ATHEMATICAL)METHODS)UNIT)1) 3 LGEBRAIC)FOUNDATIONS · Page3%of%21%...

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𝑬 ∙ 𝑑𝑨 =𝑞𝜀(

𝑩 ∙ 𝑑𝑨 = 0

𝑬 ∙ 𝑑𝑺 = −𝑑Φ. 𝑑𝒕

𝑩 ∙ 𝑑𝑺 = 𝜇(𝑖 + 𝜇(𝜀(𝑑Φ3 𝑑𝒕

MATHEMATICAL METHODS UNIT 1 CHAPTER 3 – ALGEBRAIC FOUNDATIONS Key knowledge

• Factorization patterns, the quadratic formula and discriminant, the remainder, factor and rational root theorems and the null factor law

• key mathematical content from one or more areas of study related to a given context • specific and general formulations of concepts used to derive results for analysis within a given

context

Key skills

• expand and factorise linear and simple quadratic expressions with integer coefficients by hand • express 𝑎𝑥2

+ 𝑏𝑥 + 𝑐 in completed square form where 𝑎, 𝑏, 𝑐 ∈ 𝑍 and 𝑎 ≠ 0, by hand

Chapter 3 – Set Questions

Exercise 3.2: Quadratic equations with rational roots 1ab, 2,3, 5, 6, 7acd, 8ad, 9ad, 10a, 11abef, 12acf, 13cd, 14bc, 16a Exercise 3.3: Quadratics over R 1abc, 3abcd, 4ab, 5, 6, 7a, 9a, 10, 11, 12, 14ace, 15ac, 16a, 17ace, 18bcd, 19ace, 20ac, 21a, 22a Exercise 3.4: Applications of Quadratic equations 2, 3, 4, 5b, 7ab, 8, 9, 11, 14 Exercise 3.5: Graphs of quadratic Polynomials 1, 2, 3, 4, 5ab, 6ab, 7, 8, 9ab, 11ac, 12ad, 13ad, 14ad, 15ad, 17a, 18ad, 19, 20 Exercise 3.6: Determining the rule of a quadratic polynomial from a graph 1ab, 2, 3, 4, 5a, 6ab, 7abcd, 9a, 10ab, 11, 12abc, 13a Exercise 3.7: Quadratic inequations 1, 2, 3, 4, 5, 7, 9cd, 10b, 11ac, 12b, 13b, 14abc, 15c, 16, 17b, 19 Exercise 3.8: Quadratic Models and applications 1, 2, 3, 4, 5, 6, 8, 9, 11, 12

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3.2 QUADRATIC EQUATIONS WITH RATIONAL ROOTS Quadratic equations and the null Factor law

The general quadratic equation can be written as 𝑎𝑥2

+ 𝑏𝑥 + 𝑐 = 0, where 𝑎, 𝑏, 𝑐 are real constants and 𝑎 ≠ 0. If the quadratic expression on the left-‐hand side of this equation can be factorised, the solutions to the quadratic equation may be obtained using the Null Factor Law.

The null Factor law states that, for any 𝑎 and 𝑏, if the product 𝑎𝑏 = 0 then 𝑎 = 0 or 𝑏 = 0 or both 𝑎 and 𝑏 = 0.

Applying the null Factor law to a quadratic equation expressed in the factorised form as:

(𝑥 − 𝛼)(𝑥 − 𝛽) = 0,

would mean that

(𝑥 − 𝛼) = 0 𝑜𝑟 (𝑥 − 𝛽) = 0 ∴ 𝑥 = 𝛼 𝑜𝑟 𝑥 = 𝛽

To apply the null Factor law, one side of the equation must be zero and the other side

must be in factorised form.

roots, zeros and factors

The solutions of an equation are also called the roots of the equation or the zeros of the quadratic expression. This terminology applies to all algebraic equations and not just quadratic equations.

The quadratic equation (𝑥 − 1)(𝑥 − 2) = 0 has roots 𝑥 = 1, 𝑥 = 2.

These roots are the zeros of the quadratic expression since substituting either of 𝑥 = 1, 𝑥 = 2 in the quadratic expression (𝑥 − 1)(𝑥 − 2) makes the expression equal zero.

As a converse of the null Factor law it follows that if the roots of a quadratic equation, or the zeros of a quadratic, are 𝑥 = 𝛼 and 𝑥 = 𝛽, then (𝑥 − 𝛼) and (𝑥 − 𝛽) are linear factors of the quadratic. The quadratic would be of the form (𝑥 − 𝛼)(𝑥 − 𝛽) or any multiple of this form, 𝑎(𝑥 − 𝛼)(𝑥 − 𝛽).

Example 1

a) Solve the equation 12𝑥K

− 7𝑥 = 10 b) Given that x = 3 and x = −2 are zeros of a quadratic, form its linear factors and expand the

product of these factors.

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Using the perfect square form of a quadratic

As an alternative to solving a quadratic equation by using the null Factor law, if the quadratic is a perfect square, solutions to the equation can be found by taking square roots of both sides of the equation. A simple illustration is

using square root method

𝑥K

= 9 𝑥 = ± 9 = −3 OR

using null Factor law method

𝑥K

= 9 𝑥K − 9 = 0

(𝑥 − 3)(𝑥 + 3) = 0 𝑥 = ±3

If the square root method is used, both the positive and negative square roots must be considered.

Worked Example 2

Solve the equation (2x + 3)2 − 25 = 0.

Equations which reduce to quadratic form

Substitution techniques can be applied to the solution of equations such as those of the form 𝑎𝑥4

+ 𝑏𝑥2

+ 𝑐 = 0. Once reduced to quadratic form, progress with the solution can be made.

The equation 𝑎𝑥4

+ 𝑏𝑥2

+ 𝑐 = 0 can be expressed in the form 𝑎(𝑥K)K

+ 𝑏𝑥K

+ 𝑐 = 0. letting 𝑢 = 𝑥K, this becomes 𝑎𝑢K + 𝑏𝑢 + 𝑐 = 0, a quadratic equation in variable u.

By solving the quadratic equation for 𝑢, then substituting back 𝑥K for u, any possible solutions for 𝑥 can be obtained. Since 𝑥2 cannot be negative, it would be necessary to reject negative 𝑢 values since 𝑥K =𝑢 would have no real solutions.

Example 3

Solve the equation: 2𝑥R

= 31𝑥K

+ 16

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3.3 QUADRATICS OVER R – PART 1 When 𝑥 − 4 is expressed as (𝑥 − 2)(𝑥 + 2) it has been factorised over 𝑄, as both of the zeros are rational numbers. However, over 𝑄, the quadratic expression 𝑥K

− 3 cannot be factorised into linear factors. Surds need to be permitted for such an expression to be factorised.

Factorisation over R

The quadratic 𝑥K

− 3 can be expressed as the difference of two squares𝑥K

− 3 = 𝑥K

− ( 3)K

using surds. This can be factorised over R because it allows the factors to contain surds.

𝑥2 − 3 = 𝑥2 − ( 3)2 = (𝑥 − 3)(𝑥 + 3) If a quadratic can be expressed as the difference of two squares, then it can be factorised over R. To express a quadratic trinomial as a difference of two squares a technique called ‘completing the square’ is used.

Worked Example 4c

Factorise 4𝑥2

− 11 over R:

‘Completing the Square’ technique

Expressions of the form 𝑥2

± 𝑝𝑥 + VK

K= 𝑥 ± V

K

K are perfect squares.

For example, 𝑥K + 4𝑥 + 4 = (𝑥 + 2)K.

To illustrate the ‘completing the square’ technique, consider the quadratic trinomial 𝑥K

+ 4𝑥 + 1.

If 4 is added to the first two terms 𝑥2

+ 4𝑥 then this will form a perfect square𝑥2

+ 4𝑥 + 4.

However, 4 must also be subtracted in order not to alter the value of the expression.

𝑥K

+ 4𝑥 + 1 = 𝑥K

+ 4𝑥 + 𝟒 − 𝟒 + 1 Grouping the first three terms together to form the perfect square and evaluating the last two terms,

= (𝑥K

+ 4𝑥 + 4) − 4 + 1 = (𝑥 + 2)K

− 3 By writing this difference of two squares form using surds, factors over R can be found.

= (𝑥 + 2)K − ( 3)K = (𝑥 + 2 − 3)(𝑥 + 2 + 3)

Thus, 𝑥2

+ 4𝑥 + 1 = (𝑥 + 2 − 3)(𝑥 + 2 + 3)

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‘Completing the square’ is the method used to factorise monic quadratics over R. A monic quadratic is one for which the coefficient of 𝑥K equals 1.

For a monic quadratic, to complete the square, add and then subtract the square of half the coefficient of x. This squaring will always produce a positive number regardless of the sign of the coefficient of 𝑥.

To complete the square on 𝑎𝑥2

+ 𝑏𝑥 + 𝑐, the quadratic should first be written as𝑎 𝑥2 + XYZ+ [

Z

and the technique applied to the monic quadratic in the bracket. The common factor a is carried down through all the steps in the working.

Worked Example 4a & 4b

Factorise the following over R: a) 𝑥K

− 14𝑥 − 3 b) 2𝑥K

+ 7𝑥 + 4

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3.3 QUADRATICS OVER R – PART 2 The discriminant

Some quadratics factorise over Q and others factorise only over R. There are also some quadratics which cannot be factorised over R at all. This happens when the ‘completing the square’ technique does not create a difference of two squares but instead leads to a sum of two squares. In this case no further factorisation is possible over R.

i.e., completing the square on 𝑥2

− 2𝑥 + 6 gives: 𝑥2

− 2𝑥 + 6 = (𝑥2

− 2𝑥 + 1) − 1 + 6 = (𝑥 −1)2

+ 5

As this is the sum of two squares, it cannot be factorised over R. Evaluating what is called the discriminant will allow these three possibilities to be discriminated between. To define the discriminant, we need to complete the square on the general quadratic trinomial 𝑎𝑥2

+ 𝑏𝑥 + 𝑐.

The term, 𝑏2

− 4𝑎𝑐, is called the discriminant of the quadratic. It is denoted by the Greek letter delta, Δ.

• If 𝛥 < 0 the quadratic has no real factors. • If 𝛥 ≥ 0 the quadratic has two real factors. The two factors are distinct (different) if 𝛥 > 0

and the two factors are identical if 𝛥 = 0.

For a quadratic 𝑎𝑥2

+ 𝑏𝑥 + 𝑐 with real factors and 𝑎, 𝑏, 𝑐 ∈ 𝑄:

• If 𝛥 is a perfect square, the factors are rational; the quadratic factorises over 𝑄. • If 𝛥 > 0 but not a perfect square, the factors contain surds; the quadratic factorises over 𝑅.

Completing the square will be required if 𝑏 ≠ 0. • If 𝛥 = 0, the quadratic is a perfect square.

Worked Example 5

For each of the following quadratics, calculate the discriminant and hence state the number and type of factors and whether the ‘completing the square’ method would be needed to obtain the factors.

a) 2𝑥2 + 15𝑥 + 13 b) 5𝑥2

− 6𝑥 + 9

c) −3𝑥2 + 3𝑥 + 8 d) cd

R𝑥K − 12𝑥 + de

f

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Quadratic equations with real roots

The choices of method to consider for solving the quadratic equation 𝑎𝑥2

+ 𝑏𝑥 + 𝑐 = 0 are:

• factorise over Q and use the null Factor law • factorise over R by completing the square and use the null Factor law

• use the formula 𝑥 = gX± XhgRZ[KZ

Often the coefficients in the quadratic equation make the use of the formula less tedious than completing the square. Although the formula can also be used to solve a quadratic equation which factorises over Q, factorisation is usually simpler, making it the preferred method.

Worked Example 6

Use the quadratic formula to solve the equation 𝑥(9 − 5𝑥) = 3.

Worked Example 7

a) Use the discriminant to determine the number and type of roots to the equation 15𝑥2

+ 8𝑥 − 5 = 0.

b) Find the values of k so the equation 𝑥2

+ 𝑘𝑥 − 𝑘 + 8 = 0 will have one real solution and check the answer.

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Quadratic equations with rational and irrational coefficients

The formula gives the roots of the general equation 𝑎𝑥2

+ 𝑏𝑥 + 𝑐 = 0, 𝑎, 𝑏, 𝑐 ∈ 𝑄 as

𝑥d =−𝑏 − Δ2𝑎 𝑎𝑛𝑑 𝑥K =

−𝑏 + Δ2𝑎

If Δ>0 but not a perfect square, these roots are irrational and 𝑥d 𝑎𝑛𝑑 𝑥Kare a pair of conjugate surds.

• If a quadratic equation with rational coefficients has irrational roots, the roots must occur in conjugate surd pairs.

• The conjugate surd pairs are of the form: 𝑚 − 𝑛 𝑝 and 𝑚 + 𝑛 𝑝 with 𝑚, 𝑛, 𝑝 ∈ 𝑄.

Worked Example 8(a)

Use ‘completing the square’ to solve the quadratic equation𝑥2

+ 8 2𝑥 − 17 = 0. Are the solutions a pair of conjugate surds?

Equations of the form 𝒙 = 𝒂𝒙 + 𝒃

Equations of the form 𝑥 = 𝑎𝑥 + 𝑏 could be written as 𝑥 = 𝑎 𝑥 2 + 𝑏 and reduced to a quadratic equation 𝑢 = 𝑎𝑢2

+ 𝑏 by the substitution 𝑢 = 𝑥. Any negative solution for u would need to be rejected as 𝑥 ≥ 0.

Alternatively, squaring both sides of the equation 𝑥 = 𝑎𝑥 + 𝑏 the quadratic equation 𝑥 = (𝑎𝑥 + 𝑏)2 is formed, with no substitution required. However, since the same quadratic equation would be obtained by squaring𝑥 = −(𝑎𝑥 + 𝑏), the squaring process may produce extraneous ‘solutions’ — ones that do not satisfy the original equation. It is always necessary to verify the solutions by testing whether they satisfy the original equation.

Worked Example 9

Solve the equation 3 + 2 𝑥 = 𝑥 for 𝑥.

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3.4 APPLICATIONS OF QUADRATIC EQUATIONS Quadratic equations may occur in problem solving and as mathematical models. In formulating a problem, variables should be defined and it is important to check whether mathematical solutions are feasible in the context of the problem.

Quadratically related variables

The formula for the area, A, of a circle in terms of its radius, 𝑟, is 𝐴 = 𝜋𝑟2. This is of the form 𝐴 = 𝑘𝑟2 as 𝜋 is a constant. The area varies directly as the square of its radius with the constant of proportionality 𝑘 = 𝜋. This is a quadratic relationship between A and r. For any variables 𝑥 and 𝑦, if 𝑦 is directly proportional to 𝑥2, then 𝑦 = 𝑘𝑥2 where 𝑘 is the constant of proportionality.

Also, y could be the sum of two parts, one part of which was constant and the other part of which was in direct proportion to 𝑥2 so that 𝑦 = 𝑐 + 𝑘𝑥2.

Also, y could be the sum of two parts, one part of which varied as 𝑥 and another as 𝑥2, then 𝑦 = 𝑘1𝑥 +𝑘2𝑥

2 with different constants of proportionality required for each part.

The quadratic relation 𝑦 = 𝑐 + 𝑘1𝑥 + 𝑘2𝑥2 shows y as the sum of three parts, one part constant,

one part varying as 𝑥 and one part varying as 𝑥2.

Example Question 1

The owner of a fish shop bought x kilograms of salmon for $400 from the wholesale market. At the end of the day all except for 2 kg of the fish were sold at a price per kg which was $10 more than what the owner paid at the market. From the sale of the fish, a total of $540 was made. How many kilograms of salmon did the fish shop owner buy at the market?

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3.5 GRAPHS OF QUADRATIC POLYNOMIALS A quadratic polynomial is an algebraic expression of the form 𝑎𝑥K + 𝑏𝑥 + 𝑐 where each power of the variable 𝑥 is a positive whole number, with the highest power of 𝑥 being 2. It is called a second-‐degree polynomial.

A linear polynomial of the form 𝑎𝑥 + 𝑏 is a first degree polynomial since the highest power of 𝑥 is 1.

The graph of a quadratic polynomial is called a parabola.

The graph of 𝒚 = 𝒙𝟐 and transformations

The simplest parabola has the equation 𝑦 = 𝑥K. Key features of the graph of 𝑦 = 𝑥K:

• it is symmetrical about the y-‐axis

• the axis of symmetry has the equation x = 0

• the graph is concave up (opens upwards)

• it has a minimum turning point, or vertex, at the point (0, 0).

Making the graph wider or narrower

The graphs of 𝑦 = 𝑥K for 𝑎 = du, 1 and 3 are drawn on the same

set of axes. Comparison of the graphs with 𝑦 = 𝑥K shows the graph will be narrower if 𝑎 > 1, and wider if 0 < 𝑎 < 1.

The coefficient of 𝑥2, a, is called the dilation factor. It measures the amount of stretching or compression from the x-‐axis.

Translating the graph up or down

The graphs of 𝑦 = 𝑥2

+ 𝑘 for 𝑘 = −2, 0 and 2 are drawn on the same set of axes.

Comparison shows the graphs have a turning point at (0, 𝑘) move the graph of 𝑦 = 𝑥2 vertically upwards by 𝑘 units if 𝑘 > 0. move the graph of 𝑦 = 𝑥2 vertically downwards by 𝑘 units if 𝑘 < 0.

The value of 𝑘 gives the vertical translation. For the graph of 𝑦 = 𝑥2

+ 𝑘, the graph of 𝑦 = 𝑥2 has been translated by 𝑘 units from the x-‐axis.

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Translating the graph left or right

The graphs of 𝑦 = (𝑥 − ℎ)2 for ℎ = −2, 1 and 4 are drawn on the same set of axes. Comparison shows that the graph 𝑦 = (𝑥 − ℎ)K will have a turning point at (ℎ, 0) and move the graph of 𝑦 = 𝑥2 horizontally to the right by ℎ units if ℎ > 0 and move the graph horizontally to the left by ℎ units if ℎ < 0

Reflecting the graph in the x-‐axis

The graph of y = −x2 is obtained by reflecting the graph of y = x2 in the x-‐axis.

Key features of the graph of y = −x2:

• it is symmetrical about the y-‐axis • the axis of symmetry has the equation x = 0 • the graph is concave down (opens downwards) • it has a maximum turning point, or vertex, at the point (0,

0).

A negative coefficient of x2 indicates the graph of a parabola is concave down.

Regions above and below the graph of a parabola

The regions that lie above and below the parabola y = x2 are illustrated in the diagram.

If the region is closed, the points on theboundary parabola are included in the region.If the region is open, the points on theboundary parabola are not included in the region.

A point can be tested to confirm which side of the parabola to shade.

Sketching parabolas from their equations

The key points required when sketching a parabola are:

• the turning point • the y-‐intercept • any x-‐intercepts.

The axis of symmetry is also a key feature of the graph.

The equation of a parabola allows this information to be obtained but in differing ways, depending on the form of the equation.

We shall consider three forms for the equation of a parabola:

• general form • turning point form • x-‐intercept form.

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1. The general, or polynomial form, 𝑦 = 𝑎𝑥K

+ 𝑏𝑥 + 𝑐

If 𝑎 > 0 then the parabola is concave up and has a minimum turning point.

If 𝑎 < 0 then the parabola is concave down and has a maximum turning point.

The methods to determine the key features of the graph are as follows.

• Substitute x = 0 to obtain the y-‐intercept (the y-‐intercept is obvious from the equation). • Substitute y = 0 and solve the quadratic equation ax2 + bx + c = 0 to obtain the x-‐intercepts.

There may be 0, 1 or 2 x-‐intercepts, as determined by the discriminant. • The equation of the axis of symmetry is 𝑥 = − X

KZ

• the turning point lies on the axis of symmetry so its x-‐coordinate is 𝑥 = − XKZ .Substitute this

value into the equation to find the y-‐coordinate of the turning point.

Worked Example 13

Sketch the graph of 𝑦 = dK𝑥K − 𝑥 − 4 and label the key points with their coordinates.

Turning point form, 𝒚 = 𝒂(𝒙 − 𝒉)𝟐

+ 𝒌

Since h represents the horizontal translation and k the vertical translation, this form of the equation readily provides the coordinates of the turning point.

• The turning point has coordinates (h, k). o If a > 0, the turning point is a minimum o If a < 0 it will be a maximum.

• Find the y-‐intercept by substituting x = 0. • Find the x-‐intercepts by substituting y = 0 and solving the equation 𝑦 = 𝑎(𝑥 − ℎ)K + 𝑘.

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Worked Example 14

Express 𝑦 = 3𝑥K

− 12𝑥 + 18 in the form 𝑦 = 𝑎(𝑥 − ℎ)K

+ 𝑘 and hence state the coordinates of its vertex.

Factorised, or x-‐intercept, form 𝒚 = 𝒂(𝒙 − 𝒙𝟏)(𝒙 − 𝒙𝟐)

This form of the equation readily provides the x-‐intercepts.

• The x-‐intercepts occur at 𝑥 = 𝑥1 and 𝑥 = 𝑥2. • The axis of symmetry lies halfway between the x-‐intercepts and its equation, 𝑥 = Y~�Yh

K, gives the

𝑥 −coordinate of the turning point. • The turning point is obtained by substituting 𝑥 = Y~�Yh

K into the equation and calculating the y-‐

coordinate. • The 𝑦 −intercept is obtained by substituting 𝑥 = 0.If the linear factors are distinct, the graph

cuts through the 𝑥 −axis at each 𝑥 −intercept.

If the linear factors are identical making the quadratic a perfect square, the graph touches the 𝑥 −axis at its turning point.

Worked Example 15

Sketch the graph of 𝑦 = − dK(𝑥 + 5)(𝑥 − 1).

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The discriminant and the x-‐intercepts

The zeros of the quadratic expression 𝑎𝑥2

+ 𝑏𝑥 + 𝑐, the roots of the quadratic equation 𝑎𝑥K

+ 𝑏𝑥 + 𝑐 = 0 and the x-‐intercepts of the graph of a parabola with rule 𝑦 = 𝑎𝑥K

+ 𝑏𝑥 + 𝑐 all have the same 𝑥 −values; and the discriminant determines the type and number of these values.

• If 𝛥 > 0, there are two x-‐intercepts. The graph cuts through the 𝑥 −axis at two different places.

• If 𝛥 = 0, there is one x-‐intercept. The graph touches the 𝑥 −axis at its turning point. • If 𝛥 < 0, there are no x-‐intercepts. The graph does not intersect the 𝑥 −axis and lies entirely

above or entirely below the 𝑥 −axis, depending on its concavity.

If 𝑎 > 0, 𝛥 < 0, the graph lies entirely above the 𝑥 −axis and every point on it has a positive 𝑦 −coordinate. 𝑎𝑥K + 𝑏𝑥 + 𝑐 is called positive definite in this case. If 𝑎 < 0, 𝛥 < 0, the graph lies entirely below the 𝑥 −axis and every point on it has a negative y-‐coordinate. 𝑎𝑥K + 𝑏𝑥 + 𝑐 is called negative definite in this case.

When 𝛥 ≥ 0 and for 𝑎, 𝑏, 𝑐 ∈ 𝑄, the 𝑥 intercepts are rational if 𝛥 is a perfect square and irrational if 𝛥 is not a perfect square.

Worked Example 16

Use the discriminant to: determine the number and type of x-‐intercepts of the graph defined by

𝑦 = 64𝑥K + 48𝑥 + 9

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3.6 DETERMINING THE RULE OF A QUADRATIC

POLYNOMIAL FROM A GRAPH Whether the equation of the graph of a quadratic polynomial is expressed in𝑦 = 𝑎𝑥K + 𝑏𝑥 + 𝑐 form, 𝑦 = 𝑎(𝑥 − ℎ)K

+ 𝑘 form or 𝑦 = 𝑎(𝑥 − 𝑥1)(𝑥 − 𝑥2) form, each equation contains 3 unknowns. Hence, 3 pieces of information are needed to fully determine the equation. This means that exactly one parabola can be drawn through 3 non-‐collinear points.

If the information given includes the turning point or the intercepts with the axes, one form of the equation may be preferable over another.

Worked Example 17

Determine the rules for the following parabolas.

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Using simultaneous equations

In Worked example 17b three points were available, but because two of them were key points, the x-‐intercepts, we chose to form the rule using the y = a(x − x1)(x − x2) form. If the points were not key points, then simultaneous equations need to be created using the coordinates given.

Worked Example 18

Determine the equation of the parabola that passes through the points (1, −4), (−1, 10) and (3, −2).

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3.7 QUADRATIC INEQUATIONS Sign Diagrams of Quadratics

If ab > 0, this could mean a > 0 and b > 0 or it could mean a < 0 and b < 0. To assist in the solution of a quadratic inequation, either the graph or its sign diagram is a useful reference.

A sign diagram indicates the values of x where the graph of a quadratic polynomial is above, on or below the x-‐axis. It shows the x-‐values for which 𝑎𝑥K + 𝑏𝑥 + 𝑐 > 0, the x-‐values for which 𝑎𝑥K + 𝑏𝑥 +𝑐 = 0 and the x-‐values for which 𝑎𝑥K + 𝑏𝑥 + 𝑐 < 0.

Worked Example 19

Draw the sign diagram of (4 − 𝑥)(2𝑥 − 3).

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Solving quadratic inequations

To solve a quadratic inequation:

• rearrange the terms in the inequation, if necessary, so that one side of the inequation is 0 (similar to solving a quadratic equation)

• calculate the zeros of the quadratic expression and draw the sign diagram of this quadratic • read from the sign diagram the set of values of x which satisfy the inequation.

Worked Example 20b

Find {𝑥: 𝑥K ≥ 3𝑥 + 10}

Intersections of lines and parabolas

The possible number of points of intersection between a straight line and a parabola will be either 0, 1 or 2 points.

Simultaneous equations can be used to find any points of intersection and the discriminant can be used to predict the number of solutions. To solve a pair of linear-‐quadratic simultaneous equations, usually the method of substitution from the linear into the quadratic equation is used.

Worked Example 21

a) Calculate the coordinates of the points of intersection of the parabola 𝑦 = 𝑥2

− 3𝑥 − 4 and the line 𝑦 − 𝑥 = 1.

b) How many points of intersection will there be between 𝑦 = 2𝑥 − 5 and 𝑦 = 2𝑥2 + 5𝑥 + 6?

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Quadratic inequations in discriminant analysis

The need to solve a quadratic inequation as part of the analysis of a problem can occur in a number of situations, an example of which arises when a discriminant is itself a quadratic polynomial in some variable.

Worked Example 22

For what values of 𝑚 will there be at least one intersection between the line 𝑦 = 𝑚𝑥 + 5 and the parabola 𝑦 = 𝑥2

− 8𝑥 + 14?

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3.8 QUADRATIC MODELS AND APPLICATIONS Maximum and minimum values

The greatest or least value of the quadratic model is often of interest. A quadratic reaches its maximum or minimum value at its turning point. The y-‐coordinate of the turning point represents the maximum or minimum value, depending on the nature of the turning point.

• If a<0, a(x−h)2 +k ≤ k so the maximum value of the quadratic is k.

• If a>0, a(x−h)2 + k ≥ k so the minimum value of the quadratic is k.

Worked Example 22

A stone is thrown vertically into the air so that its height h metres above the ground after t seconds is given by ℎ = 1.5 + 5𝑡 − 0.5𝑡2.

a) What is the greatest height the stone reaches? b) After how many seconds does the stone reach its greatest height? c) When is the stone 6 metres above the ground? Why are there two times? d) Sketch the graph and give the time to return to the ground to 1 decimal place.

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TABLE OF CONTENT

Key knowledge .................................................. 1

Key skills ............................................................ 1

CHAPTER 3 – SET QUESTIONS ...................................... 1

3.2 QUADRATIC EQUATIONS WITH RATIONAL ROOTS ................................................................... 2

QUADRATIC EQUATIONS AND THE NULL FACTOR LAW ...... 2

ROOTS, ZEROS AND FACTORS ....................................... 2 Example 1 ......................................................... 2

USING THE PERFECT SQUARE FORM OF A QUADRATIC ....... 3

Worked Example 2 ............................................ 3

EQUATIONS WHICH REDUCE TO QUADRATIC FORM .......... 3

Example 3 ......................................................... 3

3.3 QUADRATICS OVER R – PART 1 ........................ 4

FACTORISATION OVER R .............................................. 4

Worked Example 4c .......................................... 4

‘COMPLETING THE SQUARE’ TECHNIQUE ........................ 4

Worked Example 4a & 4b ................................. 5

3.3 QUADRATICS OVER R – PART 2 ........................ 6

THE DISCRIMINANT .................................................... 6

Worked Example 5 ............................................ 6

QUADRATIC EQUATIONS WITH REAL ROOTS .................... 7

Worked Example 6 ............................................ 7

Worked Example 7 ............................................ 7

QUADRATIC EQUATIONS WITH RATIONAL AND IRRATIONAL COEFFICIENTS ............................................................ 8

Worked Example 8(a) ....................................... 8

EQUATIONS OF THE FORM 𝒙 = 𝒂𝒙 + 𝒃 .................... 8 Worked Example 9 ............................................ 8

3.4 APPLICATIONS OF QUADRATIC EQUATIONS ..... 9

QUADRATICALLY RELATED VARIABLES ............................ 9

Example Question 1 .......................................... 9

3.5 GRAPHS OF QUADRATIC POLYNOMIALS ......... 10

THE GRAPH OF 𝒚 = 𝒙𝟐 AND TRANSFORMATIONS ........ 10

MAKING THE GRAPH WIDER OR NARROWER ................. 10

TRANSLATING THE GRAPH UP OR DOWN ...................... 10

TRANSLATING THE GRAPH LEFT OR RIGHT ..................... 11

REFLECTING THE GRAPH IN THE X-‐AXIS ......................... 11

REGIONS ABOVE AND BELOW THE GRAPH OF A PARABOLA ............................................................................ 11

SKETCHING PARABOLAS FROM THEIR EQUATIONS .......... 11

1. The general, or polynomial form, 𝑦 = 𝑎𝑥2

+ 𝑏𝑥 + 𝑐 ............................................. 12

Worked Example 13 ........................................ 12

TURNING POINT FORM, 𝒚 = 𝒂(𝒙 − 𝒉)𝟐

+ 𝒌 ......... 12

Worked Example 14 ........................................ 13

FACTORISED, OR X-‐INTERCEPT, FORM 𝒚 = 𝒂(𝒙 − 𝒙𝟏)(𝒙 − 𝒙𝟐) ..................................................... 13

Worked Example 15 ........................................ 13

THE DISCRIMINANT AND THE X-‐INTERCEPTS .................. 14

Worked Example 16 ........................................ 14

3.6 DETERMINING THE RULE OF A QUADRATIC POLYNOMIAL FROM A GRAPH .............................. 15

Worked Example 17 ........................................ 15

USING SIMULTANEOUS EQUATIONS ............................. 16

Worked Example 18 ........................................ 16

3.7 QUADRATIC INEQUATIONS ............................. 17

SIGN DIAGRAMS OF QUADRATICS ............................... 17

Worked Example 19 ........................................ 17

SOLVING QUADRATIC INEQUATIONS ............................ 18

Worked Example 20b ...................................... 18 INTERSECTIONS OF LINES AND PARABOLAS .................... 18

Worked Example 21 ........................................ 18

QUADRATIC INEQUATIONS IN DISCRIMINANT ANALYSIS ... 19

Worked Example 22 ........................................ 19

3.8 QUADRATIC MODELS AND APPLICATIONS ...... 20

MAXIMUM AND MINIMUM VALUES ............................ 20

Worked Example 22 ........................................ 20

TABLE OF CONTENT .............................................. 21

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