Week 4Functions and Graphs

Objectives At the end of this session, you will be able to:

Define and compute slope of a line. Write the point-slope equation of a line. Write and graph the slope-intercept equation of the line. Recognize the equation for horizontal and vertical lines. Recognize and use the general form of a line’s equation. Find slopes and equations of parallel and perpendicular lines. Find the distance between two points using the distance formula. Find the midpoint of a line segment. Define a circle and write the standard form of a circle’s equation. Give the center and radius of a circle whose equation is in

standard form. Convert the general form of the circle’s equation to standard form.

IndexLines and SlopesPoint slope form of a equation of a lineSlope intercept form of a equation of a lineEquation of horizontal and vertical lineFinding the slope and y-intercept from the general equation of the lineParallel linesPerpendicular linesDistance formulaMidpoint formulaCircleStandard form of equation of a circleGeneral form of equation of a circle

Lines and slopesLet us recall some important points:

Line: Line is a collection of all the points (coordinates) that satisfy a linear equation.Graph of an equation: Let (x1, y1) and (x2, y2) be two solutions of an equation. On plotting these points on a graph paper and joining these points, we obtain a line, called the graph of the given equation.Intercepts are points where a line crosses the coordinate axes.x- intercept is the point where the line crosses the x-axis.

For instance, for equation y= 2x + 1, if y = 0 and we solve for x, we get

x = -1/2 . The point (-1/2 , 0 ) is the x-intercept.

y- intercept is the point where the line crosses the y-axis. For instance, for the same equation if x = 0, y = 1 The point (0,1) is the y-intercept If we just plot these points and use a ruler connect them, we get the

line for the equation.  

Let us recall how we graph an equation

Let us graph an equation y = 2x + 1  Now we find certain values of y corresponding to x which will

help us plot the graph.  For instance, if we assume x = 0, then our equation becomes,

y = 2 ( 0 ) + 1 (substituting the value of x in the given linear equation)

y= 0 + 1 (simplify the equation) y = 1

So, we get one of the coordinate points as (0,1) Now, let us assume x = 1, then

y = 2 (1) + 1 ( substituting the value of x in the given linear equation)

y = 2 + 1 ( simplifying the equation) y = 3

So, our next coordinate point (1,3)

Graph of equation  (Cont…)Similarly assume some more values of x and compute y for those values and get a set of coordinate points as follows:

  

When we plot these points on the rectangular coordinate axis and connect these points, we get a straight line. Any point on this line will satisfy the given equation.

X(Assumed

values)

-1 0 1

y (Computed values)

-1 1 3

1 2 3 4 5 6-1-2-3-4-5-6

1

2

3

4

5

6

-2

-4

-5

-6

-1

(1, 3)

(0, 1)

(-1, -1)-3

X

Y

y = 2x + 1

Slope Slope of a line is a measure of the steepness of a line.Moving from one fixed point P ( x1, y1) to another fixed point Q ( x2, y2), slope compares the vertical change ( rise) i.e. y2 – y1 to the horizontal change (run) i.e. x2 – x1.Definition: The Slope (m) of the line through the distinct points ( x1, y1) and ( x2, y2) is

m = Change in y = Rise Change in x Run = where x2 – x1 0

It is a common notation to let letter ‘m’ represent the slope of a line.

(x2, y2)

P

Q

(x1, y1)

x1 x2

y1

y2

Runx2 – x1

Risey2 – y1

Y

X2 1

2 1

y yx x

Calculating Slope of LineLet us find the slope of the line passing through each pair of points: (-3, -1) and (-2, 4)

  Let (x1, y1) = (-3, -1) and (x2, y2) = (-2, 4) Substituting the values in the equation of the

slope: m = we get, m = 4 – (-1) -2 - (-3)  = 4 + 1 (minus . minus = plus) -2 + 3 = 5/1 (simplifying the equation)

The slope of the line is 5, indicating that there is a vertical change or a rise of 5 units for each horizontal change, a run of 1 unit. The slope is positive, and the line rises from left to right.

-5 1 2 3 4 5-1-2-3-4

1

2

3

4

5

6

-2

-3

-4

-5

-6

-1

(-2, 4)

(-3, -1)

2 1

2 1

y yx x

Important Points (Slope)Positive Slope

Positive slope moves upwards to the right (rises).

In the figure (a), line rises from left to right.

Negative Slope Negative slope moves

downwards to the right (falls). In the figure (b), line falls from

left to right.

m>0

m<0

(a)

(b)

X

Y

X

Y

Important Points (Slope) (Cont…)Zero Slope

A horizontal line has no change in y, so the slope is zero.

In figure (a), line is horizontal, so m = 0.

Undefined Slope A vertical line has no

change in x, so the slope is infinite.

In figure (b), line is vertical, so m is undefined.

m=0

X

Y

m is undefined

X

Y

(a)

(b)

Point Slope Form of Equation of a Line

The point slope equation of a nonvertical line with slope m that passes through the point ( x1 , y1 ) is

y – y1 = m ( x – x1 ) Example: Let us find the point slope form of the equation of a line passing through (2, -5) with slope 6.

Point slope form of the equation is : y – y1 = m (x – x1) In this case we have m = 6, x1 = 2, y1 = -5 Substituting in the point slope form of equation, we get

y – (-5) = 6 (x – 2)y + 5 = 6 ( x - 2) ( minus . minus = plus)We solve this equation for y,y + 5 = 6x - 12 (applying distributive property on the right side)y = 6x - 12 – 5 (subtract 5 from both sides)y = 6x - 17 (simplify)

This is the point slope equation of a line passing through point (2, -5) with slope 6.

  

Slope Intercept Form of Equation of Line

The slope intercept equation of a nonvertical line with slope m and y- intercept b is y = m x + b

Some Important Points:

If b = 0, the equation y = m x + b reduces to y = m x, which is the equation of the line through the origin.

If m = 0, b 0, then the equation y = m x + b reduces to y = b, which is the equation of the line parallel to x-axis at a distance b from it.

If m = 0, b = 0, then the equation becomes y = 0, which represents the x-axis.

Slope Intercept Form of Equation of Line (Cont…)Now let us plot a graph of a line y = (3/5) x + 1 The equation of the line is in the form y = m x + b We find that Slope, m = 3/5 (m = coefficient of x) Next, we plot the y intercept.

Take x = 0 Then, y = (3/5) . 0 + 1 = 1 ( Substituting the

value of x in the equation and solving for y) Point ( 0, 1) is the y intercept. We plot this point on the graph.

Now we have to find a second point using the slope. Write the slope in the form of a fraction i.e. m = 3/5. Here 3 is the rise and 5 is the run. From point ( 0, 1) we move 3 units up (rise) and 5

units to the right (run). This gives us the second point (5, 4).We plot this

point on the graph. Now using a straight edge to draw a line through

the two points.

1 2 3 4 5 6-1-2-3

1

2

3

4

5

6

-2

-3

-4

-5

-6

-1

(5, 4)

(0, 1)

-1

Equations of Horizontal and Vertical lines.Horizontal Line

Equation of a horizontal line is y = b. ( Figure a)

The point (0, b) is the y-intercept.

Vertical LineEquation of a vertical line is x = a. (Figure b)The point (a, 0) is the x-intercept.

(0,b)

(a)

(b)

(a, 0)

X

X

Y

Y

Finding the slope and y-intercept from the general equation of a line

Now let us find the slope and y-intercept of a line whose equation is given in the general form A x + B y + C = 0 where A, B and C are real numbers, and A and B 0

As we have to find the y-intercept, so we isolate the y term and reduce the given equation to the form y = m x + b

Example: 3x + 6y - 12 = 0. 6y - 12 = -3x (Subtract 3x from both sides) 6y = -3x + 12 ( Add 12 to both sides) y = (-3/6) x + 12/6 (Divide both sides by 6) y =(-1/2) x + 2

The coefficient of x, -1/2, is the slope and the constant term, 2, is the y-intercept. 

Parallel LinesTwo non intersecting lines that lie in the same plane are parallel.

 Slope and Parallel lines

Slopes of parallel lines are equal.i.e.If slope of a line is m then the slope of a line parallel to this line will also be m.

We can say that if m1 and m2 are the slopes of two parallel lines then

m1 = m2.If two nonvertical lines are parallel, then they have the same slope. (Figure a)Conversely, if two distinct nonvertical lines have the same slope, then they are parallel. (Figure a)Two distinct vertical lines, both with undefined slopes are parallel. (Figure b)

Y

X

(a)

X

Y

(b)

Perpendicular LinesTwo lines that intersect at a right angle (90 degrees) are called perpendicular lines.

Slope and Perpendicular LinesSlopes of perpendicular lines are negative inverses. i.e. If slope of one line is m then the slope of the line perpendicular to this line is –1/m. If two nonvertical lines are perpendicular, then the product of their slopes is –1.Conversely, if the product of the slopes of the two lines is –1, the lines are perpendicular.One line is perpendicular to another line if its slope is negative reciprocal of the slope of the other.

X

Y

900

Distance FormulaDistance between two given points

The distance between two given points in a plane is the length of the line segment joining them.

Distance Formula The distance, d, between the

points (x1,y1) and ( x2,y2) in the rectangular coordinate system is

d

(x2, y2)

(x1, y1)

|y2 – y1|

|x2 – x1|

X

Y

2 22 1 2 1( ) ( )d x x y y

Distance Formula (Cont…)Let us find the distance between (2, -2) and (5, 2)

Here we assume that (x1,y1) = ( 2, -2 ) and (x2, y2) = ( 5, 2 ) Using the distance formula, we get,

= [ 5 – 2]2 + [2 – (-2)]2

= (5 – 2 )2 + ( 2 + 2 )2 ( minus . minus = plus)= 32 + 4 2 ( performing the resulting operations)

= 9 + 16 (square 3 and 4)

= 25 (add)

= 5 ( 25 = 5 . 5 = 5 )

1 2 3 4 5-1-2-3-4-5

1

2

3

4

5

-2

-3

-4

-5

-1

(2, -2)

(5, 2)

X

Y

2 22 1 2 1( ) ( )d x x y y

Midpoint formula The midpoints of a line segment joining two points (x1,y1) and

( x2,y2) divides the segment in the ratio 1:1. The coordinates of midpoint are

(x1, y1)

(x2 , y2, )

1 2 1 2( )2 2,x x y y

1 2 1 2( )2 2,x x y y

X

Y

CircleDefinition: A Circle is the set of all points in a plane that are equidistant from a fixed point, called center. The fixed distance from the circle’s center to any point on the circle is called the radius.

The figure shows a circle with center (h, k) and radius r.

(h, k)Radius(r)

X

Y

Standard Form of Equation of a Circle Let us now obtain the equation of a circle. We have placed the circle in the rectangular

coordinate system. Center of the circle is (h, k) and radius is r. Suppose (x, y) represents the coordinates of any point on the circle.

Let us find the distance between the two points (h, k) and (x, y) using distance formula,

Substituting the values, (x1, y1) = (h, k) and (x2, y2) = (x, y), we get,

Squaring both sides,

Reversing the sides we obtain the standard form of the equation of a circle as

(x - h)2 + (y - k)2 = r2

(h,k)

(x.y)

r

X

Y

2 2( ) ( )r x h y k

2 2 2( ) ( )r x h y k

2 22 1 2 1( ) ( )d x x y y

Using the Standard Form of Equation of Circle to Graph a Circle Procedure:• Standard Equation of the circle is : (x – h)2 + (y - k)2 = r2

• In order to graph a circle, we need to know its center (h, k) and radius r.

• We can find the values for h, k, and r by comparing the given equation to the standard form of the equation of a circle.

• To graph a circle, we first plot the center (h, k) and then locate the radius according to the coordinates of the center and locate at least four points by going right, left, up and down from the center.

• Then, using the points obtained by the radius, we can graph the circle.

 

(h,k)

(x.y)

r

X

Y

General Form of Equation of a Circle The general form of the equation of the circle is

x2 + y2 + D x + E y + F = 0

Converting the general form of the circle’s equation to standard form and graphing the circle.We have the equation of the circle in general form,

x2 + y2 + D x + E y + F = 0.Group the x-terms , y-terms and the constant terms separately.

i.e., ( x2 + D x ) + ( y2 + E y ) = - F Then add and subtract terms, say P2 and Q2 to complete the square on x and y and perform the same operations on the right side of the equation as well.

i.e. ( x2 + D x + P2) + ( y2 + E y + Q2) = - F + P2 + Q2

Simplify the terms obtained on the both sides of the equation, ( x + P)2 + ( y + Q)2 = r2

We can rewrite this equation as: [ x –(-P)]2 + [y –(-Q)]2 = r2

Now, reduce the equation to its standard form: (x – h)2 + (y – k) 2 = r2 (where h = -P and k = -Q)

SummaryLet us recall what we have learnt so far:

The slope of the line through the distinct points ( x1, y1) and ( x2,y2) is Change in y (m) = , where x2 – x1 0 Change in x

Equations of lines: Point-slope form y – y1 = m (x – x1) Slope-intercept form y = m x + b General form Ax + By + C = 0 Horizontal line y = b Vertical line x = a

Parallel lines have equal slopes and perpendicular lines have slopes that are negative reciprocals.

Distance formula: The distance, d, between the points (x1, y1) and ( x2, y2) in the rectangular coordinate system is

2 22 1 2 1( ) ( )d x x y y

2 1

2 1

y yx x

Summary (Cont…)Midpoint formula: The midpoints of a line segment joining two points (x1,y1) and

( x2, y2) divides the segment in the ratio 1:1. The coordinates of midpoint are

Standard form of equation of a circle with center (h, k) and radius r is (x - h)2 + (y - k)2 = r2

General form of equation of a circle is x2 + y2 + D x + E y + F = 0

1 2 1 2( )2 2,x x y y