Real Functions (4

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Real Functions (4.1 – 4.4)

4.1 Dependent and independent variables. Functional notation. Range and domain.

4.2 The graph of a function. Simple examples.

4.3 Algebraic representation of geometrical relationships. Locus problems.

4.4 Region and inequality. Simple examples.” (syllabus)

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LOCUS: CIRCLE

The word “Locus” comes from the Latin word meaning place. In mathematics when we use the term “locus” we refer to a path traced out by a movable point P(x, y) as it satisfies a set of conditions.

Suppose you are standing at a particular point and want to identify all points at a distance of 3 metres from you. These points would trace out a circle of radius 3 metres with you standing at the centre. We can say that this circle is the locus of points that are 3 metres from where you are standing.

Consider a circle with centre at the origin and radius r. P(x, y) is a point on the circumference.

From our distance formula d = we get r =

r =

By squaring both sides we get the equation for a circle with its centre at the origin and radius r: x² + y² = r²

Now consider a circle with centre C(h, k) and radius r.

Applying the distance formula d = we get

r =

By squaring both sides we get the equation for a circle with centre (h, k) and radius r:

(x-h)² + (y-k)² = r²

Centre origin, radius r Centre (h, k), radius r

x² + y² = r² (x-h)² + (y – k)² = r²

Example 1. Find the locus of a point that moves so that its distance from the origin is 4 units.

Answer 1. Let the point be P(x, y)

4 = 4² = x² + y² x² + y² = 16

Example 2. Find the locus of a point that moves so that its distance from the point (2, 3) is 5 units.

Let the point be P(x, y)

5 = squaring both sides gives:

(x-2)² + (y-3)² = 25

Exercise. Find the locus of a point that moves so that its distance from:

(i) the origin is 10 units (ii) the origin is 13 units

(iii)the point (0,4) is 6 units (iv) the point (-1, 3) is 7 units

Answers: (i) x² + y² = 100 (ii) x² + y² = 169 (iii) x² + (y- 4)² = 36

(iv) (x + 1)² + (y -3)² = 49

LOCUS: STRAIGHT LINE

Remember that the term “locus” refers to a path traced out by a movable point P(x, y) as it satisfies a set of conditions

The set of conditions may describe a straight line.

Example 1. Determine the locus of a point that moves such that its distance from the x-axis is twice its distance from the y-axis.

Answer 1. The distance from the x-axis (the y-coordinate) is twice the distance from the y-axis (the x-coordinate)

Hence the equation is: y = 2x

Example 2. Determine the locus of a point that moves such that it is equidistant from the points A(2, 4) and B( 6, 8).

Answer 2. The most obvious point that is equidistant from A & B is the mid-point

(4, 6)

However any point that is on the perpendicular bisector of AB is also equidistant from A & B.

Let the point be P(x, y) so that PA = PB

Using the distance formula d = we get

PA = and PB =

Since PA = PB then =

Squaring both sides gives: (x-2)² + (y-4)² = (x-6)² + (y-8)²

Expanding both sides; x² – 4x + 4 + y² – 8y + 16 = x² – 12x + 36 + y² – 16y + 64

Bringing all terms to the left gives:

x² – x² + y² – y² – 4x + 12x – 8y + 16y + 4 + 16 – 36 - 64 = 0

Simplifying: 8x + 8y – 80 = 0 and dividing by 8 gives the equation:

x + y – 10 = 0

Exercise1. Find the locus of the point P(x, y) that moves so that it is

(i) 2 units below the x-axis

(ii) 5 units to the right of the y-axis

(iii) equidistant from the x and y axes

(iv) 4 units from the y-axis

(v) 2 units further from the x-axis than it is to the y-axis

(vi) equidistant from the origin and the point Q(3, 3)

(vii) equidistant from the points S(1, 2) and T(7, 6)

(viii) equidistant from the points U(-4, -4) and V(4, 4)

Exercise 2. Show that the locus of a point that moves so that it is twice as far from the origin as it is from W(6, 6) is not a straight line.

Answers 1. (i) y = – 2 (ii) x = 5 (iii) x = y or x – y = 0 & x = -y or x = y = 0

(iv) x = 4 and x = – 4 (v) y = x + 2 & -y = x + 2 (vi) x + y – 3 = 0 (vii) 3x + 2y – 20 = 0 (viii) x = - y or x + y = 0

Answer 2. = 2

x² – 12x + 36 + y² – 12y + 36 = 4(x² + y²)

3x² + 12x + 3y² + 12y – 72 = 0

This is the equation of a circle and not a straight line.

LOCUS: PARABOLA

A parabola is the locus of a point that moves such that its distance from a fixed point (the focus) is equal to its perpendicular distance from a straight line (the directrix).

The important features of a parabola are shown below.

Consider a parabola symmetrical about the y-axis, with its vertex at the origin and having a focal length a.

The focus is the point F(0,a) and the directrix is the line y = -a.

The point P(x, y) is a point on the parabola.

The distance of P from the focus is given by the distance formula:

d = d =

The perpendicular distance of P from the line y = -a is y + a

So from the definition of a parabola: = y + a

Squaring both sides gives: x² + y² – 2ay + a² = y² + 2ay + a²

Cancelling the y² and a² on both sides gives x² – 2ay = y + 2ay i.e. x² = 4ay

Exercise: Find the locus of a point that moves so that it is equidistant from the point

(i) F(0, 2) and the line y = -2 (ii) F(0, 6) and the line y = -6

(iii) F(0, -4) and the line y = 4 (iv) F(0, -10) and the line y = 10

Answers: (i) x² = 8y (ii) x² = 24y (iii) x² = -16y (iv) x² = -40y

LOCUS: PARABOLA 2

The equation of a parabola of focal length a and vertex at (h, k) is: (x – h)² = 4a(y-k)

Example: Determine the locus of a point that moves such that it is equidistant from the point (2, 6) and the line y = 2.

Answer:

Let the point be (x, y)

Distance from (2, 6) is

Distance from y = 2 is y – 2

Equating gives = y – 2

Squaring both sides: (x – 2)² + (y-6)² = (y – 2)²

Expanding: x² – 4x + 4 + y² – 12y + 36 = y² – 4y + 4

Simplifying: x² – 4x – 8y + 36 = 0

Since this equation contains an x² and a y but no y² you would recognise it as a parabola.

Getting the equation into the parabola form: x² – 4x + 4 = 8y – 32

(x – 2)² = 4 x 2(y – 4)

This is a parabola with vertex (2, 4) and focal length 2.

Second method:

Since it is equidistant from a point and a line it is a parabola.

The first step is to find the vertex.

The x coordinate of the vertex is the same as the x coordinate of the focus.

The y coordinate of the vertex is midway between the y coordinate of the focus and the directrix. i.e. = 4 vertex is (2, 4)

The focal length is the distance from the focus to the vertex = 6 – 4 = 2

Substituting in (x – h)² = 4a (y – k) gives (x – 2)² = 4 x 2(y – 4) = 8(y-4)

Exercise:

Find the locus of a point that moves so that is equidistant from

(i) the point (3, 7) and the line y = 1 (ii) the point (8, 5) and the line y = -1

(iii) the point (-1, 4) and the line y = 2 (iv) the point (-3, -4) and the line y = 4

Answers:

(i) (x – 3)² = 4 x 3(y – 4) = 12 (y – 4) or x² – 6x + 57 = 12y

(ii) (x – 8)² = 4 x 3 (y-5) = 12(y-2) or x² – 16x + 88 = 12y

(iii) (x+1)² = 4 x 1(y – 4) = 4 (y-3) or x² + 2x + 13 = 4y

(iv) (x + 3)² = 4 x -4 y = -16y or x² + 6x + 9 = - 16y

SIDEWAYS PARABOLAS

Parabolas of the type y² = 4ax pass through the origin and are symmetrical about the x-axis.

Example: Find the focus and directrix of the parabola y² = 12x

Answer: y² = 12x is y² = 4x3y focal length is 3

Focus is (3,0) and directrix is x = -3.

Exercise1: Complete the following table for parabolas with vertex at the origin:

	Equation	Focus	Directrix
(i)	y² = 20x
(ii)	y² = 6x
(iii)		(4, 0)
(iv)			x = -6
(v)	y² = 8x
(vi)	y² = x

Answers 1:

	Equation	Focus	Directrix
(i)	y² = 20x	(5, 0)	x = -5
(ii)	y² = 6x	(1.5, 0)	x = - 1.5
(iii)	y² = 16x	(4, 0)	x = -4
(iv)	y² = 24x	(6, 0)	x = -6
(v)	y² = 8x	(2, 0)	x = -2
(vi)	y² = x	(0.25, 0)	x = -0.25

Parabolas that do not have their vertex at the origin are of the form: (y –k)² = 4a(x-h)

Example: Find the equation of the parabola with focus (6, 2) and directrix x = 2

Answer: The vertex is midway between the focus and directrix so the vertex is (4, 2) and the focal length is half the distance between the focus and directrix so

a = 2

Equation of parabola is (y – 2)² = 8(x-6)

Exercise 2: Complete the following table:

	Focus	Directrix	Focal Length	Vertex	Equation
(i)	(4, 8)	x = 1
(ii)	(10, 5)	x = 4
(iii)					(y – 6)² = -12(x – 1)
(iv)					(y + 6)² = 6(x + 6)

Answers 2:

	Focus	Directrix	Focal Length	Vertex	Equation
(i)	(4, 8)	x = 1	1.5	(2.5, 8)	(y – 8)² = 6(x – 2.5)
(ii)	(10, 5)	x = 4	3	(7, 5)	(y – 5)² = 12(x – 7)
(iii)	(-2, 6)	x = 4	-3	(1, 6)	(y – 6)² = -12(x – 1)
(iv)	(-4.5, -6)	x = - 7.5	1.5	(-6, -6)	(y + 6)² = 6(x + 6)

The Hyperbola

The simplest type of hyperbola is represented by equations of the type xy = c where c is a constant.

They approach but never touch the x and y axes. The axes are called asymptotes because the graph will get very, very close to them but never touch them.

Consider the equation xy = 1. A table of values is shown below.

x	1	1/2	2	1/4	4	1/8	8
y	1	2	1/2	4	1/4	8	1/8

A similar table of values can be made by using all negative numbers.

x	-1	-1/2	-2	-1/4	-4	-1/8	-8
y	-1	-2	-1/2	-4	-1/4	-8	-1/8

Consequently the graph is drawn in the first and third quadrants. The two sections of the graph are reflections of one another.

If x = 0 then y = , but does not have a definite value and is not a number. Consequently x cannot be zero. The graph is said to be discontinuous at x = 0.