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Blue Book 16 Quadratic Graphs
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Quadratic Graphs Contents Page Completing the Square.
Determining the coordinates and nature of the turning point. Lines of symmetry (Part 1) Determining the y-intercept Sketching a parabola (Part 1) Determining where a graph cuts the x-axis Line of symmetry (Part 2) Sketching a parabola (Part 2) Blue Book 16 Quadratic Graphs Contents Page
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Completing the Square.
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Completing the Square. Express π₯ 2 +8π₯+3 in the form π₯+π 2 +π
First - Multiply out π₯+π 2 +π to give π₯ 2 +2ππ₯+ π 2 +π Second - Compare terms and coefficients π₯ 2 +2ππ₯+ π 2 +π 2π=8 π=4 π₯ 2 +8π₯+3 π₯ 2 +2ππ₯+ π 2 +π π 2 +π=3 π₯ 2 +8π₯+3 4 2 +π=3 π=β13 Answer π₯ 2 +8π₯+3= π₯+4 2 β13
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Completing the Square. Example Express π₯ 2 β4π₯β1 in the form π₯βπ 2 +π
π₯βπ 2 +π= π₯ 2 β2ππ₯+ π 2 +π π₯ 2 β2ππ₯+ π 2 +π β2π=β4 π=2 π₯ 2 β4π₯β1 π₯ 2 β2ππ₯+ π 2 +π π 2 +π=β1 π₯ 2 β4π₯β1 2 2 +π=β1 π=β5 Answer π₯ 2 +8π₯+3= π₯β2 2 β5
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2. Determining the coordinates and nature of the turning point.
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2. Determining the coordinates and nature of the turning point.
Given π¦= π₯+π 2 +π then the coordinates of the turning point of the parabola will be βπ, π Example 1 π¦= π₯+4 2 β13 will have a turning point at β4, β13
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Determining the coordinates and nature of the turning point.
Example 2 π₯β will have a turning point at 2, 5
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Determining the coordinates and nature of the turning point.
The nature of the turning point (whether it is a maximum or minimum) can be determined by the coefficient of π₯ 2 . If the coefficient of π₯ 2 is positive the parabola will have a minimum turning point If the coefficient of π₯ 2 is negative the parabola will have a maximum turning point π¦= π₯ 2 β2π₯β3 π¦= βπ₯ 2 β2π₯β3
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Determining the coordinates and nature of the turning point.
Example 3 Determine the coordinates and nature of the turning point of the following parabolas (a) π¦= (π₯+2) 2 +7 Answer a minimum turning point at (β2, 7) The coefficient of π₯ 2 would be positive if we multiplied out the brackets so the turning point is a minimum.
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Determining the coordinates and nature of the turning point.
(b) π¦= 4β(π₯β3) 2 Answer a maximum turning point at (3, 4) (3, 4) The coefficient of π₯ 2 would be negative if we multiplied out the brackets so the turning point is a maximum.
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Determining the coordinates and nature of the turning point.
Answer a minimum turning point at (β2, β1) (d) π¦= β2β(π₯+1) 2 Answer a maximum turning point at (β1, β2) (e) π¦= (π₯β5) 2 +1 Answer a minimum turning point at (5, 1) (f) π¦= β(π₯+2) 2 +8 Answer a maximum turning point at (β2, 8) (g) π¦= (π₯β7) 2 β3 Answer a minimum turning point at (7, β3) (h) π¦= 5β(π₯β5) 2 Answer a maximum turning point at (5, 5)
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3. Lines of symmetry (Part 1)
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3. Lines of symmetry (Part 1)
Given π¦= π₯+π 2 +π then the parabolaβs line of symmetry will have the equation π₯=βπ. The line of symmetry must go through the turning point. π₯=β2 Example 1 π¦= π₯+2 2 β3 will have a a line of symmetry with equation π₯=β2.
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Lines of symmetry (Part 1)
Example 2 For each parabola below, state the equation of the line of symmetry (a) π¦= β2β(π₯+1) 2 Answer π₯=β1 (b) π¦= (π₯β5) 2 +1 Answer π₯=5 (c) π¦= β(π₯+2) 2 +8 Answer π₯=β2 (d) π¦= (π₯β7) 2 β3 Answer π₯=7 (e) π¦= 5β(π₯β5) 2 Answer π₯=5
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4. Determining the y-intercept
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4. Determining the y-intercept
The π¦-intercept is the point where a graph cuts the π¦-axis. Every point on the π¦-axis has a π₯-coordinate = 0. We can find the π¦-intercept by substituting π₯=0 into its equation. Example 1 Find the coordinates of the π¦-intercept of the parabola π¦= (π₯+2) 2 β1 . π¦= (0+2) 2 β1 π¦=4β1 π¦=3 The coordinates of the π¦-intercept are (0, 3)
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Determining the y-intercept
When π₯=0, it follows that ππ₯=0, π₯ 2 =0, and ππ₯ 2 =0 Given π¦= ππ₯ 2 +ππ₯+π the π¦-intercept will have coordinates 0, π . Example 2 π¦=π₯ 2 +2π₯β3 will have its π¦-intercept at the point 0, β3 .
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Determining the y-intercept
Example 3 For each parabola below, state the coordinates of the π¦-intercept. (a) π¦= (π₯+1) 2 +2 Answer (0, 3) (b) π¦= π₯ 2 β2π₯+1 Answer (0, 1) (c) π¦= (π₯+2) 2 β8 Answer (0, β4) (d) π¦= π₯ 2 +5π₯β3 Answer (0, β3) (e) π¦= 2β(π₯β1) 2 Answer (0, 1)
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5. Sketching a parabola (Part 1)
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Cuts the axes Turning point Axes Annotate
5. Sketching a parabola (Part 1) When sketching a parabola you should clearly annotate your neatly drawn graph with all the important points of the graph following the steps shown below. Cuts the axes Turning point Axes Annotate Find where the graph cuts the π-axes. It will cut the π-axis when π=π. Make a note of this point. Find the turning point of the parabola and decide whether the turning point is a minimum or a maximum. State the equation of the axis of symmetry. Start by drawing the π-axis. When drawing the axis consider the maximum and minimum values. Draw the axis of symmetry using a dotted line. Annotate the coordinates of the turning point. Draw the parabola making sure it is symmetrical and the nature is correct. Annotate where the parabola cuts the π-axis.
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Cuts the axes Turning point Axes Annotate
Sketching a parabola (Part 1) Example 1 Sketch the graph of π¦= π₯β Cuts the axes Turning point Axes Annotate
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Sketching a parabola (Part 1)
Find where the graph cuts the π-axis. Cuts the π¦-axes when π₯=0 π¦= π₯β π¦= 0β π¦=25+2 π¦=27 Cuts the π¦-axes at (0, 27)
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Sketching a parabola (Part 1)
Find the turning point, its nature and the equation of the axis of symmetry π¦= π₯β Minimum turning point at (5, 2) Equation of line of symmetry π₯=5
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5. Sketching a parabola (Part 1)
π₯=5 π¦ Annotate and draw the graph π₯
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Sketching a parabola (Part 1)
π₯=5 π¦ Annotate and draw the graph (5, 2) π₯
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Sketching a parabola (Part 1)
π₯=5 π¦ Annotate and draw the graph (5, 2) π₯
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Sketching a parabola (Part 1)
π₯=5 π¦ Annotate and draw the graph (0, 27) (5, 2) π₯
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Sketching a parabola (Part 1)
Example 1 Sketch the graph of π¦= (π₯β2) 2 +3 π¦ π₯=2 (0, 7) (2, 3) π₯
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Sketching a parabola (Part 1)
Example 2 Sketch the graph of π¦= β(π₯+3) 2 β4 π¦ π₯=β3 π₯ (β3, β4) (0, β13)
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6. Determining where a graph cuts the x-axis
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6. Determining where a parabola cuts the π-axis
Factorised Given π¦=(π₯βπ)(π₯βπ) then the coordinates of where the graph cuts the π₯-axis will be π, 0 and (π, 0) Example 1 Determine where π¦=(π₯β2)(π₯+4) cuts the π₯-axis Answer β4, 0 and (2, 0)
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Determining where a parabola cuts the π-axis
Not Factorised Example 2 Determine where π¦= π₯ 2 β5π₯+6 cuts the π₯-axis Factorise RHS π¦=(π₯β2)(π₯β3) Answer 2, 0 and (3, 0)
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Determining where a parabola cuts the π-axis
Example 3 For each parabola below, state the coordinates of the points where it cuts the π₯-axis (a) π¦=(π₯β3)(π₯+4) Answer (β4, 0) and (3, 0) (b) π¦=(π₯+1)(π₯β7) Answer(β1, 0) and (7, 0) (c) π¦=(π₯β1)(π₯+2) Answer(β2, 0) and (1, 0) (d) π¦= π₯ 2 β3π₯β10 Answer(β2, 0) and (5, 0) (e) π¦= π₯ 2 βπ₯β12 Answer(β3, 0) and (4, 0)
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7. Line of symmetry (Part 2)
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7. Line of symmetry (Part 2)
π₯=1 Where a parabola cuts the π₯-axis, the line of symmetry is always halfway between the points where the parabola cuts the π₯-axis. Example 1 The graph shows part of the parabola π¦=(π₯+2)(π₯β4). State the equation of the line of symmetry. β2 4 ππ§π¬π°ππ« π₯= β2+4 2 =1 So the equation of the line of symmetry is π₯=1
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Line of symmetry (Part 2)
Example 2 The graph shows part of the parabola π¦=(π₯β1)(π₯+5). State the equation of the line of symmetry and the coordinates of the turning point. π₯=β2 β5 1 ππ§π¬π°ππ« π₯= β5+1 2 =β2 So the equation of the line of symmetry is π₯=β2 π¦= β2β1 β2+5 =β9 so the turning point is (β2, β9) (β2, β9)
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Line of symmetry (Part 2)
Example 3 For each parabola below, state the equation of the line of symmetry and the coordinates of the turning points (a) π¦=(π₯β1)(π₯+3) Answer π₯=β1, Turning Point (β1, β4) (b) π¦=(π₯+1)(π₯β7) Answer π₯=3, Turning Point (3, β16) (c) π¦=(π₯β2)(π₯+2) Answer π₯=0, Turning Point (0, β4) (d) π¦= π₯ 2 β3π₯β10 Answer π₯= 3 2 , Turning Point ( 3 2 , β 49 4 ) (e) π¦= π₯ 2 βπ₯β12 Answer π₯= 1 2 , Turning Point ( 1 2 , β 49 4 )
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8. Sketching a parabola (Part 2)
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Cuts the axes Turning point Axes Annotate
5. Sketching a parabola (Part 1) When sketching a parabola you should clearly annotate your neatly drawn graph with all the important points of the graph following the steps shown below. Cuts the axes Turning point Axes Annotate It will cut the π-axis when π=π. Make a note of this point. It will cut the π-axis when y=π. Make a note of these point where they exist. Find the turning point of the parabola and decide whether the turning point is a minimum or a maximum. State the equation of the axis of symmetry. Start by drawing the π-axis. When drawing the axis consider the maximum and minimum values. Draw the axis of symmetry using a dotted line. Annotate the coordinates of the turning point. Draw the parabola making sure it is symmetrical and the nature is correct. Annotate where the parabola cuts the π-axis.
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Cuts the axes Turning point Axes Annotate
Sketching a parabola (Part 1) Example 1 Sketch the graph of π¦= π₯ 2 +2π₯β8 Cuts the axes Turning point Axes Annotate
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Sketching a parabola (Part 1)
Find where the graph cuts the π-axis. π¦= π₯ 2 +2π₯β8 π¦=(π₯β2)(π₯+4) Cuts the π₯-axes at β4, 0 and 2, 0 . Find where the graph cuts the π-axis. π¦= (0)β8 Cuts the π¦-axes at 0, β8 .
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Sketching a parabola (Part 1)
Find the turning point, its nature and the equation of the axis of symmetry Cuts the π₯-axes at β4, 0 and 2, 0 . π₯= β4+2 2 =β1 So the equation of the line of symmetry is π₯=β1 π¦= β1β2 β1+4 =β9 So there will be a minimum turning point at (β1, β9)
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5. Sketching a parabola (Part 1)
π₯=β1 π¦ Annotate and draw the graph π₯
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5. Sketching a parabola (Part 1)
π₯=β1 π¦ Annotate and draw the graph π₯ (β1, β9)
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5. Sketching a parabola (Part 1)
π₯=β1 π¦ Annotate and draw the graph π₯ (β1, β9)
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5. Sketching a parabola (Part 1)
π₯=β1 π¦ Annotate and draw the graph π₯ (β4, 0) (2, 0) (β1, β9)
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5. Sketching a parabola (Part 1)
π₯=β1 π¦ Annotate and draw the graph π¦= π₯ 2 +2π₯β8 π₯ (β4, 0) (2, 0) (0, β8) (β1, β9)
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Quadratic Graphs THE END
Blue Book 16 Quadratic Graphs THE END
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