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I can plot a quadratic graph
I can draw up a table from the equation with no problems (Yes/need more work/No) I can fill up a table by substituting values of x in each row and then adding the columns to find the corresponding values of y From the table, I know which two values I need to plot I know how to plan where to draw the axes according to my values of x and y (Yes/need more work/No) I know how to plot the points and join to form the curve
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Notice the βsymmetryβ in the numbers
π= π π +πβπ π -4 -3 -2 -1 1 2 3 π π +π -6 π = π π +πβπ (β4) 2 16 9 (β3) 2 4 (β2) 2 (β1) 2 1 (0) 2 1 (1) 2 4 (2) 2 9 (3) 2 β4 β3 β2 β1 1 2 3 12 6 2 2 6 12 β6 β6 β6 β6 β6 β6 β6 β6 6 β4 β6 β6 β4 6 Notice the βsymmetryβ in the numbers
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Notice the βsymmetryβ in the numbers
π=πβ π π π -3 -2 -1 1 2 3 π β π π π = πβ π π 6 6 6 6 6 6 6 β (β1) 2 β (β0) 2 β92 β(β3) 2 β42 β (β2) 2 β12 02 β12 β (1) 2 β42 β (2) 2 β92 β (3) 2 β3 2 5 6 5 2 β3 Notice the βsymmetryβ in the numbers
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Notice the βsymmetryβ in the numbers
π= ππ π +ππβπ π -3 -2 -1 1 2 π π π +ππ -9 π = 2 π π +ππβπ 18 2(9) 2(β3) 2 8 2(4) 2(β2) 2 2(1) 2 2(β1) 2 2(0) 2(0) 2 2 2(1) 2(1) 2 8 2(4) 2 (2) 2 2(0) β6 2(β3) β4 2(β2) β2 2(β1) 2 2(1) 2(2) 4 4 4 12 12 β9 β9 β9 β9 β9 β9 3 β5 β9 β9 β5 3 Notice the βsymmetryβ in the numbers
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Notice the βsymmetryβ in the numbers
π= βπ π βππ+ππ π -5 -4 -3 -2 -1 1 2 3 β π π βππ +10 π = β π π βππ+ππ β25 β(β5) 2 β16 β(β4) 2 β9 β(β3) 2 β4 β(β2) 2 β1 β(β1) 2 β(0) 2 β1 β(1) 2 β4 β(2) 2 β(3) 2 β9 10 β2(β5) 8 β2(β4) 6 β2(β3) 4 β2(β2) 2 β2(β1) β2(0) β2 β2(1) β4 β2(2) β6 β2(3) +10 +10 +10 +10 +10 +10 +10 +10 +10 β5 2 7 10 11 10 7 2 β5 Notice the βsymmetryβ in the numbers
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To be able to plot a system of graphs
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Plotting a linear graph & a quadratic graph
When a straight line graph and a quadratic graph are plotted on the same axes, we call it a system of linear and quadratic graphs They can meet at: 2 points 1 point No point The points where they meet are called points of intersection
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Points of Intersection
Points of intersection are those points where the graphs are equal to each other In this example, the graphs drawn are: π= π π βππ+π π=ππ+π At the points of intersection (1,3) and (6,13) the graphs are equal π π βππ+π=ππ+π π₯=1 ππ π₯=6 Let us see an example on Geogebra You can solve a complicated looking equation by drawing two graphs and reading their POIs
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CW/HW A) Plot these two graphs: Linear Graph ο π=π
Quadratic Graph ο π= π π +ππ (π values : -5 to 2) Write down the points of intersection B) Plot these two graphs: Linear Graph ο π=πβπ Quadratic Graph ο π= π π βππβπ (π values : -2 to 6) C) Plot these two graphs: Linear Graph ο π=βπ+π Quadratic Graph ο π= π π +πβπ (π values: -4 to 3) CW/HW
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