Parabolic functions

Contents

6.1 Solving quadratic equations

6.2 Algebraic solution of quadratics

6.3 Sketching quadratic functions

6.4 Finding simultaneous solutions Chapter summary Chapter review

Syllabus subject matter

Introduction to functions ■ Concepts of function, domain and range ■ Ordered pairs, tables, graphs and equations as representations of functions and relations ■ Graphs as a representation of the points whose coordinates satisfy an equation ■ General shapes of functions, including: polynomials up to degree 4; reciprocal functions; absolute value functions ■ Practical applications: polynomials up to degree 2; reciprocal functions; absolute value functions ■ Solutions to simultaneous equations in two variables: graphically, using technology; algebraically (linear and quadratic equations only) Quantitative concepts and skills ■ Calculation and estimation with and without instruments ■ Basic algebraic manipulations ■ Plotting points using Cartesian coordinates ■ Solutions of a quadratic equation Syllabus

Guide

Chapter 6

Parabolic functions

6.1

Solving quadratic equations

You have examined linear functions in Chapter 3 of this textbook. Linear functions are one of

a family of functions known as polynomial functions.

A linear function has the standard (general) formf(x) =ax+b and is a polynomial of degree 1.

A linear equation has the standard form ax+b= 0. The graph of a linear function f(x) =ax+b

is a straight line and is equivalent to the graph of the corresponding linear equation y=ax+b.

A root or solution of an equation is a value of the variable that makes the equation true.

If a graph of a function is drawn, the root is the value of x where the graph crosses the x-axis,

i.e. f(0). Because the roots of a function are the values for which f(x) = 0, they are also known as

the zeros of the function. The zeros are also the x-intercepts of the function.

It is clear that a linear function can at most have one root.

A quadratic function is a polynomial of degree 2.

Quadratic functions are also known as quadratics.

Quadratic functions and equations arise from the motion of projectiles near the Earth. When a stone is thrown, its path is parabolic. Determining where and when it will land involves the solution of quadratic equations. Quadratic equations arise from the solution of many area problems. The speeds of chemical reactions are quadratic functions, the resistance of a wire is a quadratic function of temperature and the energy of a moving object varies as a quadratic with speed. In this chapter, you will learn how to solve quadratic equations and the essential features of quadratic functions.

Polynomial functions

A polynomial function in x may be defined as:

f(x) = anxn+ an− 1xn − 1+ an − 2xn − 2+ … + a1x1+ a0

For this definition, n is known as the degree of a polynomial and is the greatest power to which the variable (x) is raised.

!

x Linear functions

Roots y = f(x)

Standard form of a quadratic equation and function

A quadratic equation involves the second power of the variable (and no higher power). The standard (general) form of the quadratic equation is

ax2+ _bx + _c = _{0 where a} ≠ _0.

A quadratic equation can be written in the form of a quadratic function as

f (x) = ax2+ _bx + _c

!

The shape of the graph of the quadratic in

Example 1 is known as a parabola. The graphs of

all quadratics have this distinctive shape, so

quadratics are also known as parabolic functions.

A quadratic may have as many as two roots (zeros).

A distinctive feature of a parabola of the form

f(x) = ax2 _{+ bx + c is that it is symmetrical about a}

vertical axis passing through its turning point.

Use a graph to find the roots of the quadratic x2− _3x − ₅ = _0.

Solution

Construct a table of values for the function.

Now use the values to sketch the function, joining points with a smooth curve.

The roots of the quadratic are f(0), i.e. the values of x where y = f(x) = 0.

Using the graph, we can estimate that f (x) = 0 at about x =−1.2 and x = 4.2.

Write the equation. x2− _3x − ₅ = ₀

Express as a function. f(x) = x2− _3x − ₅

We want to graph the function, so let y = f(x). y = x2− _3x − ₅

x −3 −2 −1 0 1 2 3 4 5 6

y 13 5 −1 −5 −7 −7 −5 −1 5 13

State the result. The roots of x2− _3x − ₅ = _{0 are x} ≈ _{–1.2 and x} ≈ _4.2.

−4

−1 4

y

12

8

4

−2

−3 2 3 5 x

−8

y = x2− _3x − ₅

10

6

2 14

−6

1 6

Example

1

x y = f(x)

No roots

Two roots

One root

x f(x)

Turning point

Turning point Axis of

symmetry

When plotted on a Cartesian plane, values of f(x) are positive above the x-axis and negative below the x-axis.

All graphics calculators have a function. The following example shows how this

function may be used to quickly find approximate values for the zeros of a function.

y

x f(x) > 0

f(x) < 0

y = f(x)

TRACE

Use a graphics calculator to find approximate zeros of the function

y = f (x) = x2− _7x + ₉

Solution

To find the zeros we can use the fact that f (x) changes sign from positive to negative, or negative to positive, when it crosses the x-axis.

Enter the function Y = X2− _7X + _{9 in Y} 1.

Set the (or V-Window) to minimum and maximum domain values of 0 and 6

and range values of −4 and 4.

the function. Looking at the graph,you can see that there are zeros near 2 and 5.

Select the function and use the and

keys to move the cursor close to the zero (root) near 2.

When you get very close to the zero, you will notice that the value of Y changes from positive to negative (or vice versa depending on the direction in which you are moving).

The screen dumps show how Y changes from0.111…

to −0.061… with a single press of the key. (These

values may vary slightly depending on the calculator used.) This means that the zero occurs at a value of X somewhere between 1.666 … and 1.714 … If required, a more exact

value can be found by zooming in and again using the function.

Note: When using the function in this example, the cursor of the calculator appears to ‘jump’ along the curve in small increments or steps. This is because most calculators use discrete (but small) increments to calculate the points on a curve. We know that the function is continuous, but the calculator draws it using discrete

increments. Because the increments are small, it doesn’t really matter for most situations. The same procedure may now be used to locate

the other zero at X ≈ 5.3.

The zeros of f (x) = x2− _7x + _{9 are}

about 1.7 and 5.3.

WINDOW

GRAPH

TRACE

TRACE

TRACE

Most graphics calculators have inbuilt functions that can be used to find the roots of equations.

Use a graphics calculator to draw a graph and hence find approximate solutions for 3x2+ _x = _8.

Solution

Express in function form: f(x) = 3x2+ _x − _{8. Enter the function y} = _3x2+ _x − _{8 in Y} 1.

Set the (or V-Window) to the default minimum and maximum domain and

range values of −10 and 10.

Casio fx-9860G AU

Select GRAPH from the main menu and select DRAW to graph the function.

As can be seen from the display, the zeros are near −2 and 1.5.

Change the V-Window so that Xmin =−2 and Xmax = 2.

GRAPH the function again.

Select G-Solv and then select ROOT to find the zeros of the function more exactly.

Texas Instruments TI-84

the function. As can be seen from the display, the zeros are near −2 and 1.5.

Change the so that Xmin =−2 and Xmax= 2.

the function again.

Select 2: zero from the CALC menu to find the zeros more exactly.

Type −2 for the left bound and −1 for the right bound, then

for the guess. This gives a value of X ≈−1.8081.

Sharp EL-9900

See the instructions given on the CD-ROM. This gives a value of X ≈−1.8081. Similarly, the other zero is X ≈ 1.4748.

The solutions of 3x2+ _x − ₈ = ₀

are about −1.81 and 1.47.

Repeating the procedure, with left and right bounds of say 1 and 2 respectively, gives the other zero as X ≈ 1.4748.

The solutions of 3x2+ _x − ₈ = ₀

are about −1.81 and 1.47.

WINDOW

GRAPH

WINDOW

GRAPH

ENTER

Example

3

You can also find solutions to some quadratic equations by systematic trial and error. This is the simplest numerical method for solution. It involves the process of systematically trying different values of x until a solution is found.

In some cases, systematic trial and error will not give an exact solution. If so, we must decide how accurate we want the solution. This can be a tedious process ‘by hand’. Using a graphics calculator, a spreadsheet or a computer algebra system helps to make the calculations easier.

Find the solutions to 5x2− _x − ₄₈ = _{0 by systematic trial and error.}

Solution

Calculate 5x2− _x − _{48 for a range of values of x. Look for x values so that 5x}2− _x − ₄₈ = _0.

We know the general shape of the graph. Make a rough sketch.

Because the value of 5x2− _x − _{48 changes from}

negative to positive between x = 3 and x = 4, there must be another solution between 3 and 4.

x −4 −3 −2 −1 0 1 2 3 4 5x2− _x − _{48 36 0} −₂₆ −₄₂ −₄₈ −₄₄ −₃₀ −_{6 28}

One solution is immediately clear. Write it down. x =−3

Write down the function. f(x) = 5x2− _x − ₄₈

Substitute an x value between 3 and 4, say 3.4. f(3.4) = 5 × (3.4)2− _3.4 − ₄₈

= 6.4 3.4 was too high. Looking at the graph, try a

smaller value, say 3.2.

f(3.2) = 5 × (3.2)2− _3.2 − ₄₈

= 0

This is the second solution. x = 3.2

Write the answer. The solutions are x =−3 and x = 3.2.

f (x) 40

20

−3

−4 1 2 3 x

−30

−50

f(x) = 5x2_{− x − 48}

30

10

−40

4

−2

−20

−10

Example

4

Use a spreadsheet to help find solutions to 3x2= _6x + _{12 correct to 2 decimal places.}

Solution

Change the equation so that it is in the standard form: 3x2− _6x − ₁₂ = _0.

In the spreadsheet, type ‘Start’ into cell A1, ‘Increment’ into A2, −4 into B1, 1 into B2, ‘x’ into A3 and ‘Value’ into B3, then type in the following formulas:

Type the formula =B1 into cell A4. Type the formula =A4+$B$2 into cell A5.

Type the formula =3*A4^2-6*A4-12 into cell B4 to enter the function.

Your spreadsheet should look like this:

Now copy cell A5 to cells A6:A14 and B4 to cells B5:B14.

One solution must lie between x values −2 and −1 and the other between 3 and 4.

First root

To find the solution between −2 and −1 more exactly, type −2 into B1 and 0.1 into B2. Then part of your spreadsheet should look like this:

The solution must lie between −1.3 and −1.2. Type −1.3 into B1 and 0.01 into B2.

The solution must be between −1.24 and −1.23. Type −1.24 into B1 and 0.001 into B2.

Exercise 6.1

Solving quadratic equations

1 In each of the following, use the graph to estimate the roots of the function correct to

1 decimal place.

a b

c d

2 Plot graphs of the following quadratics to find the approximate roots of each.

a f (x) = x2+ _x − _{12 b y} = ₅ − _4x − _x2 _{c y} = _x2− _4x − ₆

d f(x) = 3x + 7 − 2x2 _{e f (x)} = _2x2+ _3x − _{8 f y} = − _4x − ₇

3 Use a graphics calculator to draw a graph of each of the following and hence find the

approximate roots.

a f (x) = 2x2− _5x − _{10 b f (x)} = ₈ − − _{4x c y} = _5x2− _14x − ₁₁

d y = 6x + 31 − 3x2 _{e y} = _4x2− _8x − _{26 f f(x)} = + _7x − ₁₅

Second root

Similarly, work on the other root, between 3 and 4. You should obtain:

The solution is closer to 3.24 than 3.23. The solutions to 3x2= _6x + _{12, correct to}

2 decimal places, are −1.24 and 3.24.

Additional Exercise

6 .1

2

−3 −2 −1 1 x 10

−30

−25

−20 5

−5

−10 y 15

−15

4 2

−2 x

5

−20

−5

−10 y 10

−15

6

−8

−6 4 2

−2 y 8

−4

6

−4 −2 2 4 x

−4 −2 2 4 x 10

2 8 6 4 y 12

2

x2

2

---x2

2

---x2

---4 A spreadsheet is being used to find the approximate roots of quadratic functions in a similar process to that used in Example 5. For each of the following spreadsheet screen dumps, state two values between which the root lies.

a b

c d

5 Find the zeros, correct to 2 decimal places, of each of the following quadratics using

systematic trial and error.

a f (x) = 3x2− _8x − _{3 b f (x)} = _3x2− _2x − _{15 c y} = ₁₁ + _3x − _6x2 d y = 13 − 2x − 2x2 _{e f (x)} = _x2+ _5x − _{7 f y} = _x2− _2x − ₁₀

6 Use a spreadsheet and systematic trial and error to find the roots of each of the following

quadratics correct to 2 decimal places.

a f (x) = 2x2− _9x − _{13 b y} = _4x − _3x2+ _{17 c y} = + _3x − ₁

d y = + 3x + 15 e f (x) = 9x2− _5x − _{17 f f (x)} = ₁₇ − _5x − _3x2

6.2

Algebraic solution of quadratics

You saw in the previous section that solving quadratic equations using graphical and trial-and-error methods can be tedious and may not provide exact answers. We can also solve quadratic equations algebraically. The first step in this process requires the quadratic expression to be factorised.

Before we see how to factorise quadratic expressions, it is helpful to review the process for multiplying binomial expressions.

x2

3

---x2

---Factorising quadratics is a matter of reversing the product of the binomials that result in a

quadratic expansion as shown in Example 6. Simple quadratics in the form ax2+ _bx + _{c may}

be factorised using either the decomposition method or the cross method.

The decomposition method is the reverse of the distributive law method of quadratic expansion. We ‘decompose’ the middle term into two terms.

■ The coefficients of the two new terms must add to give the coefficient of the quadratic’s

middle term.

■ Their product must be the same as the product of the coefficients of the first and last terms.

■ The middle term is decomposed and the factorisation completed using the grouping method.

The cross method is the reverse of the FOIL method of quadratic expansion.

■ List factors of the x2 _{term underneath each other on the left.}

■ List the factors of the last term under each other on the right.

Expand the following products.

a (x + 4)(2x + 7) b (3x − 5)(4x + 9)

Solution

Method 2 (FOIL)

Using the FOIL method, we multiply the First, Outer, Inner and Last terms of the product as shown.

First terms are x and 2x.

Outer terms are 4 and 7.

Inner terms are 4 and 2x.

Last terms are 4 and 7.

a Method 1 (distributive law) Write the product. Apply the distributive law a(b + c) = ab + ac.

(x + 4)(2x + 7)= x(2x + 7) + 4(2x + 7)

Apply the distributive law again. = 2x2+ _7x + _8x + ₂₈

Simplify. = 2x2+ _15x + ₂₈

Now multiply the terms. (x + 4)(2x + 7)= x × 2x + x × 7 + 4 × 2x + 4 × 7

= 2x2+ _7x + _8x + ₂₈

Simplify. = 2x2+ _15x + ₂₈

b Method 1 (distributive law) Write the product. Apply the distributive law a(b + c) = ab + ac.

(3x − 5)(4x + 9)= 3x(4x + 9) − 5(4x + 9)

Apply the distributive law again. = 12x2+ _27x − _20x − ₄₅

Simplify. = 12x2+ _7x − ₄₅

Method 2 (FOIL)

Multiply the terms. (3x − 5)(4x + 9)= 3x × 4x + 3x × 9 − 5 × 4x − 5 × 9

= 12x2+ _27x − _20x − ₄₅

Simplify. = 12x2+ _7x − ₄₅

First Outer

Inner Last

(x + 4)(2x + 7)

■ Use cross products to obtain the outer and inner products.

■ Add the outer and inner products to check whether the factors are correctly listed to produce

the middle term of the quadratic.

■ If the middle term is incorrect, rearrange the factors or try new factors until the correct middle

term is obtained.

■ The factors on the left will be the coefficients of the first terms in the binomial factors of the

quadratic. The factors on the right will be the coefficients of the last terms in the binomial factors.

■ Form the binomial factors.

It doesn’t matter which method you choose. However, it is probably best to use only one of these methods, as using both tends to be confusing. Your teacher may have a preference for one of the methods. Some people choose the method according to the coefficient of the first term: they use the cross method if it is 1, and the decomposition method for other cases.

Factorise these quadratics.

a x2− _11x + _{24 b 20x}2− _31x − ₉

Solution

a Decomposition method

Consider the quadratic x2− _11x + _{24. The coefficient of the middle term is} −_11.

The product of the coefficients of the first and last terms is +24.

We want two numbers that add to −11 and multiply to +24. Since the product is positive, the numbers must have the same sign. Since they add to give −11, they must both be negative.

Find numbers whose product is 24. 24 = 24 × 1= 12 × 2 = 3 × 8 = 6 × 4

Select the pair that add to −11. −3 +−8=−11

Now write the expression. x2− _11x + ₂₄

Decompose −11x using −3 + −8 = −11. = x2− _3x + −_8x + ₂₄

Factorise. Be careful with signs! = x(x − 3) − 8(x − 3)

Factorise again: common factor is (x − 3). = (x − 8)(x − 3)

Cross method

Consider the quadratic x2− _11x + _24.

Possible factors of 1 are 1 × 1.

Possible factors of 24 are 24 × 1, −24 ×−1, 8 × 3, and so on.

Set up the ‘cross’ with 1 × 1 on the left, and try the factors of the second term. on the right.

The coefficients of the first terms are 1 and 1. The coefficients of the last terms are −8 + −3.

Now write the result. x2 _{– 11x + 24} = _{(x – 8)(x} − ₃₎

1

1

−6

−4

−10

−8

−3

−11 That’s it!

1

1

−6

−4

−10

−8

−3

−11

(1x + −8)

(1x + −3)

Once a quadratic expression is factorised, it can be solved by using the null factor law.

b Decomposition method

Consider the quadratic 20x2− _31x − _{9. The coefficient of the middle term is} −_31.

The product of the coefficients of the first and last terms is −180.

We want two numbers that add to −31 and multiply to −180. Since the product is negative, the numbers must have opposite signs: one positive and one negative.

Cross method

Consider the quadratic 20x2− _31x − _9.

Possible factors of 20 are 20 × 1, 10 × 2 and 5 × 4. Possible factors of −9 are 9 ×−1, −9 × 1 and 3 ×−3.

List products. 180= 10 × 18 = 9 × 20 = 3 × 60

= 5 × 36 = 2 × 90 = 1 × 180

Select the pair that add to −31. 5 + −36=−31

Write the expression. 20x2− _31x − ₉

Decompose −31x using 5 +−36 = −31. = 20x2+ _5x +−_36x − ₉

Factorise. = 5x(4x + 1) − 9(4x − 1)

Factorise again: common factor is (4x − 1). = (5x − 9)(4x + 1)

Set up a ‘cross’ with 10 × 2 on the left, and try factors of the second term on the right.

Try a new ‘cross’ with 5 × 4 on the left. The binomials are (5x + −9) and (4x + 1).

Write the result. 20x2− _31x − ₉ = _(5x − _9)(4x + ₁₎

10

2

9

−1 8

−1

9 88

−9

1

−8 1

−9

−88 3

−3

−24

−3

3

24 Nothing works!

5

4

9

−1 31

−9

1

−31 That’s it!

Null factor law

If ab = 0, then either a = 0, b = 0 or a = b = 0.

!

Solve the following quadratic equations.

a (x − 3)(x + 4) = 0 b 5p2+ _2p − ₇₀ = _2p2+ _14p − ₇

Solution

a Write down the equation. (x − 3)(x + 4) = 0

Apply the null factor law. (x − 3) = 0 or (x + 4) = 0

Solve the resulting equations. x = 3 or x =−4

Completion of the square is an alternative algebraic method of solving quadratic equations. Completion of the square is a useful technique because it also works in many cases where

factorisation fails. This method relies on the factorisation x2+ _2ax + _a2= _(x + _a)2 _{to make a}

perfect square of the variable parts of the equation. The steps are as follows.

b Write down the equation. 5p2+ _2p − ₇₀= _2p2+ _14p − ₇

Express in standard form. 3p2− _12p − ₆₃= ₀

Take out the common factor. 3(p2− _4p − ₂₁₎= ₀

Factorise the quadratic. 3( p − 7)( p + 3)= 0

Solve using the null factor law. p= 7 or p =−3

Solve x + 5 =− .

Solution

Write down the equation. x + 5 =−

Multiply by the denominator to clear the fraction. x2+ _5x =−₆

Express in standard form. x2+ _5x + ₆ = ₀

Factorise. (x + 2)(x + 3) = 0

Solve using the null factor law. x =−2 or x =−3

6

x

---6

x

---Example

9

Completion of the square

1 Start with the standard form of the quadratic equation.

2 If necessary, divide to make the coefficient of x2 _{equal to 1.}

3 Put the constant term on the right-hand side.

4 Find a (from the pattern x2+ _2ax + _a2= _(x + _a)2_{) by halving the linear coefficient,}

i.e. the coefficient of x.

5 Add a2 _{to both sides.}

6 Write the left-hand side as a perfect square.

7 Take the square root, writing two equations to show the positive and negative cases.

8 Find the final solutions.

Completion of the square is a technique that can be applied to any quadratic equation. This means we can derive a formula for the general quadratic equation using this method. The general derivation is shown on the CD-ROM.

Solve x2+ _6x − ₁₆ = _{0 by completing the square.}

Solution

Write down the equation. x2+ _6x − ₁₆ = ₀

Move the constant term to the RHS. x2+ _6x = ₁₆

The coefficient of x is 6, so a = 3.

Add a2 _{to both sides. x}2+ _6x + ₉ = ₁₆ + ₉

Use the pattern x2+ _2ax + _a2= _(x + _a)2 _{to write}

the LHS as a perfect square.

(x + 3)2 = ₂₅

Take the square root of both sides. x + 3 = 5 or x + 3 = −5

Solve the resulting equations. x = 2 or x =−8

Example

10

Solve 3p2= ₁₂ − _{15p correct to 2 decimal places by completing the square.}

Solution

Write down the equation. 3p2= ₁₂ − _15p

Express with constant term only on the RHS. 3p2+ _15p= ₁₂

Divide throughout by the common factor 3. p2+ _5p= ₄

Now a = 5 2 = , so a2= _.

Add a2 _{to both sides. p}2+ _5p+ = ₄+

Use x2+ _2ax + _a2= _(x + _a)2 _{to factorise the LHS,}

and simplify the RHS. =

Take the square roots. p + = ±

Solve the resulting equations. p= or

Evaluate correct to 2 decimal places. ≈ 0.70 or −5.70

Write the solutions. p≈ 0.70 or p ≈−5.70

5 2

--- 25

4

---25 4

--- 25

4

---p 5

2 ---+

⎝ ⎠

⎛ ⎞2 ₄₁

4

---5 2

--- 41

2

---41–5

2

--- − 41–5 2

---Example

11

Extra Material

The quadratic formula

Quadratic formula

For any quadratic equation ax2+ _bx + _c = _{0 where a} ≠ _{0, the solutions are given by}

x =−b± b2–4ac

2a

For f(x) = ax2+ _bx + _{c, the expression b}2− _{4ac, called the discriminant, is a useful tool for}

determining the number of roots a quadratic equation has, and whether the roots (if any) are rational or irrational numbers. The discriminant is explored on the CD-ROM.

Exercise 6.2

Algebraic solution of quadratics

1 Factorise these quadratics.

a 3x2 − _11x + _{10 b x}2+ _6x − _{16 c 7x}2+ _x − ₆

d 2x2− _11x + _{12 e 4x}2− _13x − _{12 f 8x}2+ _18x + ₉

g 5x2+ _36x + _{7 h 3x}2− _14x − _{49 i 7x}2− _52x − ₃₂

j 5x2− _44x + _{32 k 20x}2− _37x − _{18 l 25x}2− _30x − ₁₆

2 Solve the following by factorisation where necessary.

a 2x(x − 12) = 0 b (x + 2)(x − 5) = 0 c x2− _8x = ₀

d 4x2+ _6x = _{0 e x}2+ _9x + ₁₄ = _{0 f x}2− _8x + ₁₅ = ₀

Solve 3x2= _4x + _{12 using the quadratic formula.}

Solution

Write the equation. 3x2 = _4x + ₁₂

Express in standard form. 3x2− _4x − ₁₂ = ₀

Compare with ax2+ _bx + _c = _{0. a}= _{3, b} =−_{4, c} =−₁₂

Write the quadratic formula. x=

Substitute for a, b and c. =

Evaluate. =

Simplify. =

Simplify the surd. =

=

Factorise. =

Simplify. =

Write as two solutions. = or

If approximate solutions are needed, evaluate. ≈ 2.77 or −1.44

Write the approximate solutions. x≈ 2.77 or x ≈−1.44

−b± b2_–4ac

2a

---4

– ± ( )–4 2_–4×₃×_–12

–

2×3

---4 ± 16+144

6

---4 ± 160

6

---4 ± 16× 10

6

---4 ± 4 10

6

---4 1( ± 10)

6

---2 1( ± 10)

3

---2 1( + 10)

3

--- 2 1( – 10) 3

---Example

12

Extra Material

The discriminant

Additional Exercise

3 Solve by factorisation.

a 3x2+ _5x − ₁₂ = _{0 b 2x}2− _24x + ₂₂ = _{0 c 3x}2+ _15x + ₁₂ = ₀

d 4x2− _28x − ₃₂ = _{0 e 2x}2+ _14x + ₂₀ = _{0 f x}2− _2x = ₃

4 Solve by factorisation.

a x − 4 = b x2− _x = _{6 c x}2− _x = ₁₂

d x2− _6x =−_{9 e 2x}2+ _5x + ₂ = _{0 f 3x}2+ _10x + ₃ = ₀

5 Solve by factorisation.

a 5a2+ _8a − ₄ = _{0 b 2v}2− _19v + ₄₂ = _{0 c n}2− ₆₄ = ₀

d 4c2− _28c + ₄₉ = _{0 e 14e}2+ _33e + ₁₈ = _{0 f 9p}2+ _10p − ₁₆ = ₀

6 Solve by factorisation.

a 2b2= _11b − _{14 b 10d}2+ _10d + ₁₀ = _3d2− _9d

c 6x2− _5x − ₅₀ = _5x2− _5x − _{1 d 5a}2= _4(a2+ _2a − ₄₎

e 4g(g + 6) = g − 15 f 16(x2− ₁₎ = _14x − ₁₉

7 Solve by factorisation.

a x2− _2x + ₂ = _b + ₅ = − _2x

c + 2x = 1 − d + = +

e 16x = f 4x − = x + 2

8 Solve by completing the square. Express irrational answers correct to 2 decimal places.

a k2+ _4k − ₅ = _{0 b v}2+ _2v − ₃₅ = ₀

c j2= _9j − _{8 d x}2= _x + ₁₃₂

e m2= − _{f b}2+ _4b − ₆ = ₀

9 Solve the following equations using the quadratic formula. Where appropriate, give both

exact and approximate answers correct to 2 decimal places.

a x2+ _14x + ₄₅ = _{0 b x}2+ _20x + ₅₁ = ₀

c x2= _7x + _{294 d x}2+ _4x = ₇

e x2− _6x = _{21 f x}2= _13x + ₁₄₀

10 Solve the following using the quadratic formula, correct to 2 decimal places.

a 3f2+ _f − ₁₄ = _{0 b 5t}2= _9t + ₁₈

c 6x2= _5x + _{1 d x}2− _x − ₅ = ₀

e 9p2= _15p + _{5 f 2d}2+ _3d = ₆

11 Solve the following equations for x, using the most appropriate method.

a x2+ _6x − ₁₀ = _{0 b 18x}2+ _15x = ₁₂

c x2− _4x − ₁₃ = _{0 d x}2− _6.146x + _9.328 = ₀

e 1 − + = 0 f − =

g x2+ _29x + ₆ = _{0 h 3x}2− _5.6x + _0.56 = ₀

5

x

---1+3 x2

4

--- x2+9 x

4

--- 4 x+1 2

---3 x2

2

--- 1

3

--- x

3

--- x2

4

--- x

3

--- x2

6

--- x+2

4

---2 x–54

x–4

--- 48

x–3

---8 9

--- 2m

3

---x

2

--- x2

24

--- 1

x+1 --- 4

x

--- 1

---Modelling and problem solving

12 The length of a rectangular lawn is 0.9 m more than its width, and its area is 16.2 m2_{. Write}

an equation and solve it to find the dimensions of the lawn.

13 A square room is 3 m high and 10 L of paint is needed to paint all the walls and ceiling.

If 1 L of the paint will cover 16 m2_{, write an equation and find the length of the room.}

14 The length of a closed box is three times its height, and the box is 4 cm wide. If the total

surface area is 88 cm2_{, find the dimensions of the box.}

15 Marnie can walk 1 km/h faster than Jon. She completes a 20 km hike 1 hour before him.

Write an equation and solve it to find their walking speeds.

16 A bath toy bought for $x is sold for $24,

making x% profit. What was the cost of the toy?

17 A tank can be filled in 6 hours using two

pipes. The larger pipe alone would fill it 5 hours sooner than the smaller pipe alone. How long would each pipe alone take?

18 A cyclist travelled at a certain speed to a

town 30 km away, then reduced speed by 4 km/h and came back. If the cyclist had travelled at a steady 8 km/h for the whole trip, the total time would have been half an hour less. What was the initial speed of the cyclist?

19 The total resistance R of two resistors R₁ and R₂ in parallel is given by

= +

A particular resistor is placed in parallel with one 20 ohms greater in resistance. The total resistance is 24 ohms. What are the resistances of the two resistors?

16.2 m2

1

R

--- 1

R₁

--- 1

R₂

---6.3

Sketching quadratic functions

In this chapter you have already seen that the graph of a quadratic functions has a distinctive parabolic shape, which is symmetrical about a vertical axis drawn through the turning point. The turning point is so named because it is the point on the curve where the slope of the graph changes from positive to negative or from negative to positive.

The turning point of a quadratic function is either a maximum or a minimum as shown in the diagram above.

The graphics calculator is well suited to allow you to see how the shape of the graph of a

quadratic function can change. We will consider the quadratic function y = f(x) = a(x − p)2+ _q.

It is important to have the function expressed in this form rather than the general form because it is easier to see any relationships that emerge.

Horizontal shift

We begin by concentrating on the behaviour of the function y = a(x − p)2+ _{q as the value of p}

changes, so let’s fix a to be 1 and q to be 0. This means we will look at the function y = (x − p)2_.

Enter the function Y = X2 _{in Y}

1. (This is Y = (X − 0)2.)

Set the (or V-Window) to minimum and maximum domain values of −6 and 6 and

range values of 0 and 10.

the function.

Now, on the same set of axes, Y = (X + 3)2 _and

Y = (X − 3)2_.

Watch the graphs carefully as they are drawn.

Repeat this procedure with Y = X2_{, Y} = _(X + ₅₎2 _{and Y} = _(X − ₅₎2_.

Comparing the graphs you have drawn with y = (x − p)2_{, you can see that the value of}

p will slide or translate the graph of y = x2 _{horizontally p units to the right or left: to}

the right if p is positive, to the left if p is negative.

You can also see that, for the function expressed in the form y = (x − p)2_{, the value of p is the}

x-coordinate of the turning point.

Vertical shift

This time, we concentrate on the behaviour of the function y = a(x − p)2+ _{q as the value of q}

changes, so let’s fix a to be 1 and p to be any number, say 2. This means we will look at the function y = (x − 2)2+ _q.

Clear all graphs and enter the function Y = (X − 2)2 _{in Y . (This means q} = _0.)

Negative slope y

x Minimum Maximum

Positive slope Negative slope

Positive slope

Axis Axis

Technology

p = 3 p = −3

WINDOW

GRAPH

Set the (or V-Window) to minimum and maximum domain values of −3 and 5 and

range values of −6 and 10.

the function.

Now, on the same set of axes, Y = (X − 2)2+ _{5 and}

Y = (X + 2)2− _5.

Watch the graphs carefully as they are drawn. Repeat this procedure with values of q other than 5.

Comparing the graphs you have drawn with y = (x − 2)2+ _{q, you can see that the value}

of q will slide or translate the graph of y = x2 _{vertically q units up or down.}

You can also see that, for the function expressed in the form y = (x − p)2+ _{q, the value of q is}

the y-coordinate of the turning point.

Dilation

This time, we concentrate on the behaviour of the function y = a(x − p)2+ _{q as the value of a}

changes, so let’s fix p and q to be any numbers, say 2 and 5 respectively. This means we will look at the function y = a(x − 2)2+ _5.

Clear all graphs and enter the function Y = (X − 2)2+ _{5 in Y}

1. (This means a = 1.)

Set the (or V-Window) to minimum and maximum domain values of −2 and 7 and

range values of 0 and 15.

the function Y = (X − 2)2+ _5.

Now, on the same set of axes, Y = 3(X − 2)2+ _5,

Y = 6(X − 2)2+ _{5 and Y} = _0.5(X − ₂₎2+ _5.

Watch the graphs carefully as they are drawn. Repeat this procedure with other positive values of a.

Comparing the graphs you have drawn with y = a(x − 2)2+ _{5, you can see that the value of}

a dilates the graph. The value of a has no impact on the turning point.

■ If a 1, the graph of y = a(x − 2)2+ _{5 is narrower than the graph of y} = _(x − ₂₎2+ _{5 (has}

an enlargement or dilation factor 1).

■ If a 1 (and a 0), the graph of y = a(x − 2)2+ _{5 is wider than the graph of}

y = (x − 2)2+ _{5 (has an enlargement or dilation factor 1).}

A value of a 0 will cause the parabola to be inverted. Try a similar investigation with negative

values of a. Other than the inversion of the parabola, you will obtain similar results.

The turning point of a quadratic function can be found without graphing by using a variation of the completing the square method. This method was used for solving quadratic equations in

section 6.2. Finding the turning point of a quadratic function y = ax2+ _bx + _{c follows similar}

steps, except that the constant term remains on the same side as the x2 _{and x terms.}

WINDOW

GRAPH q = 5

q = −5

GRAPH

WINDOW

GRAPH a = 6 a = 3 a = 1 a = 0.5

GRAPH

Shape of quadratic graphs

y = a(x − p)2+ _q

Dilation Horizontal shift Vertical shift

a 1, narrows parabola x-coordinate of y-coordinate of

a 1, expands parabola turning point turning point

We can also find the coordinates of the turning point by using the fact that the graph of a quadratic is symmetrical about an axis passing through the turning point. Consider

the graph of f (x) = ax2+ _bx + _{c as shown on the right.}

The turning point (xt, f (xt)) lies on the axis xt= k.

The graph has zeros at x1 and x2.

Using the quadratic formula, the zeros are

x=

Using symmetry k= (x1+ x2)

=

= =

But k= x

Finding the turning point using completion of the square

For the quadratic function y = ax2+ _bx + _c:

1 Factorise the RHS so that the coefficient of x2 _{is 1.}

2 Complete the perfect square by adding the square of half the coefficient of the x term, and subtracting the added term.

3 Separate out the perfect square and then factorise it.

The process changes y = ax2+ _bx + _{c to y} = _a(x − _p)2+ _{q to give the turning point (p, q).}

Maximum and minimum points

■ For positive coefficients of x2_{, the turning point is a minimum.}

■ For negative coefficients of x2_{, the turning point is a maximum.}

!

Alternative Method

Find the turning point of f(x) = 8x + 4 − 2x2 _{and state whether it is a maximum or a}

minimum.

Solution

Write the function. f (x) = 8x + 4 − 2x2

Rearrange and factorise to make the coefficient of the x2 _{term equal to 1.}

=−2(x2− _4x) + ₄

Add and subtract the square of half the coefficient of the x term.

=−2(x2− _4x + ₄ − ₄₎ + ₄

Separate out the perfect square. =−2(x2− _4x + ₄₎ +−₂ ×−₄ + ₄

Factorise the perfect square and simplify. =−2(x − 2)2+ ₁₂

Write the turning point. Turning point is (2, 12).

Coefficient of x2 _{0. Turning point is a maximum.}

Example

13

x y

x1 x2

Zero Zero

Turning point (xt, f (xt))

xt= k

f (x) = ax2_{+ bx + c}

b± b2_–4ac

– 2a

---1 2

---1 2

--- −b+ b2–4ac

2a

--- −b– b2–4ac 2a ---+

⎝ ⎛

⎠ ⎞

1 2

--- –2b

2a

---⎝ ⎠

⎛ ⎞ –b

---So xt =

∴ Turning point (xt, f(xt)) =

The following summary of the various features of quadratic graphs that will be useful when you draw the graphs of quadratic functions.

b

– 2a

---b

– 2a --- f, –b

2a ---⎝ ⎠ ⎛ ⎞

⎝ ⎠

⎛ ⎞

Turning point

The turning point of the quadratic f(x) = ax2+ _bx + _{c where a} ≠ _{0 is}

b

– 2a --- f, –b

2a ---⎝ ⎠ ⎛ ⎞

⎝ ⎠

⎛ ⎞

!

Identify the turning point of f (x) = x2+ _4x − _{21 and state its type.}

Solution

Write the function. f(x) = x2+ _4x − ₂₁

Compare with ax2+ _bx + _{c. a} = _{1, b} = _{4, c} =−₂₁

Calculate . = =−2

Calculate f . f = (−2)2+ ₄ ×−₂ − ₂₁ =−₂₅

Write the turning point. Turning point is (−2,−25).

Coefficient of x2 _{0. Turning point is a minimum.}

b

–

2a

--- –b

2a

--- –4

2

---b

–

2a

---⎝ ⎠

⎛ ⎞ –b

2a ---⎝ ⎠ ⎛ ⎞

Example

14

Quadratic graphs

The graph of a quadratic function is a parabola. The graph has the following shape:

■ If the coefficient of the x2 _{term is positive, the}

function decreases to a minimum value at the turning point, and then increases.

■ If the coefficient of the x2 _{term is negative, the}

function increases to a maximum value at the turning point, and then decreases.

The graph is symmetrical about a vertical line called the axis passing through the turning point.

If the function has zeros, these are symmetrical about the turning point.

For graphs of quadratic functions, the domain (possible x values) is the set of real numbers.

The range (possible f (x) or y values) must be greater than or equal to the minimum value, or less than or equal to the maximum value.

Zero

Minimum _Maximum

Axis Axis

Zero _x

The essential features of the graph of a quadratic are the turning point, the zeros (if they exist), the axis of symmetry and the y-intercept.

■ The turning point of a quadratic function may be found by completing the square.

■ The zeros (if they exist) may be found by factorisation, completing the square or using the

quadratic formula.

■ The axis of symmetry is the vertical line that passes through the turning point.

■ The y-intercept is the value when x = 0.

The graph of a quadratic function can be sketched using this information. If a domain is stated, then the endpoints and range also should be found.

Sketch the function y = x2− _4x − _{12 and state the range.}

Solution

Sketch the graph.

Turning point

Write the function. y = x2− _4x − ₁₂

Complete the square. = x2− _4x + ₄ − ₄− ₁₂

Factorise the perfect square and simplify. = (x − 2)2− ₁₆

State the coordinates of the turning point. Turning point is (2, −16).

Coefficient of x2 _{term 0. Turning point is a minimum.}

Zeros

Let the function be zero. x2− _4x − ₁₂= ₀

Factorise. (x − 6)(x + 2) = 0

Solve. x = 6 or x =−2

State the result. x-intercepts are −2 and 6.

Other features

Write the function. y = x2− _4x − ₁₂

Let x = 0. y =−12

State the result. y-intercept is −12.

Axis of symmetry passes through the turning point.

Axis of symmetry is x = 2.

No domain is stated, so the implied domain is the real numbers.

The range is y −16.

y

6 x

(2, −16)

−2

−12

y = x2_{− 4x − 12}

Sketch the function f (x) = 4x − 3x2+ _{5 for} −_{1 x 3 and state the range.}

Solution

Sketch the graph.

Turning point

Write the function. f (x)= 4x − 3x2+ ₅

Rearrange and factorise. =−3(x2− _x) + ₅

Complete the square. = −3(x2− _x + − ₎ + ₅

Separate out the perfect square. = −3(x2− _x + ₎ +−₃ ×− + ₅

Factorise the perfect square and simplify. = −3(x − )2+ ₆

State the result. Turning point is ( , 6 ).

Coefficient of x2 _{term 0. Turning point is a maximum.}

Zeros

Let the function equal zero. 4x − 3x2+ ₅ = ₀

Express in standard form. −3x2+ _4x + ₅= ₀

Compare with ax2+ _bx + _c = _{0. a} =−_{3, b} = _{4, c} = ₅

Use the quadratic formula. x =

Substitute for a, b and c. =

Simplify. =

State the exact solutions. =

Approximate solutions are useful for graphing. ≈−0.786 or 2.120

State the result. x-intercepts are −0.786 and 2.120.

Other features

Write the function. f (x)= 4x − 3x2+ ₅

Find f(0). f (0)= 5

State the result. y-intercept is 5.

Axis of symmetry passes through the turning point. Axis of symmetry is x = .

Investigate f(x) at extreme domain values. f (−1)=−2 f (3) =−10

State the result. Endpoints are (−1,−2) and (3,−10).

State the result. The range is −10 y 6 .

4 3 ---4 3 --- 4 9 --- 4 9 ---4 3 --- 4 9 --- 4 9 ---2 3 --- 1 3 ---2 3 --- 1 3

---b± b2–4ac –

2a

---4

– ± 42–4×–3×5

2×–3

---4± 76

– 6 –

---2± 19

3

---2 3

---( , 6 )

x y

5

2.12

(−1, −2)

−0.79

(3, −10)

2 3

1 3

y = 4x − 3x2+ ₅

1 3

We can use the graphics calculator to draw graphs of quadratic functions and to find the key points.

Draw the graph of y = x2− _5x + _{4 on a graphics calculator and find the key points.}

Solution

For all calculators

Enter the function y = x2− _5x + _{4 in Y}

1. Set the (or V-Window).

Draw the graph with a domain from −1 to 6 and a range from −3 to 3.

Casio fx-9860G AU

Once the graph is drawn, use the G-Solv function to select MIN to find the turning point.

Once again use the G-Solv function to select ROOT to find the x-intercepts. To find the other zero, press .

The minimum is at (2.5,−2.25)

and the x-intercepts are 1 and 4.

Texas Instruments TI-84

Once the graph is drawn, press to enter the

CALC (calculate) menu. Select 3: minimum.

The calculator will ask for a left bound, right bound and guess.

Since the curve is simple, press for each question.

The calculator will display the coordinates of the minimum.

To find a zero, select 2: zero from the CALC menu.

When the calculator asks for the left bound, use to

place the cursor before the zero and press .

For the right bound, use to place the cursor after the zero

and press . When it asks for guess, just press

or place the cursor close to the zero first. The calculator will

then display the coordinates of the zero. Repeat to find the other zero.

The minimum is at (2.5,−2.25)

and the x-intercepts are 1 and 4.

Sharp EL-9900

See the instructions given on the CD-ROM.

WINDOW

2nd TRACE

ENTER

ENTER

ENTER ENTER

Example

17

Exercise 6.3

Sketching quadratic functions

1 Find the position and type of turning point for each of the quadratic functions graphed below.

a b

c d

2 Use the turning point and y-intercept to determine the equation of each of the functions

shown in question 1.

3 Which of the following graphs best represents the quadratic function y = −x2− _2x − _3?

Additional Exercise

6.3

1 2 3 4 5 6 7

−1 x

2 4

−6

−4

−8

−10

−12 y

12 14

6 8

4

2

−2 y

−3 −2 −1

−4 x

10

3 3.5

1.5 2

1

0.5

−0.5 y

−3 −2 −1 x

−1 2.5

10 12

4 6

2

−2

−4 y

1 2 3

−1 x

8

−1

−2 1 3 x

−3 2

y 2

−4

−6

−1

−2 1 3 x

−3 2

y 2

−4

−6

A B

−1

−2 1 3 x

−3 2

y 2

−4

−6

−1

−2 1 3 x

−3 2

y 2

−4

−6

4 Match each of the following quadratic functions with the correct sketch.

a f (x) = x2+ _{5 b f (x)} =−_x2+ _{8x c f (x)} =−_3x2+ _4x + ₄ d f (x) = 2x2− _x − _{3 e f (x)} = _2x2+ _9x + _{5 f f (x)} =−_6x2+ _7x + ₃

5 State the position and type of turning point for each of the following quadratic functions.

a y = (x + 1)2− _{2 b y} = _3(x − ₂₎2+ _{3 c y} = ₉ + _(x − ₁₎2

d y = 1 − 2(x + 3)2 _{e y} = _5[12 − _4(x + ₂₎2_{] f y} =−_2[1 + _2(x − ₄₎2_]

6 Find the position and type of turning point for each of the following quadratic functions.

a y = x2− _3x + _{1 b y} =−_x2+ _7x + _{2 c y} = _x2− _5x + ₂

d y = x2− _3x − _{18 e y} = _x2− _7x + _{5 f y} = _3x + ₁ − _x2

7 Find the position and type of turning point for each of the following quadratic functions.

a f (x) = x2+ _6x + _{10 b f (x)} = _2x2+ _16x + _{28 c f (x)} = _10x − _x2− ₂₅ d f (x) = 16x2+ ₆₂₈ − _{208x e f (x)} = _4x2− _20x + _{41 f f (x)} =−_3x2− _24x − ₅₇ g f (x) = x2− _4x − _{12 h f (x)} =−_x2− _9x − ₁₄

8 For the quadratic function y = x2− _4x − _{12 find the:}

a y-intercept b x-intercepts (if any)

c nature and position of the turning point d axis of symmetry.

Sketch the graph and state the domain and range.

9 For the quadratic function y =−2x2− _2x + _{24 find the:}

a y-intercept b x-intercepts (if any)

c nature and position of the turning point d axis of symmetry.

Sketch the graph and state the domain and range.

10 Sketch the following quadratic functions, showing all key points, and state the ranges.

a y = (x + 3)2− _{4 b y} =−_x2− _{3x, for} −_{4 x 2}

c y = x2− _2x − _{2 d y} = _x2− _{9, for} −_{4 x 4}

e y = 9 − x2 _{f y} = ₉ − _6x − _3x2_{, for} −_{2 x 4}

g y =−3x(x − 4) h y = 4x2− _4x − ₃

i y = 8 − 10x − 3x2_{, for} −_{5 x 2 j y} =−_3x2+ _12x

k y = 16x − 4x2 _{l y} = _3x2− _12x − ₁₅

A f(x) B

6

2

−1 2 x

−4

1 3

f(x) 4

2

−1 2 x

−4

1 3

f(x) 6 4

−2 2 x

−4

−4

−6 2 4

C

D E f(x)

4 2

2 x 1 6

−2

2 x

−2 10 f(x)

2 4 6 8

0

20 f(x)

−4 16 12 8 4

2 4 6 8 x

−2

−4

F

−1

3 1

−1

11 Use your graphics calculator to find the key points of:

a y = x2+ _6x + _{2 b f (x)} = ₉ − _2x2− _5x

c y = 23x − 4x2− _{1 d f (x)} = ₇ − _5x + _2x2

12 Find the equations of the following parabolas in the form f (x) = ax2+ _bx + _c.

a y-intercept of −6 and turning point at (2,−2)

b y-intercept of −3 and x-intercepts of −1 and 2

c x-intercepts of −2 and 3 and a maximum of 5

Modelling and problem solving

13 Show that the turning point of y = ax2+ _bx + _{c occurs at by completing}

the square.

14 Find the distance between the turning points of y = 10x − x2 _{and 2y} = ₇₅ + _10x − _x2_.

15 The water solubility of a particular compound changes with temperature as follows.

Maximum amount dissolved (g/L) = 25 − T + 0.0125T2_{where T is measured in °C.}

Sketch a graph of this relationship, and hence find the temperature with the minimum solubility.

16 The diagram on the right shows a square plot of ground with a

side length of 40 m. Two areas of the plot, PAQ and CDQ,

are found to be unsuitable for planting crops. The plot owner wants to used the remaining (shaded) area for planting vegetables. Show that the area, A, available for planting vegetables is given by

A = 800 + 20x − x2

and find the value of x for which the area is a maximum.

17 Find the distance from the turning point

of 20y = x(50 − x) to the line y = 40.

18 The height of a ball thrown upwards at

25 m/s is approximately given by the equation

h(t) = 2 − 5t2+ _25t

Find the turning point and intercepts, and hence the time taken to reach the maximum height and the time taken to reach the ground.

6.4

Finding simultaneous solutions

You have already seen that simultaneous linear equations can be solved graphically by finding the intersection of the lines. A similar method can be applied to simultaneous equations where one is linear and the other quadratic.

b

– 2a

--- 4ac–b2

4a ---,

⎝ ⎠

⎛ ⎞

A

B C

D P

Q x

x 1

---Use graphs to find simultaneous solutions to y = 2x2− _x − _{10 and y} = ₄ − _4x.

Solution

Reasonably accurate graphs are needed for the solution, so are best drawn on graph paper.

Graph 1

It is useful to add some extra points to help sketch the graphs.

Graph 2

Plot both graphs on the one set of axes.

From the graph, it appears that the intersections are at about (−3.5, 18) and (2,−4).

Write the equation. y = 2x2− _x − ₁₀

Identify the features of the graph. It is a parabola with y-intercept −10.

Let y = 0 to find zeros. 0 = 2x2− _x − ₁₀

Factorise. 0 = (2x − 5)(x + 2)

Solve. 0 = (2x − 5) or 0 = (x + 2)

x = 2 or x =−2

State the result. Zeros exist at −2 and 2 .

Write the equation. y = 2x2− _x − ₁₀

Express so x2 _{term has coefficient of 1. y} =₂ − ₁₀

Complete the square. y =2 − 10

Separate out the perfect square. y =2 + 2 ×− − 10

Factorise the perfect square and simplify. y = 2(x − )2− ₁₀

State the turning point. Minimum is at ( ,−10 ).

Write the equation. y = 4 − 4x

Let y = 0. x-intercept is 1.

Let x = 0. y-intercept is 4.

The exactness of these solutions can be checked by substitution into the equations.

2(−3.5)2− ₍−_3.5) − ₁₀ = _{18 OK}

4 − 4(2) = −4 OK

Write the answer. The simultaneous solutions are

x = −3.5, y = 18 and x = 2, y =−4.

1 2 ---1 2

---x2 x

2 ---–

⎝ ⎠

⎛ ⎞

x2 x

2 ---– 1 16 --- 1 16 ---– + ⎝ ⎠ ⎛ ⎞

x2 x

2 ---– 1 16 ---+ ⎝ ⎠

⎛ ⎞ 1

16 ---1 4 --- 1 8 ---1 4 --- 1 8

---x −4 −1 4

y 26 −7 18

−5 −10 −2 −3 10 5 15 20 25 y x

1 2 3

−1

y =2x2− _x − ₁₀ y ₌

4 ₋ 4_x

Three possible cases exist for the solution of simultaneous equations with a quadratic and a linear equation:

■ two solutions (as in Example 18) ■ one solution, where the line is a tangent ■ no solutions, where the lines do not intersect. A graphics calculator is particularly useful when solving this type of problem.

One solution

Linear functions Quadratic

function

Two

No solutions y

x solutions

Use a graphics calculator to solve y = x2+ _2x − _{3 and x} − _2y − ₄ = _0.

Solution

For all calculators

Both functions must be in the form y = f(x), so we change the second function to y = − 2. Enter the functions into Y1 and Y2 respectively.

Set the (or V-Window) to a domain of [−4, 2] and a range of [−5, 1] and draw

the graphs.

Casio fx-9860G AU

Once the graphs are drawn, use the G-Solv function to select ISCT to find the points of intersection of the graphs.

The first point of intersection will then be displayed.

To find the other point of intersection, press . The second point of intersection will then be displayed.

The intersections are at (−2,−3) and (0.5,−1.75), so the

sim