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Answer the questions below about the straight line
Recap⦠Answer the questions below about the straight line How would you describe what the line looks like? What is the gradient? What is its y-intercept? Write down any coordinate on the line. Write down any coordinate not on the line. y = 3x - 2
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How did we find coordinates of points on the line?
Recapβ¦ How did we find coordinates of points on the line? y = 3x - 2 π₯ -2 -1 1 2 3 π¦
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Plotting the graph (-2,-8) (-1,-5) ( 0,-2) ( 1, 1) ( 2, 4) ( 3, 7) π₯
1 2 3 π¦ -8 -5 4 7 (-2,-8) (-1,-5) ( 0,-2) ( 1, 1) ( 2, 4) ( 3, 7)
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What is different about this equation?
π¦ = π₯ 2 β2π₯β8 π₯ -3 -2 -1 1 2 3 4 5 π¦ 7 -5 -8 -9 -8 -5 7 Some people start with zero and then the positive integers when filling in the table. Why might this be the case? What do you notice about the coordinates?
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π¦ = π₯ 2 β2π₯β8 Example 1 π₯ -3 -2 -1 1 2 3 4 5 π¦ y 1 2 3 4 -1 -2 5 6 7 8
1 2 3 4 -1 -2 5 6 7 8 -3 -4 -5 -6 -7 -8 y -9 Example 1 π¦ = π₯ 2 β2π₯β8 π₯ -3 -2 -1 1 2 3 4 5 π¦
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Make sure the curve does not have a flat bottom.
1 2 -1 -2 -3 -4 -5 3 4 5 6 7 8 -6 -7 -8 x y Example 2 π¦ = π₯ 2 +5π₯+2 π₯ -3 -2 -1 1 2 3 4 5 π¦ Make sure the curve does not have a flat bottom.
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On your whiteboardsβ¦ Write down a quadratic equation. and anotherβ¦
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Title: Plotting quadratic curves
In your books Title: Plotting quadratic curves π¦ = π₯2 + 2π₯ + 1 For each table: Complete the table of values Draw a set of axes that will fit the coordinates. Choose your scale carefully. Plot the coordinates and join them with a smooth curve. π₯ -3 -2 -1 1 2 3 π¦ π¦ = π₯2 + π₯ π₯ -3 -2 -1 1 2 3 π¦ π¦ = 2π₯2 + π₯ β 3 π₯ -3 -2 -1 1 2 3 π¦
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Check using your calculator
To draw a graph of π¦ = π₯ 2+ 2π₯ +1 1) Complete a table of values: Goes up in 1βs π₯ π¦ β3 β2 β1 1 2 3 4 1 1 4 9 16 This slide summarizes the steps required to plot a graph using a table of values. Type in 3 Type in -3 Now use you calculator to check your answers.
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Answers π¦ = π₯2 + 2π₯ + 1 π₯ -3 -2 -1 1 2 3 π¦ 4 9 16 π¦ = π₯2 + π₯ π₯ -3 -2
1 2 3 π¦ 4 9 16 π¦ = π₯2 + π₯ π₯ -3 -2 -1 1 2 3 π¦ 6 12 π¦ = 2π₯2 + π₯ β 3 π₯ -3 -2 -1 1 2 3 π¦ 12 7 18 Who got different answers to these?
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There are six quadratic graphs on the sheet.
Using what we have done today, match each parabola to one of the equations below. You will be asked to explain how you know, so write something at the side of each graph. Students should look at the x-coordinates on the graphs and substitute them into each equation to see which ones match π¦ = (π₯ β 2)2 π¦ = π₯2 + π₯ + 1 π¦ = π₯2 β 3π₯ + 1 π¦ = 2 β π₯ β π₯2 π¦ = 3 β π₯2 π¦ = π₯2 + 2π₯ β 1
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π¦ = π₯ 2 β2π₯β8 Example 1 π₯ -3 -2 -1 1 2 3 4 5 π¦ y 1 2 3 4 -1 -2 5 6 7 8
1 2 3 4 -1 -2 5 6 7 8 -3 -4 -5 -6 -7 -8 y -9 Example 1 π¦ = π₯ 2 β2π₯β8 π₯ -3 -2 -1 1 2 3 4 5 π¦
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π¦ = π₯ 2 +5π₯+2 Example 2 π₯ -3 -2 -1 1 2 3 4 5 π¦ y x 1 2 -1 -2 -3 -4 -5
1 2 -1 -2 -3 -4 -5 3 4 5 6 7 8 -6 -7 -8 x y Example 2 π¦ = π₯ 2 +5π₯+2 π₯ -3 -2 -1 1 2 3 4 5 π¦
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