Blue Book 16 Quadratic Graphs

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

Completing the Square.

Completing the Square. Express 𝑥 2 +8𝑥+3 in the form 𝑥+𝑎 2 +𝑏 First - Multiply out 𝑥+𝑎 2 +𝑏 to give 𝑥 2 +2𝑎𝑥+ 𝑎 2 +𝑏 Second - Compare terms and coefficients 𝑥 2 +2𝑎𝑥+ 𝑎 2 +𝑏 2𝑎=8 𝑎=4 𝑥 2 +8𝑥+3 𝑥 2 +2𝑎𝑥+ 𝑎 2 +𝑏 𝑎 2 +𝑏=3 𝑥 2 +8𝑥+3 4 2 +𝑏=3 𝑏=−13 Answer 𝑥 2 +8𝑥+3= 𝑥+4 2 −13

Completing the Square. Example Express 𝑥 2 −4𝑥−1 in the form 𝑥−𝑝 2 +𝑞 𝑥−𝑝 2 +𝑞= 𝑥 2 −2𝑝𝑥+ 𝑝 2 +𝑞 𝑥 2 −2𝑝𝑥+ 𝑝 2 +𝑞 −2𝑝=−4 𝑝=2 𝑥 2 −4𝑥−1 𝑥 2 −2𝑝𝑥+ 𝑝 2 +𝑞 𝑝 2 +𝑞=−1 𝑥 2 −4𝑥−1 2 2 +𝑞=−1 𝑞=−5 Answer 𝑥 2 +8𝑥+3= 𝑥−2 2 −5

2. Determining the coordinates and nature of the turning point.

2. Determining the coordinates and nature of the turning point. Given 𝑦= 𝑥+𝑎 2 +𝑏 then the coordinates of the turning point of the parabola will be −𝑎, 𝑏 Example 1 𝑦= 𝑥+4 2 −13 will have a turning point at −4, −13

Determining the coordinates and nature of the turning point. Example 2 𝑥−2 2 +5 will have a turning point at 2, 5

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 𝑥 2 . If the coefficient of 𝑥 2 is positive the parabola will have a minimum turning point If the coefficient of 𝑥 2 is negative the parabola will have a maximum turning point 𝑦= 𝑥 2 −2𝑥−3 𝑦= −𝑥 2 −2𝑥−3

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) 𝑦= (𝑥+2) 2 +7 Answer a minimum turning point at (−2, 7) The coefficient of 𝑥 2 would be positive if we multiplied out the brackets so the turning point is a minimum.

Determining the coordinates and nature of the turning point. (b) 𝑦= 4−(𝑥−3) 2 Answer a maximum turning point at (3, 4) (3, 4) The coefficient of 𝑥 2 would be negative if we multiplied out the brackets so the turning point is a maximum.

Determining the coordinates and nature of the turning point. Answer a minimum turning point at (−2, −1) (d) 𝑦= −2−(𝑥+1) 2 Answer a maximum turning point at (−1, −2) (e) 𝑦= (𝑥−5) 2 +1 Answer a minimum turning point at (5, 1) (f) 𝑦= −(𝑥+2) 2 +8 Answer a maximum turning point at (−2, 8) (g) 𝑦= (𝑥−7) 2 −3 Answer a minimum turning point at (7, −3) (h) 𝑦= 5−(𝑥−5) 2 Answer a maximum turning point at (5, 5)

3. Lines of symmetry (Part 1)

3. Lines of symmetry (Part 1) Given 𝑦= 𝑥+𝑎 2 +𝑏 then the parabola’s line of symmetry will have the equation 𝑥=−𝑎. The line of symmetry must go through the turning point. 𝑥=−2 Example 1 𝑦= 𝑥+2 2 −3 will have a a line of symmetry with equation 𝑥=−2.

Lines of symmetry (Part 1) Example 2 For each parabola below, state the equation of the line of symmetry (a) 𝑦= −2−(𝑥+1) 2 Answer 𝑥=−1 (b) 𝑦= (𝑥−5) 2 +1 Answer 𝑥=5 (c) 𝑦= −(𝑥+2) 2 +8 Answer 𝑥=−2 (d) 𝑦= (𝑥−7) 2 −3 Answer 𝑥=7 (e) 𝑦= 5−(𝑥−5) 2 Answer 𝑥=5

4. Determining the y-intercept

4. Determining the y-intercept The 𝑦-intercept is the point where a graph cuts the 𝑦-axis. Every point on the 𝑦-axis has a 𝑥-coordinate = 0. We can find the 𝑦-intercept by substituting 𝑥=0 into its equation. Example 1 Find the coordinates of the 𝑦-intercept of the parabola 𝑦= (𝑥+2) 2 −1 . 𝑦= (0+2) 2 −1 𝑦=4−1 𝑦=3 The coordinates of the 𝑦-intercept are (0, 3)

Determining the y-intercept When 𝑥=0, it follows that 𝑏𝑥=0, 𝑥 2 =0, and 𝑎𝑥 2 =0 Given 𝑦= 𝑎𝑥 2 +𝑏𝑥+𝑐 the 𝑦-intercept will have coordinates 0, 𝑐 . Example 2 𝑦=𝑥 2 +2𝑥−3 will have its 𝑦-intercept at the point 0, −3 .

Determining the y-intercept Example 3 For each parabola below, state the coordinates of the 𝑦-intercept. (a) 𝑦= (𝑥+1) 2 +2 Answer (0, 3) (b) 𝑦= 𝑥 2 −2𝑥+1 Answer (0, 1) (c) 𝑦= (𝑥+2) 2 −8 Answer (0, −4) (d) 𝑦= 𝑥 2 +5𝑥−3 Answer (0, −3) (e) 𝑦= 2−(𝑥−1) 2 Answer (0, 1)

5. Sketching a parabola (Part 1)

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 𝒚-axes. It will cut the 𝒚-axis when 𝒙=𝟎. 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 𝒙-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 𝒚-axis.

Cuts the axes Turning point Axes Annotate Sketching a parabola (Part 1) Example 1 Sketch the graph of 𝑦= 𝑥−5 2 +2 Cuts the axes Turning point Axes Annotate

Sketching a parabola (Part 1) Find where the graph cuts the 𝒚-axis. Cuts the 𝑦-axes when 𝑥=0 𝑦= 𝑥−5 2 +2 𝑦= 0−5 2 +2 𝑦=25+2 𝑦=27 Cuts the 𝑦-axes at (0, 27)

Sketching a parabola (Part 1) Find the turning point, its nature and the equation of the axis of symmetry 𝑦= 𝑥−5 2 +2 Minimum turning point at (5, 2) Equation of line of symmetry 𝑥=5

5. Sketching a parabola (Part 1) 𝑥=5 𝑦 Annotate and draw the graph 𝑥

Sketching a parabola (Part 1) 𝑥=5 𝑦 Annotate and draw the graph (5, 2) 𝑥

Sketching a parabola (Part 1) 𝑥=5 𝑦 Annotate and draw the graph (5, 2) 𝑥

Sketching a parabola (Part 1) 𝑥=5 𝑦 Annotate and draw the graph (0, 27) (5, 2) 𝑥

Sketching a parabola (Part 1) Example 1 Sketch the graph of 𝑦= (𝑥−2) 2 +3 𝑦 𝑥=2 (0, 7) (2, 3) 𝑥

Sketching a parabola (Part 1) Example 2 Sketch the graph of 𝑦= −(𝑥+3) 2 −4 𝑦 𝑥=−3 𝑥 (−3, −4) (0, −13)

6. Determining where a graph cuts the x-axis

6. Determining where a parabola cuts the 𝒙-axis Factorised Given 𝑦=(𝑥−𝑎)(𝑥−𝑏) then the coordinates of where the graph cuts the 𝑥-axis will be 𝑎, 0 and (𝑏, 0) Example 1 Determine where 𝑦=(𝑥−2)(𝑥+4) cuts the 𝑥-axis Answer −4, 0 and (2, 0)

Determining where a parabola cuts the 𝒙-axis Not Factorised Example 2 Determine where 𝑦= 𝑥 2 −5𝑥+6 cuts the 𝑥-axis Factorise RHS 𝑦=(𝑥−2)(𝑥−3) Answer 2, 0 and (3, 0)

Determining where a parabola cuts the 𝒙-axis Example 3 For each parabola below, state the coordinates of the points where it cuts the 𝑥-axis (a) 𝑦=(𝑥−3)(𝑥+4) Answer (−4, 0) and (3, 0) (b) 𝑦=(𝑥+1)(𝑥−7) Answer(−1, 0) and (7, 0) (c) 𝑦=(𝑥−1)(𝑥+2) Answer(−2, 0) and (1, 0) (d) 𝑦= 𝑥 2 −3𝑥−10 Answer(−2, 0) and (5, 0) (e) 𝑦= 𝑥 2 −𝑥−12 Answer(−3, 0) and (4, 0)

7. Line of symmetry (Part 2)

7. Line of symmetry (Part 2) 𝑥=1 Where a parabola cuts the 𝑥-axis, the line of symmetry is always halfway between the points where the parabola cuts the 𝑥-axis. Example 1 The graph shows part of the parabola 𝑦=(𝑥+2)(𝑥−4). State the equation of the line of symmetry. −2 4 𝐀𝐧𝐬𝐰𝐞𝐫 𝑥= −2+4 2 =1 So the equation of the line of symmetry is 𝑥=1

Line of symmetry (Part 2) Example 2 The graph shows part of the parabola 𝑦=(𝑥−1)(𝑥+5). State the equation of the line of symmetry and the coordinates of the turning point. 𝑥=−2 −5 1 𝐀𝐧𝐬𝐰𝐞𝐫 𝑥= −5+1 2 =−2 So the equation of the line of symmetry is 𝑥=−2 𝑦= −2−1 −2+5 =−9 so the turning point is (−2, −9) (−2, −9)

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) 𝑦=(𝑥−1)(𝑥+3) Answer 𝑥=−1, Turning Point (−1, −4) (b) 𝑦=(𝑥+1)(𝑥−7) Answer 𝑥=3, Turning Point (3, −16) (c) 𝑦=(𝑥−2)(𝑥+2) Answer 𝑥=0, Turning Point (0, −4) (d) 𝑦= 𝑥 2 −3𝑥−10 Answer 𝑥= 3 2 , Turning Point ( 3 2 , − 49 4 ) (e) 𝑦= 𝑥 2 −𝑥−12 Answer 𝑥= 1 2 , Turning Point ( 1 2 , − 49 4 )

8. Sketching a parabola (Part 2)

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 𝒚-axis when 𝒙=𝟎. Make a note of this point. It will cut the 𝒙-axis when y=𝟎. 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 𝒙-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 𝒚-axis.

Cuts the axes Turning point Axes Annotate Sketching a parabola (Part 1) Example 1 Sketch the graph of 𝑦= 𝑥 2 +2𝑥−8 Cuts the axes Turning point Axes Annotate

Sketching a parabola (Part 1) Find where the graph cuts the 𝒙-axis. 𝑦= 𝑥 2 +2𝑥−8 𝑦=(𝑥−2)(𝑥+4) Cuts the 𝑥-axes at −4, 0 and 2, 0 . Find where the graph cuts the 𝒚-axis. 𝑦= 0 2 +2(0)−8 Cuts the 𝑦-axes at 0, −8 .

Sketching a parabola (Part 1) Find the turning point, its nature and the equation of the axis of symmetry Cuts the 𝑥-axes at −4, 0 and 2, 0 . 𝑥= −4+2 2 =−1 So the equation of the line of symmetry is 𝑥=−1 𝑦= −1−2 −1+4 =−9 So there will be a minimum turning point at (−1, −9)

5. Sketching a parabola (Part 1) 𝑥=−1 𝑦 Annotate and draw the graph 𝑥

5. Sketching a parabola (Part 1) 𝑥=−1 𝑦 Annotate and draw the graph 𝑥 (−1, −9)

5. Sketching a parabola (Part 1) 𝑥=−1 𝑦 Annotate and draw the graph 𝑥 (−1, −9)

5. Sketching a parabola (Part 1) 𝑥=−1 𝑦 Annotate and draw the graph 𝑥 (−4, 0) (2, 0) (−1, −9)

5. Sketching a parabola (Part 1) 𝑥=−1 𝑦 Annotate and draw the graph 𝑦= 𝑥 2 +2𝑥−8 𝑥 (−4, 0) (2, 0) (0, −8) (−1, −9)

Quadratic Graphs THE END Blue Book 16 Quadratic Graphs THE END