Download presentation
Presentation is loading. Please wait.
1
Plotting Equations of Proportionality
Slideshow 27, Mathematics Mr Richard Sasaki
2
Objectives Review key graph elements
Naming and showing co-ordinate points Be able to draw and name simple linear graphs.
3
Graph Elements Where is this? Origin (3.5, 3) Title (for example π¦=ππ₯)
π₯-axis π¦-axis Gridlines Values of π¦ Values of π₯
4
Graph Quadrants πΌπΌ πΌ πΌπΌπΌ πΌπ π₯β€0 π¦β₯0 π₯β₯0 π¦β₯0 π₯β€0 π¦β€0 π₯β₯0 π¦β€0
Note: Try to remember this! Letβs look at the ranges of values for π₯ and π¦ in each quadrant of the graph anticlockwise. Each of the quadrants (sections out of four) are namedβ¦. π₯β€0 π¦β₯0 π₯β₯0 π¦β₯0 πΌπΌ πΌ π₯β€0 π¦β€0 π₯β₯0 π¦β€0 πΌπΌπΌ πΌπ
5
Graph Elements When plotting points on graph, you may need to show which is where with labels. For this, we need some new notation. If we want to name this point, P, how would we do that? P (4, 2) π₯ co-ordinate π¦ co-ordinate We can now give different points, different names.
6
Plotting Points Letβs plot some named co-ordinates on the graph below.
π΄ β πΊ β πΉ(2, 3) πΊ(0, 8) π» β π»(β3, 6) πΌ(β6, β7) π΅ β πΉ β Write down the points in red. π· β π΄(β6, 9) π΅(4, 4) πΆ β πΈ β πΆ(5, β2) π·(0, 2) πΈ(β1, β4) πΌ β
7
Answers β Part 1 πΉ β πΊ β π΅ β πΈ β π΄ β π· β π½ β πΆ β πΌ β π» β 2 2 16 4
2 2 16 4 β14β2 β β6 β20 16 β16 2 5 11 5 1 β11
8
Answers β Part 2 (3, 3), (3, β3), (β3, β3), (β3, 3)
(0, 3), (3, 0), (0, β3), (β3, 0) 2.5 (π ππ’πππ π’πππ‘π ) 9π 4 (π ππ’πππ π’πππ‘π ) 27 2 (π ππ’πππ π’πππ‘π )
9
Drawing Graphs When we plot a linear graphβ¦
How many points (minimum) do we actually need? Itβs a straight lineβ¦so only two! If thereβs only two, it should be easy to draw without plotting points.
10
Drawing Graphs When drawing graphs in the form π¦=ππ₯, the line always passes through the origin The graph shown is π¦=3π₯ This means as π₯ increases by 1, π¦ increases by . 3 3 1 This is the case anywhere along the line. Note: π¦=ππ₯ is not common with graphs.
11
Drawing Graphs So if we draw some graph π¦=ππ₯, it passes through the origin, and if we increase π₯ by 1, we increase π¦ by . π β Letβs draw π¦=5π₯. We know the line passes through (0, 0)β¦ β And if we increase π₯ by 1, we increase π¦ by . 5 Now we can draw our line!
12
2π₯ β3π₯ π₯ 4 The relationship is not linear. The rate of change changes. A horizontal line. Almost vertical, very, very steep and positive.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.