1. Graphs of Straight line 7/5/14
LI: to draw graphs by plotting the points on X-Y axis.
SC:P EX 13.02#1,2
HWK Beta P 368 Ex 27.4

2. We call this line the graph of y = 2x.
13.02
Complete this table for values of co-ordinates that fit the rule y = 2x and draw the graph. y –2 –1 1 2 3
Co-ordinates
y (y = 2x) x 6 4 2 4 6
(3, 6) 4
(2, 4) 2
(1, 2)
(0, 0) –1
(–1,–2) –4
(–2, –4)
We call this line the graph of y = 2x.

3. Gradients=Slope SC: P182 EX 13.03 – 13-06 #odds
LI: 1. to understand gradients from lines.
Positive and negative slopes.
2. to draw gradients
vertical(Y)
Horizontal(X)
Gradient =
SC: P182 EX – #odds

4. Who is this on Baldwin Street on 20th Jan 2011

5. Photo (c) Rob Ainsley, shown here sliding gently down Baldwin Street, New Zealand
CTC Magazine, June/July 2001
“Unfortunately the people planning it were in Britain -- 12,000 miles away -- and their maps weren't quite contoured up to OS standards. That main road ran along the bottom of a steep valley, and the side road, laid faithfully according to the plans, ran straight up a hillside. It resulted in Baldwin St being the steepest paved road in the world.”
Source

6. Source http://www. bikereader
What do you think the gradient is of this part of the street? What is the angle of rise?

7. ^http://home.intranet.org/~miche/jamesdig nan/travelpages/ot1.html
Which way is up? Chris on Baldwin Street. (Photo taken by Rick.17 Sept 03)
< bum62.html
1.266 1
The famous Baldwin Street hill, is noted in the Guinness Book of World Records as the steepest road in the world at 1:1.266, or an astounding 38 degrees.
^ nan/travelpages/ot1.html

8. Write down the gradients of the lines shown on the grids below.
In both cases, choose points on the line on the background grid. a b
The gradient is
The gradient is 1
Note: These are positive gradients.
Line slants to right

9. What is the gradient of the line shown below?
13.05
What is the gradient of the line shown below?
Look for two points on the line on the background grid.
First point
These give:
A sideways ‘run’ of 6 units
Second point
A ‘rise’ of 4 units downwards
so we take the rise as being –4
Negative gradient.
Line slants to left.
The gradient of the line is

10. Draw a line that has a gradient of 3.
13.03B
Draw a line that has a gradient of 3.
Gradient of 3 means
Start from any point at all on a grid
Go along 1 unit
Go up 3 units
This gives another point on the line
Join the two points, extending the line in both directions

11. What is the gradient of the line given by the rule
The gradient is the coefficient of (number in front of) x.
Gradient =

12. Graphs of Straight line 13/5/14
LI: to draw graphs by plotting the points on X-Y axis.
SC:P EX 13.02#1
EX #all
Ex 13.04#1,2 odds
Finish for homework.
Other h/w -HOTMATHS

13. Gradient(m) from Equation
Equation of type:y=mx
m= gradient, m is the number in front of x.
X is the x axis variable
Line will always pass through (0,0)
LI: to draw graph quickly from rule y=mx
Using the gradient(m).

14. SLO: Structured Learning Outcome
1. can draw the gradient from (0,0)
2. Can draw graph of y=mx
3.Can identify that y=mx always passes through (0.0)

15. What is the gradient of the line given by the rule
The gradient is the coefficient of (number in front of) x.
Gradient =

16. Draw the line through (0, 0) with this gradient.
13.04B
Draw the graph of y = 4x
The gradient is or 4
the coefficient of x 4 2 2 4 x y
Draw the line through (0, 0) with this gradient.

17. 13.04B
Draw the graph of y = 4x 4 2 2 4 x y

18. The graph goes through (0, 0) because the rule is like y = mx
13.06
Draw the line for y = –3x
The graph goes through (0, 0) because the rule is like y = mx
Its gradient is –3, which we write as 4 2 2 4 x y
From (0, 0)
Go 1 unit along
then 3 units down (the gradient is negative!)

19. The graph goes through (0, 0) because the rule is like y = mx
13.06
Draw the line for y = –3x
The graph goes through (0, 0) because the rule is like y = mx
Its gradient is –3, which we write as 4 2 2 4 x y
From (0, 0)
Go 1 unit along
then 3 units down (the gradient is negative!)

20. 13.07
Write down the y-intercept and draw each of these lines a y = x b y = x – 2
a The y-intercept is 1
b The y-intercept is –2
The gradient is 1
The gradient is 1 x y y x 1 2

21. Drawing any line 14/5/14 LI: to draw graphs of any equation of type
Y=mx+c
Any straight line equation can be written as y=mx+c
The gradient (Slope) = m
The y axis intercept= c
To draw the graph use gradient and y- intercept together.

22. Method for y=mx+c Eg. Draw the graph of y=2x-4 1. write what m= and c=
2. Draw x-y axis. Label the axis -5 to 5
3. mark y-intercept (c)
4. from c draw gradient of m.
5. join the two points using a ruler.
6. Eureka. Admire the graph.

23. Draw the line given by the rule y = 2x – 4
The gradient is 2
The y-intercept is –4 x
Place a mark on the y-axis at –4 4
This gives a point which will be on the line.
From this point, draw a line with a gradient of 2.
Note
This gives another point on the line.
Draw a line that goes through both points
, so go across 1 unit, and up 2 units.

24. Draw the line given by the rule y = 2x – 4
The gradient is 2
The y-intercept is –4 x
Place a mark on the y-axis at –4 4
This gives a point which will be on the line.
From this point, draw a line with a gradient of 2.
Note
This gives another point on the line.
Draw a line that goes through both points
, so go across 1 unit, and up 2 units.

25. 13.07
Write down the y-intercept and draw each of these lines a y = x b y = x – 2
a The y-intercept is 1
b The y-intercept is –2
The gradient is 1
The gradient is 1 x y y x 1 2

26. Draw the line given by the rule
The gradient is
The y-intercept is 1 1
Mark 1 on the y-axis x
From this point, draw in a line of gradient

27. Draw the line given by the rule
The gradient is
The y-intercept is 1 1
Mark 1 on the y-axis x
From this point, draw in a line of gradient

28. Special Lines X= number is a vertical line.
Y= number is a horizontal line
Beta P EX P190

29. Draw a y = 2 b x = 3 a Horizontal line b Vertical line
13.09
Draw a y = b x = 3
a Horizontal line
Rule is y = number b
Vertical line
Rule is x = number x y 2 2 2 1 1 2 3 x y 4

30. Topic 1:Graphs 91257 AS 2.2 Internal Assessment Credits= 4 EOY
Polynomials (curved graph)
1. Parabola General equation y= ax2+ bx +c
2. Cubics y= (x+a)(x+b)(x+c)
3. Simple Hyperbolas y=ax/b
4. Exponential
5. Functions-Domain and Range
6. Transformations

31. Quadratic-Parabola Beta P 196 Ex 13.11 and 13.12
LI: to draw graphs of families of Parabola
SC: can draw basic Y=x2
Can identify its properties
Can carry out all transformations
Can draw graphs of y=(x+a)(x+b)
(open NZD11-demo14)
Do now
Beta P 196 Ex and 13.12

32. Graphs of Parabola LI: to be able to sketch a parabola.
These are U or ∩ shaped curve.
The equation is of the type:
Y= x2 or y=x2+- a number or Y=ax2+bx +c
Gamma P 182 EX 15.01all#
HWK: Glossy gamma Pages 75-80

33. Complete the table of values and then draw y = x2 – 3
13.11
Complete the table of values and then draw y = x2 – 3 2 1 –1 –2
Co-ordinates
y (Note y = x2 – 3) x
(–2)2 – 3 =
–2  –2 – 3 =
4 – 3 = 1
(–2, 1)
(–1)2 – 3 =
–1  – 1 – 3 =
1 – 3 =
– 2
(–1, –2)
(0)2 – 3 =
0  0 – 3 =
0 – 3 =
– 3
(0, –3)
(1)2 – 3 =
1  1 – 3 =
1 – 3 =
– 2
(1, –2)
(2)2 – 3 =
2  2 – 3 =
4 – 3 = 1
(2, 1) y 2 2 2 x 2

34. Calculate points from this table and plot the points as you go.
CAS
Calculate points from this table and plot the points as you go. x y
The basic parabola is shown on the graph, dotted green.
Draw a parabola through the points.
Adding 2 to x2 moves the basic parabola
up by 2 units.
How?
Gamma Mathematics Workbook © Pearson Education New Zealand 2007 34

35. Calculate points from this table and plot.
CAS x y
Calculate points from this table and plot.
The basic parabola is shown on the graph, dotted green.
Draw a parabola through the points.
Adding 4 to x and then squaring moves the basic parabola
How?
to the left by 4 units.
Gamma Mathematics Workbook © Pearson Education New Zealand 2007 35

36. Calculate points from this table & plot.
CAS x y
Calculate points from this table & plot.
Draw a parabola through the points.
The graph shows the parabola cuts the x-axis at 3 and –1.
Continued on next slide.
Gamma Mathematics Workbook © Pearson Education New Zealand 2007 36

37. The x-intercepts can be obtained from the two factors:y
The x-intercepts can be obtained from the two factors:
At x-intercepts, y = 0.
Solve. x
Coordinates of the x-intercepts are (3,0) and (1,0).
Gamma Mathematics Workbook © Pearson Education New Zealand 2007 37

38. Graphs of Parabola LI: to be able to sketch a parabola.
These are U or ∩ shaped curve.
The equation is of the type:
Y= x2 or y=x2+- a number or Y=ax2+bx +c
Gamma P 182 EX 15.01all#
Beta P 382 EX 27.14

39. Revision on Graphs
Beta pp
Hwk Beta pp78-84

40. Revision on Graphs Speed=Distance/Time
Ferrari F430
Max Speed=330km/h
Speed is the main factor in stopping distance. At 100km/h
S.D= 96m.
How much for Ferrari at Max speed.
SC: P EX 27.2

41. We call this line the graph of y = 2x.
13.02
Complete this table for values of co-ordinates that fit the rule y = 2x and draw the graph. y –2 –1 1 2 3
Co-ordinates
y (y = 2x) x 6 4 2 4 6
(3, 6) 4
(2, 4) 2
(1, 2)
(0, 0) –1
(–1,–2) –4
(–2, –4)
We call this line the graph of y = 2x.

42. Gradients=Slope SC: P371 EX 27.6
LI: 1. to understand gradients from lines.
Positive and negative slopes.
2. to draw gradients
vertical(Y)
Horizontal(X)
Gradient =
SC: P371 EX 27.6

43. Drawing any line LI: to draw graphs of any equation of type Y=mx+c
Any straight line equation can be written as y=mx+c
The gradient (Slope) = m
The y axis intercept= c
To draw the graph use gradient and y- intercept together.

44. Method for y=mx+c Eg. Draw the graph of y=2x-4 1. write what m= and c=
2. Draw x-y axis. Label the axis -5 to 5
3. mark y-intercept (c)
4. from c draw gradient of m.
5. join the two points using a ruler.
6. Eureka. Admire the graph.

45. Draw the line given by the rule y = 2x – 4
The gradient is 2
The y-intercept is –4 x
Place a mark on the y-axis at –4 4
This gives a point which will be on the line.
From this point, draw a line with a gradient of 2.
Note
This gives another point on the line.
Draw a line that goes through both points
, so go across 1 unit, and up 2 units.

46. Draw the line given by the rule y = 2x – 4
The gradient is 2
The y-intercept is –4 x
Place a mark on the y-axis at –4 4
This gives a point which will be on the line.
From this point, draw a line with a gradient of 2.
Note
This gives another point on the line.
Draw a line that goes through both points
, so go across 1 unit, and up 2 units.

47. Draw the line given by the rule
The gradient is
The y-intercept is 1 1
Mark 1 on the y-axis x
From this point, draw in a line of gradient

48. Special Lines X= number is a vertical line.
Y= number is a horizontal line
Beta P EX 13.09#1, 2
.P191 Ex #1, 2,3,5, 6

49. Draw a y = 2 b x = 3 a Horizontal line b Vertical line
13.09
Draw a y = b x = 3
a Horizontal line
Rule is y = number b
Vertical line
Rule is x = number x y 2 2 2 1 1 2 3 x y 4

50. Graphs of Parabola LI: to be able to sketch a parabola.
These are U or ∩ shaped curve.
The equation is of the type:
Y= x2 or y=x2+- a number or Y=ax2+bx +c
Gamma P 116 EX 9.01
Beta P 382 EX 27.14

51. Complete the table of values and then draw y = x2 – 3
13.11
Complete the table of values and then draw y = x2 – 3 2 1 –1 –2
Co-ordinates
y (Note y = x2 – 3) x
(–2)2 – 3 =
–2  –2 – 3 =
4 – 3 = 1
(–2, 1)
(–1)2 – 3 =
–1  – 1 – 3 =
1 – 3 =
– 2
(–1, –2)
(0)2 – 3 =
0  0 – 3 =
0 – 3 =
– 3
(0, –3)
(1)2 – 3 =
1  1 – 3 =
1 – 3 =
– 2
(1, –2)
(2)2 – 3 =
2  2 – 3 =
4 – 3 = 1
(2, 1) y 2 2 2 x 2

52. Calculate points from this table and plot the points as you go.
CAS
Calculate points from this table and plot the points as you go. x y
The basic parabola is shown on the graph, dotted green.
Draw a parabola through the points.
Adding 2 to x2 moves the basic parabola
up by 2 units.
How?
Gamma Mathematics Workbook © Pearson Education New Zealand 2007 52

53. Revision on Graphs
Beta pp
Hwk Beta pp78-84