The graphs of all quadratic equations have a vertical axis of symmetry. For quadratic equations that intercept with the x-axis the axis of symmetry is the vertical line that passes though the midpoint of the solutions. In this section, quadratic equations have only been investigated by plotting the function by calculated many points on the line. Later sections will go over a complete theory on the solutions, axis of symmetry, and the vertex. The formalism shall facilitate the a way to design a quadratic equation with chosen solutions, axis of symmetry and vertex coordinate.

# Chapter 3: Introduction to Quadratic Transformations

The previous chapter investigated graphs of the general quadratic equation:

All the graphs had the same basic shape with a minimum or maximum point and a vertical axis of symmetry. However, these features were not related to the constants a,b, and c. This chapter will study the role of the constants a and c in determining the graph. Hence b=0, therefore

A transformation of quadratic equation involves a change or a movement, and the first transformation that shall be studies is vertical translate. The most fundamental or basic quadratic equation is shown below:

Fig 3.1 The simplest quadratic equation where a=1, c=0, and b=0.

## Vertical Shift

The graph of the quadratic equation:

can be moved in the vertical direction by change the value of c and the graph for c=8 is shown in Fig 3.2

Fig 3.2 A vertical transformation of 8 units in the vertical direction.

The graphs have been generated using a table of value where c=8. The y values have increased by 8 units. The transformed quadratic equation does not intersect the x-axis and has no solution. The graph can be moved down using a negative value of c as shown in Fig 3.3:

Fig 3.3 A vertical transformation of -16 units.

## Exercise 3.1

Fill in the table and then plot the transformed quadratic equation:

## Vertical Stretch and Reflection

The graph of

can be stretched by changing the value of a. The stretch graph for a=2 is shown in Fig 3.4:

Fig 3.4: The quadratic equation for a=2 has been stretch by a factor of 2.

A value of a>1 will make the graph increase more rapidly with x value. The graph for a negative x-value will also be reflected along the x-axis as shown in Fig 3.5:

Fig3.5 The quadratic equation with a=-5 has been stretched and reflected, compared to the equation with a=1.

## Exercise 3.2

Fill in the table and then plot the stretched quadratic equation:

## Multiple Transformation

Multiple transformations can be applied starting with a simple base quadratic equation:

Fig 3.6 After multiple transformation the quadratic equation is concave down.

The transformed quadratic equation shown in Fig 3.6 has solutions since it intersects the x-axis. The final graph is concave down. The transformations parameters can be varied to obtain a different quadratic equation:

Fig 3.7: After multiple transformations, the quadratic equation is below the x-axis.

The axis of symmetry for all the above quadratic equations is x=0. Later in the book horizontal transformation shall be discussed where the axis of symmetry is not necessarily x=0.

## Exercise 3.3

The following quadratic equation have been transformed. Determine the stretch factor, the vertical shift value, and if the quadratic has been reflected along the x-axis:

The next chapter will discuss calculating the solutions of the type of quadratic equation studied in this chapter.

# Chapter 4 Intersection With x and y Axis

This chapter expands on the previous chapter and considers the intersection of the quadratic equation:

with the y-axis and the x-axis. Generally, a quadratic equation may intersect the x-axis or can be above or below the x-axis. Fig 4.1 show the graph of a quadratic equation that intersects the x-axis

Fig 4.1 A quadratic graph that intersects the x-axis.

All points on the y-axis have x coordinate equal to zero. The y-intercept is found by substituting x=0 in the equation and then solving for y:

All points on the x-axis, y=0. The x-intercepts are found by setting y=0 and then solving for x

Fig 4.1 show the graph of a quadratic equation that does not intersect the x-axis

Fig 4.2 Graph of quadratic equation that does not intersect the x-axis.

The intercept as before is found by setting x=0

There is no x-intercept however it’s instructive to set y=0 and try and solve for x:

The above analysis is noteworthy, when there is no x-intercept, solving for x runs into the problem of the square root of a negative number.

## Exercise 4.1

For the following quadratic equation, determine the y-intercept coordinate, and the x coordinate intercept. Sketch the graph of the functions on a graph paper.

## Difference of Two Square

A quadratic equation with the form

is literally the difference of two squares, where the smaller square with sides of length k is taken away from the larger square with sides of length x as shown in Fig 4.3:

Fig 4.3: A small square is cut out of the big square.

Fig 4.4: A rectangle is cut as shown and then placed as shown to create a rectangle with dimensions (x+k) and (x-k).

Fig 4.4 shows how a rectangle with dimension (x+k) and (x-k) is created, therefore:

The brackets can be expanded to recover the original form of the equation. The solutions are the x values for which y=0 i.e. when the area of the rectangle in Fig4.4 is zero. The solutions are x=k and x=-k. Example, the difference of two square method is used to find the solutions of the quadratic equation:

This is a difference of two square, hence

The solutions are x=7 and x=-7.

## Exercise 4.2

Solve the follow quadratic equations using the difference of two squares method:

# Chapter 5 Factorization of Quadratic Equations with c=0.

This chapter considers quadratic equations of the form

where c=0 and the presence of the bx term moves the axis of symmetry away from the y-axis. The solutions can be found by first factorizing the equation. Factorization is simply the reverse of expanding brackets as shown in Fig 5.1:

Fig 5.1 Factorization is the reverse of expanding brackets.

Some practice in expanding brackets will lay the ground work for master factorization.

## Exercise 5.1

Expand the following bracket:

The first step in factorizing the equation below

is to note that a=14 and b=2, and the highest common divisor (HCD) of a and b is 2. The convention when factorizing is to place HCD the outside the bracket as shown below:

## Exercise 5.2

For the following equations first identify the constants a and b, and then determine the HCD of a and b, then factorize with the HCD outside the brackets

The cases where a is negative it is customary to place the negative sign outside the bracket, for example

Is factorized to:

The next exercise is good practice for factorizing equations with a<0.

## Exercise 5.3

Factorize the follow equations and place the negative sign outside the bracket:

The first step in graphing quadratic equation such as

involves finding the solution from the factorized form

The factorized equation can represent a rectangle with area y as shown below:

Fig 5.2 A rectangle with height 2x and width x+4.

The solutions, by definition are the x values for which y=0 i.e. a rectangle with zero area. The area is the product of the width w and height h of the rectangle. The area will equal zero when h=0 and this occurs when 2x=0, which is when x=0. The width w=0 when x+4=0, which is when x=-4. The solutions are therefore x=0 and x=-4 and these are plotted in the graph below as the points (0,0) and (0,-4).

The axis of symmetry is the vertical line which passes through the midpoint of the two solution:

x=(-4+0)/2=-2

and this has been plotted on the graph below. The vertex has the x-coordinate of the axis of symmetry, and this x value is substituted into the quadratic equation to find vertex coordinate:

Fig 5.4 A smooth curve has been drawn through the solutions and vertex when graphing the quadratic equation.

Exercise 5.4

For the following quadratic equations, find the solutions, the axis of symmetry, and the vertex, and then sketch the quadratic equation on a suitable graph.

## Discussion

The analysis has been done for integer values of the constants a and b, a more general treatment is a good pedagogic exercise and should yield some general results. The starting point is the equation

Which is then factorized to the equation:

The solutions are when:

solving the second equation for x the other solution is when

The axis of symmetry passes through the midpoint of the solutions and is therefore the vertical line:

The graph of y=x(ax+b) is schematically shown below:

Fig 5.4 The graph of y=x(ax+b). The graph will be concave down for a<0. The solutions and axis of symmetry are given in terms of a and b.

The formula for the axis of symmetry is important since it hold for the general quadratic equation

this will be proved in later chapters.

# Chapter 6 Factorization

The factorization considered in the last chapter was straight forward and this chapter builds on the factorization narrative and considers more complicated cases. Quadratic equations of the form

Where a=1 can be factorized to the form

where m and n are integers. This type of factorization can only be performed for specific values of b and c.

## Expanding Brackets

Setting m=-5 and n=-9 yields the equation:

The brackets can represent the sides of a rectangle as shown below:

Fig 6.1 A rectangle with dimensions (x+5) and (x-9)

The area of the 4 smaller rectangle is shown in the figure. The areas are added and then simplified:

The a,b, and c values of the simplified equations are apparent from the final equation as shown above. The reader should note that the contribution to b comes two rectangles, and the c value is just the product of m and n. The exercise below will reinforce this idea.

## Exercise 6.1

Question 1

For the following equations, expand the brackets using the rectangle and determine the values of a,b, and c for the final simplified equation.

Exercise 6.2

You are required to think backwards, and determine the missing area and/or integer for the following

Exercise 6.3

For the exercises below determine the values of the missing areas and integer, given the value of b. These exercises will reinforce the concept that two rectangles contribute to the value of b.

Factorization involves finding n and m given a,b, and c, for example factorize

The factorized form is therefore y=(x+9)(x-2). Another example factorize:

The factorized for is therefore, y=(x+9)(x-8)

Exercise 6.4

Complete the table below given the values of a,b, and c, then write down the factorized equation.

## Graphing y=(x+m)(x+n)

As an example, the steps involved in graphing

will be shown.

## Solutions

The quadratic equation can represent the sides of a rectangle, and the solutions are the x-values for which the area y of the rectangle equals zero

Fig 6.2: Rectangle where the area is a quadratic equation.

The area is product of the height and width, which means the area equals zero when the height or width equals zero, both don’t have to equal zero. The height will be zero when x=-3, and the width is zero when x=7. The solutions are therefore x=-3 and x=7. Note that the solutions have the opposite sign to the values of n and m in the equation y=(x+m)(x+n). The solutions are plotted on the graph as shown below:

Fig 6.3 The solution of the quadratic equation y=(x+3)(x-7)

The solutions can be checked by substituting the values into the original equation:

## Axis of Symmetry

The axis of symmetry passes through the midpoint of the solutions:

The axis is symmetry is a vertical line as shown below:

Fig 6.4 The axis of symmetry lies halfway between the solutions.

## Vertex

The vertex coordinate is found by substituting the x value of the axis of symmetry into the quadratic equation

The vertex is plotted and a smooth curve is drawn through the solutions and vertex as shown below:

Fig 6.5 A curve is drawn through the solutions and the vertex.

Exercise 6.5

Graph the following quadratic equations by first calculating the solutions, axis of symmetry, and vertex.

# Chapter 7: Designing a Quadratic Graph.

The general form of a quadratic equation is

where a,b, and c are constants. The quadratic graph can be designed with chosen coordinates of the solutions and vertex, and this entails finding appropriate values of the a,b, and c. The placement of the vertex is made easier when b and c are integers and are multiples of a.

The starting point to place the solution is a quadratic equation with a=1 which has been factorized to the form y=(x+m)(x+n). The graph below shows a quadratic equation of the form y=(x+m)(x+n)

Fig 7.1 A quadratic equation of the form y=(x+m)(x+n)

The solutions are the x-values for which the graph intersects the x-axis; the solutions after inspection are x=-6 and x=3 respectively. When these x values are substituted into y=(x+m)(x+n) will result in y=0; this means m and n will have the opposite sign of the solutions. The above graph is of the quadratic equation y=(x+6)(x-3) after expanding this becomes:

This means a quadratic equation with a=1, b=3, and c=-18 intersects the x-axis when x=-6 and x=3.

The above method can be used to designed a quadratic equation that intersects for chosen x values, say for example x=2 and x=7. The values of n and m have opposite signs to their corresponding solutions, hence the required equation has the factorized form y=(x-2)(x-7). The figure below shows the expanded form of the quadratic equation and also it’s graph:

Fig 7.2 graph of quadratic equation with solutions x=2 and x=7.

## Exercise 7.1

For the following quadratic graphs determine the quadratic equation in the form:

(e) A quadratic equation has a=1 and passes through the points (3,0) and (4,0), find the values of b and c.

(f) A quadratic equation has a=1 and passes through the points (0.5,0) and (-0.1,0), find the values of b and c.

The above exercises have been done with numbers, however it’s interesting to consider a quadratic equation with solutions x=-m and x=-n where j and k are real numbers. The factorized form of the quadratic equation is then y=(x+m)(x-n), which  can be expanded and simplified to

and after comparing with

shows that b=m+n and c=mn.

## Placing the Vertex

The starting point for placing the vertex is a quadratic equation with form

Where b and c are multiples of a. The factorization of this type of quadratic equation involves just one extras step as shown below, which is factoring the a term:

Here is another example but with a negative a value

The quadratic equations have been factorized to the form y=a(x+m)(x+n)

## Exercise 7.2

Find the solutions of the following by factorizing to the form y=a(x+m)(x+n)

The value of a in the equation y=a(x+m)(x+n) allows the placement of the vertex, for example consider the graph of y=a(x+5)(x-10). The solutions are x=-5 and x=10 and the axis of symmetry passes through the midpoint of the solution.

The vertex’s x-coordinate is the same value as the axis of symmetry. The y value is calculated for a=3

The vertex coordinate depends on a, and the graph of y=a(x+5)(x-10) for several positive a values is shown below

Fig 7.3 The graph shown for a>0 are all concave up.

The graphs are all concave up for a>0 and they all have the same solutions and axis of symmetry, the vertex however depends on the a value. Graphs are shown for a<0 below:

Fig 7.4: The graph shown for a>0 are all concave up.

The graphs are all concave down for a<0. The vertex can be placed by choosing its y-coordinate and working backwards to calculate the appropriate a value. The next exercise requires working backward, and the solutions with all the steps are given in the answer section.

## Exercise 7.3

For the following questions find the quadratic equation with the indicated solutions and vertex coordinate. Start the analysis with the from y=a(x+m)(x+n) and then first find the values of m and n. Determine the value of a using the coordinate of the vertex.

Factorization can only be humanly performed for integer values. There is no reason why the constants a,b, and c should be integers, and they can be any real number. The chapters will outline a method known as completing the square which can be used to find the solutions of any quadratic equation.

# Chapter 8: A Perfect Square and Transformations.

Completing the square can be used to solve any quadratic equation, unlike the factorization method. Visually completing the square is exactly that, finding the missing piece to complete a square. The precursor of completing the square is to understand what exactly is a perfect square.

## A Perfect Square and Horizontal shift

A perfect square is the product of two linear terms, where for the factorized form y=(x+m)(x+n) m=n hence

However, it is more convenient to consider the form

Where h is an integer. The minus sign has been introduced, this can be confusing but it's there so that positive values of h corresponds to positive solutions. The value of h will be shown is actually a horizontal shift.

The brackets (x-h) and (x-h) can be thought of representing the sides of a square, this is shown below for two h values:

Fig 8.1 Two perfect quadratic squares with h=-5 and h=7.

The area of 4 smaller rectangles is also shown in the above figure. The areas of the four smaller rectangles are added up to find the area of the square. Two rectangles contribute to the value of b, and just one rectangle contributes to the value of c. There is a pattern between the values of b and c, and exercise below will help the reader to spot this pattern.

## Exercise 8.1

Fill in the table below. Can you spot connection between the b and c values.

The diagram below shows the connection between b and c

Fig 8.2 The geometric connection between b and c.

Dividing b by 2 and then squaring yields the value of c.

## Exercise 8.2

Determine which of the following equations are perfect squares. Factorize the perfect squares and find the solution.

The graph of a perfect quadratic square touches the x-axis at one point which is the solution. The axis of symmetry is passes through the solution; hence the vertex has the same coordinate as the solution. The figure below shows the graph a perfect square and for reference y=x2

Fig 8.3: The horizontal transformation of y=x2 into y=(x-12)2 by 12 units.

In the above figure h=12 and the solution has a positive x-value and horizontal transformation indicated is in the positive direction. The figure below shows a transformation in the other direction

Fig 8.4: The horizontal transformation of y=x2 into y=(x+14)2 by -14 units.

In the above figure h=-14 and the solution has a negative x-value and horizontal transformation indicated is in the negative direction.

## Exercise 8.3

(a) write down the factorized form of the perfect squares shown in the graphs

(b) A horizontal transformation of 7 units has been applied to  y=x2 determine the transformed equation in the form y=ax2+bx+c.

(c) A horizontal transformation of -1 units has been applied to  y=x2 determine the transformed equation in the form y=ax2+bx+c.

(d) The quadratic equation y=x2+2x+1 has been transformed to y=x2-20x+100. Determine the transformation that has been applied.

(e) The quadratic equation y=x2-4x+4 has been transformed to y=x2-8x+16. Determine the transformation that has been applied.

The horizontal transformation studied in this chapter and the transformations studied in chapter can be applied to any quadratic equation. The simplest quadratic equation y=x2 can be transformed to

The figure below shows the application of multiple transformation to y=x2

Fig 8.5: Transformations to y=x2: vertical stretch factor 4, horizontal shift -5 units, and vertical shift -17 units. The final quadratic equation has solutions. The vertex has coordinates (-5,-17).

Transformations can also be applied so that the final quadratic functions have no solutions:

Fig 8.6: Transformations to y=x2: reflection, horizontal shift 8 units, and vertical shift -10 units. The final quadratic equation has no solutions. The vertex has coordinates (8,-10).

The solutions for the final equation in Fig 8.6 can be found by setting y=0 and rearranging the quadratic equation:

A square root of a negative -10 has no real values hence no value for x can be found.

The above examples show the quadratic equation with the form y=a(x-h)2+k has axis of symmetry x=h and vertex (h,k) as shown in the figure below:

Fig 8.7: The vertex and axis of symmetry can be found easily when a quadratic equation is in the transformed form.

The stretch factor a does not change the vertex coordinate or axis of symmetry. In contrast to the factorized form y=a(x+m)(x+n) the solutions can be found in one step and some algebraic manipulation is needed.

## Exercise 8.4

(a) The following transformation have been applied to y=x2:

Reflection through the x-axis, vertical stretch factor 2, horizontal shift 2 units, and vertical shift 19 units.

Write down the final equation, and write the coordinate of the vertex and the equation of the axis of symmetry. Calculate the value of the solution to 2 decimal places. Finally make a sketch of the quadratic equation on graph paper.

(b) y=x2 has been transformed to y=3(x+4)2-10. Write the transformation that have been applied. Write the coordinate of the vertex and the equation of the axis of symmetry. Calculate the value of the solution to 2 decimal places. Finally make a sketch of the quadratic equation on graph paper.

(c)  y=x2 has been transformed to y=-2(x-2)2+9. Write the transformation that have been applied. Write the coordinate of the vertex and the equation of the axis of symmetry. Calculate the value of the solution to 2 decimal places. Finally make a sketch of the quadratic equation on graph paper.

(d) Show algebraically that the quadratic equation below has no solution:

Write down the coordinate of the vertex and the equation of the axis of symmetry.

# Chapter 9: Completing the Square and Transformations

The “completing the square” method can be used to solve any quadratic equation, let’s consider an example: solve y=x2+4x+75 by completing the square and then sketch the graph of the equation:

Fig 9.1 A geometric representation of y=x2+4x+75

The above figure shows a geometric representation, there is a square with area x2, a rectangle with area 4x, and a rectangle with area 75. The rectangle with area 4x is now cut in half:

Fig 9.2 The rectangle is cut in half to obtain two rectangles with area 2x.

The two slices are now placed next to the square with area x2

Fig 9.3 Almost a square.

The shape on the left is almost a square with area (x+2)2.

Fig 9.4: The square is completed by taking some area from the other rectangle.

The equation y=x2+4x+75 has been transformed to y=(x+2)2+71. The vertex has coordinates (see chapter 8) (-2,71) and the equation of the axis of symmetry is x=-2. The solution can be found by setting y=0 and then solving for x:

0=(x+2)2+71

-71=(x+2)2

The step is to take the square root of both sides:

The square root of a negative number does not have any real values, so this indicated that the quadratic equation has no solutions and does not intersect the x-axis. The graph can now be sketched and is shown below:

Fig 9.5: A sketch of the quadratic equation to y=(x+2)2+71.

The above diagrams help in understanding the fundamental concept behind completing the square, and this is how the method was discovered historically. Diagram are not needed to complete the square, as an example completing the square of y=x2+6x-55 is shown below, first note b=6 for this equation.

The graph of the quadratic equation has an axis of symmetry x=-3 and vertex coordinate (-3,-64). The solutions are found by setting y=0 and then solving for x:

0=(x+3)2-64

64=(x+3)2

taking square root of both sides

±8=x+3

The solutions are therefore x=5 and x=-11. A sketch of the graph of the quadratic equation is shown below:

Fig 9.6: Sketch of the equation y=(x+3)2-64.

Completing the square is relatively easy for quadratic equations of the form y=x2+bx+c where b is an even number. The method can still be applied odd values of b. Consider the equation y=x2+7x+12. The solutions are found by completing the square:

The vertex has coordinates (-7/2,-1/4) and the axis of symmetry is x=-1/4.

Completing the square with odd values of b can be mastered by practicing many problems to get used to mathematics with fractions.

## Exercise 9.1

Complete the square for the following equations and find the solutions, axis of symmetry, and vertex.

(a) y=x2+8x+7 (b) y=x2+15x+50 (c) y=x2-10x+9 (d) y=x2-11x+30

## Completing the Square for a≠1and connection to Transformations

Completing the square can be performed for the general quadratic equation y=ax2+bx+c, and equation is transformed to the form y=a(x-h)2+k. The latter form can be interpreted as successive transformations of the simplest quadratic equation y=x2. The example below shows how to complete the square for a quadratic equation where a≠1

The above analysis show that to transform y=x2 into y=4x2-64x+239 the following transformations can be applied: vertical stretch factor 4, horizontal shift 8 units, vertical shift -17 units. This means that the equation y=4(x-8)2-17 is the same as y=4x2 provided the origin of the coordinate system is moved to the vertex point (8,-17), and this is explicitly shown below:

Fig 9.7: The quadratic equations have the same shape.

The quadratic equation y=ax2+bx+c is identical to y=ax2 provide the coordinate system is moved to the vertex; this is a remarkable but undisputable conclusion from the analysis.

Completing the square can also be performed for a<0. The example below shows how to deal with the negative value of a

The quadratic equation has a vertex at (7,8), and axis of symmetry x=7. The solutions can be found by set y=0 and solving for x. The graph of y=-2(x-7)2+8 is shown below:

Fig 9.8: The graph of y=-2(x-7)2+8 is concave down since a=-2, whereas y=x2 is concave up.

## Exercise 9.2

(a) y=x2 has been transformed to y=3x2+18x+19. Determine the transformations that have been applied, and the vertex coordinate of the transformed equation.

(b) y=x2 has been transformed to y=-4x2+27x-27. Determine the transformations that have been applied, and the vertex coordinate of the transformed equation.

(c)  y=x2 has been transformed to y=2x2-12x-1. Determine the transformations that have been applied, and the vertex coordinate of the transformed equation.

(d)   y=x2 has been transformed to y=-5x2-30x-26. Determine the transformations that have been applied, and the vertex coordinate of the transformed equation.

# Chapter 10: The Quadratic Formula and Discriminant.

The quadratic formula is used to calculate the solutions for any quadratic equation. The formula is derived by applying completing the square to a general quadratic equation which has the form y=ax2+bx+c. Performing algebra to an equation which has no numbers can seem to be difficult at first, but in principle is not more complicated than applying the method with numbers. The first step is to set y=0, and the equation y=ax2+bx+c becomes

0=ax2+bx+c

The next step is to divide all terms by a, the equation can be represented geometrically as shown below:

Fig 10.1 The geometric representation of the equation. There is a square with area x2, and two rectangles.

Fig 10.2 The next step in the derivation.

The rectangle in the middle is next sliced into two equal rectangles. The height of the new rectangles is x as shown in Fig 9.2

Fig 10.3 Almost a square.

The sliced rectangles are moved as shown in Fig 9.3 to define a shape which is almost a square.

Fig 10.4 add and subtracting the missing square as shown.

The next step shown in Fig 10.4 is to add and subtract the missing square as shown. The first three terms are then recognized to be a perfect square and is then factorized. To recap the steps so far:

The final step is to make x the subject:

The equation in the box is the famous “quadratic formula”. Admittedly the derivation is somewhat involved, but the reader is encouraged to go through the derivation many times.

The quadratic formula can be used without understanding its derivation. For example, find the solutions of y=2x2-12x+7. The first step is to set y=0, hence 0=2x2-12x+7

The axis of symmetry and vertex must be calculated before sketch the graph of the equation

The graph of the equation is shown below:

Fig 10.5 The graph of y=2x2-12x+7

The solution of the above equation are not integers and could not have been found using the factorization method.

## Exercise 10.1

For the following equations find the solutions using the quadratic formula and then find the solution using factorization.

(a) y=3x2-12x+9 (b) y=-4x2-16x-16 (c) y=7x2-7x-392 (d) y=-5x2+40x-75.

For following calculate the solutions to two decimal places, then determine the equation of the axis of symmetry, and the vertex coordinate; finally sketch the graph of the equation:

(e) y=-2x2+20x-37

(f) y=2x2-4x-1

## The Discriminant

The expression inside the square roots of the quadratic formula is known as the discriminant

for a very good reason because its determines the number of solutions. When b2-4ac<0 no solutions are possible, since it’s possible to take the square root of a negative number (unless one considers imaginary solutions).

The figure below shows the graphs of the perfect square y=x2-10x+25 and two other quadratic equations

Fig 10.6 graph of three quadratic equations.

The perfect square in Fig 10.6 has only one solution and touches x-axis. The discriminant indicates how many solutions to expect without having to actually calculate the value of the solutions:

y=x2-10x+25 where a=1, b=-10 and c=25

b2-4ac=(-10)2-4×1×25=100-100=0

b2-4ac=0 one solution

y=x2-10x+25 where a=1, b=-10 and c=30

b2-4ac=(-10)2-4×1×30=100-120=-120

b2-4ac<0 no solution

y=x2-10x+21 where a=1, b=-10 and c=21

b2-4ac=(-10)2-4×1×21=100-100=16

b2-4ac>0 two solution

## Exercise 10.2

For following equations calculate the value of the discriminant and determine the number of solutions:

(a) y=-3x2+30x-63             (b) y=-x2-10x-30

(c) y=2x2+16x+32              (d) y=-3x2-6x+4

For the following determine an expression for the determinant in terms of k, and find the value of k for which the equation has only one solution.

(e) y=-x2+kx-81

(f) y=3x2-12x+k

(g) y=kx2-12x+12

# Chapter 11: Intersections with Lines and Quadratic Graphs

Finding the intersection points between a straight line and a quadratic equation involves solving a simultaneous equation. For example, the intersection points of the line y=2x+2 and y=x2+x-4 can be estimates by graphing:

Fig 11.1 The graphs of y=2x+2 and y=x2+x-4.

For the graph the intersection points are estimated to be (-2,-2) and (3,8). It’s understandable if the reader thinks that these are the intersection points, but it could a trick of the scale. To calculate the intersections points requires solving these simultaneous equations:

y=2x+2 (A)

y=x2+x-4 (B)

substituting for y in equation (B) using equation (A) results in an equation with only the x variable:

2x+2=x2+x-4

subtract 2x from both sides

2=x2-x-4

subtracting 2 from both sides

0=x2-x-6

The equation can be solve using the quadratic formula, but for this case it is much faster to factorize to the form:

0=(x-3)(x+2)

The solutions are x=-2 and x=3. To find one intersection point substitute x=-2 into either equation (A) or (B). It’s much easy to use equation (A)

y=2×(-2)+2

y=-2

hence one intersection point is (-2,-2). Substituting x=3 into equation (A) y=8 so the other intersection point is (3,8). The intersection points can be found for this case by graphing since the intersection points have integer values. There is no reason why the intersection points should have integer values.

# Chapter 12: Applications of Quadratic Equations

## Word Problems

The key to solving word problems is to translate the words into a quadratic equation and then solve the quadratic equation. In some case the solutions can be found by factorization, and for all problems one can use completing the square or the quadratic formula.

## Example 12.1

The product of two consecutive integers is 30, what are these numbers?

This question can be done mentally but that’s the point of the example. Let the smallest number (which is unknown) be x, so the other number is x+1. Therefore:

x(x+1)=30

This equation can be solved by first expanding the brackets and re-arranging:

x2+x=30

x2+x-30=0

(x+6)(x-5)=0

The solutions are x=5 and x=-6. The two consecutive numbers are therefore 5 and 6 or -6 and -5.

## Example 12.2

A rectangle’s length is 3 units longer than its width, and the area of the rectangle is 70. Determine the dimensions of the rectangle.

The unknown width is x, hence:

(x+3)x=70

x2+3x=70

x2+3x-70=0

(x+10)(x-7)=0

The solutions are x=7 and x=-10. The negative value can be discarded since a rectangle cannot have a negative length. The answer is width=7 and length=10.

## Exercise 12.1

(a) The length of the base of a triangle is three times its height. Given that the area of the triangle is 63 units calculate the height of the triangle.

(b) The outer radius of a ring shown in the figure below is 1 unit greater than the inner radius. Given that the area of the ring is 100 units calculate the inner and outer radii.

(c) The product of two consecutive number even number is 360, determine the possible values of these numbers.

(d) A square has smaller square of area 4 units cut from one of it’s corners. Given that the area of this shape is now 60 units determine the dimensions of the original square.

(e) A window of area 200 units consists of a square region and a semicircle as show in the figure below. Determine the perimeter of the window.

## Optimization problems

Exercise 1

Exercise 2

Exercise 2.1

(a)

The y values: 27,16,7,0,-5,-8,-9,-8,-5,0,7,16,27

(b)

The y values: -18, -10,-4,0,2,2,0,-4,-10,-18

(c)

(d)

Exercise 2.2

Exercise 3.1

(a)

(b)

Exercise 3.2

Exercise 3.3

Exercise 4.1

Exercise 4.2

(a)

y=(x+10)(x-10), solutions x=10 and x=-10

(b)

y=(x+5)(x-5), solutions x=5 and x=-5

(c)

y=(x+1)(x-1), solutions x=1 and  x=-1

(d)

y=(2x+3)(2x-3), solutions x=-3/2 and x=3/2

(e)

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

(a)

(b)

(c)

(d)

Exercise 6.1

Exercise 6.2

Exercise 6.3

Exercise 6.4

Exercise 6.5

Exercise 7.1

Exercise 7.2

Exercise 7.3

Exercise 8.1

Exercise 8.2

Exercise 8.3

(a) top y=(x-8)2 and bottom y=(x+8)2. (b) y=(x-7)2 expanding y=x2-14x+49.

(c)  y=(x+1)2 expanding y=x2+2x+1.

(d) y=(x+1)2 transformed to  y=(x-10)2 which is a horizontal transformation 11 units.

(e) y=(x-2)2 transformed to  y=(x-4)2 which is a horizontal transformation 2 units.

Exercise 8.4

Exercise 9.1

Exercise 9.2

Exercise 10.1

Exercise 10.2

(a) discriminant=144 two solutions  (b) discriminant=-52 no solutions (c) discriminant=0 one solution (d) discriminant=84 no solutions.

(e) b2-4ac=k2-4×(-1) ×(-81)

k2-4×(-1) ×(-81)=0

k2-324=0

k2=324 hence k=±18

(f)  b2-4ac=(-12)2-4×3×k

144-12k=0

12k=144 hence k=12

(g) b2-4ac=(-12)2-4×k×12

144-48k=0

48k=144 hence k=3