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Ib st sl quadratics part ii

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• Ib st sl quadratics part ii

1. 1. Quadratic Functions IB Studies, Syllabus ref 4.3 Lesson 2
2. 2. Vertex or Turning point
3. 3. Vertex or Turning point The vertex of a parabola is where the parabola “turns around”.
4. 4. Vertex or Turning point The vertex of a parabola is where the parabola “turns around”. In Lesson 1 we learnt that this occurs on the axis of symmetry.
5. 5. Vertex or Turning point The vertex of a parabola is where the parabola “turns around”. In Lesson 1 we learnt that this occurs on the axis of symmetry. Hence the x-coordinate of the vertex will be the same as the AoS and the y- coordinate can be found by substitution or in function notation is:
6. 6. Vertex or Turning point The vertex of a parabola is If the value of ‘a’ is positive where the parabola “turns then the parabola is concave around”. up and the vertex would be a minimum turning point. In Lesson 1 we learnt that this occurs on the axis of symmetry. Hence the x-coordinate of the vertex will be the same as the AoS and the y- coordinate can be found by substitution or in function notation is:
7. 7. Vertex or Turning point The vertex of a parabola is If the value of ‘a’ is positive where the parabola “turns then the parabola is concave around”. up and the vertex would be a minimum turning point. In Lesson 1 we learnt that this occurs on the axis of symmetry. If the value of ‘a’ is negative then the parabola is concave down and the vertex would Hence the x-coordinate of be a maximum turning the vertex will be the same point. as the AoS and the y- coordinate can be found by substitution or in function notation is:
8. 8. Vertex or Turning point The vertex of a parabola is If the value of ‘a’ is positive where the parabola “turns then the parabola is concave around”. up and the vertex would be a minimum turning point. In Lesson 1 we learnt that this occurs on the axis of symmetry. −b If the value of ‘a’ is negative x= then the parabola is concave 2a down and the vertex would Hence the x-coordinate of be a maximum turning the vertex will be the same point. as the AoS and the y- coordinate can be found by substitution or in function notation is:
9. 9. Vertex or Turning point The vertex of a parabola is If the value of ‘a’ is positive where the parabola “turns then the parabola is concave around”. up and the vertex would be a minimum turning point. In Lesson 1 we learnt that this occurs on the axis of symmetry. −b If the value of ‘a’ is negative x= then the parabola is concave 2a down and the vertex would Hence the x-coordinate of be a maximum turning the vertex will be the same point. as the AoS and the y- coordinate can be found by substitution or in function notation is: ⎛ −b ⎞ y = f ⎜ ⎟ where f ( x ) = ax 2 + bx + c ⎝ 2a ⎠
10. 10. Vertex or Turning point The vertex of a parabola is If the value of ‘a’ is positive where the parabola “turns then the parabola is concave around”. up and the vertex would be a minimum turning point. In Lesson 1 we learnt that this occurs on the axis of symmetry. −b If the value of ‘a’ is negative x= then the parabola is concave 2a down and the vertex would Hence the x-coordinate of be a maximum turning the vertex will be the same point. as the AoS and the y- coordinate can be found by substitution or in function notation is: ⎛ −b ⎞ y = f ⎜ ⎟ where f ( x ) = ax 2 + bx + c ⎝ 2a ⎠
11. 11. Vertex or Turning point The vertex of a parabola is If the value of ‘a’ is positive where the parabola “turns then the parabola is concave around”. up and the vertex would be a minimum turning point. In Lesson 1 we learnt that this occurs on the axis of symmetry. −b If the value of ‘a’ is negative x= then the parabola is concave 2a down and the vertex would Hence the x-coordinate of be a maximum turning the vertex will be the same point. as the AoS and the y- coordinate can be found by substitution or in function notation is: ⎛ −b ⎞ y = f ⎜ ⎟ where f ( x ) = ax 2 + bx + c ⎝ 2a ⎠
12. 12. Example 1 - Determine the coordinates of thevertex of y = 2x − 8x + 1 2
13. 13. Example 1 - Determine the coordinates of thevertex of y = 2x − 8x + 1 2Firstly, match the equation to the general formand write down the values of a, b, and c.y = ax 2 +bx + cy = 2x 2 − 8x + 1hence a = 2 , b = −8 , c = 1 −bnow axis of symmetry equation from x = 2a
14. 14. Example 1 - Determine the coordinates of thevertex of y = 2x − 8x + 1 2Firstly, match the equation to the general formand write down the values of a, b, and c.y = ax 2 +bx + cy = 2x 2 − 8x + 1hence a = 2 , b = −8 , c = 1 −bnow axis of symmetry equation from x = 2a − ( −8 ) 8so x = = =2 2×2 4and remember this matches the x-coord of the vertexso now all we need is the y-coord
15. 15. Example 1 - Determine the coordinates of thevertex of y = 2x − 8x + 1 2Firstly, match the equation to the general formand write down the values of a, b, and c.y = ax 2 +bx + cy = 2x 2 − 8x + 1hence a = 2 , b = −8 , c = 1 −bnow axis of symmetry equation from x = 2a − ( −8 ) 8so x = = =2 2×2 4and remember this matches the x-coord of the vertexso now all we need is the y-coordby substituting x = 2, in to the parabola 2 y = 2 (2) − 8 (2) + 1 = 8 − 16 + 1 = −7
16. 16. Example 1 - Determine the coordinates of thevertex of y = 2x − 8x + 1 2Firstly, match the equation to the general formand write down the values of a, b, and c.y = ax 2 +bx + c Hence the vertexy = 2x 2 − 8x + 1 has coordinateshence a = 2 , b = −8 , c = 1 −b (2,-7)now axis of symmetry equation from x = 2a − ( −8 ) 8so x = = =2 2×2 4and remember this matches the x-coord of the vertexso now all we need is the y-coordby substituting x = 2, in to the parabola 2 y = 2 (2) − 8 (2) + 1 = 8 − 16 + 1 = −7
17. 17. Example 1 - Determine the coordinates of thevertex of y = 2x − 8x + 1 2Firstly, match the equation to the general formand write down the values of a, b, and c.y = ax 2 +bx + c Hence the vertexy = 2x 2 − 8x + 1 has coordinateshence a = 2 , b = −8 , c = 1 −b (2,-7)now axis of symmetry equation from x = 2a − ( −8 ) 8so x = = =2 2×2 4and remember this matches the x-coord of the vertexso now all we need is the y-coordby substituting x = 2, in to the parabola Now would this be 2 y = 2 (2) − 8 (2) + 1 a maximum or a = 8 − 16 + 1 minimum turning = −7 point???
18. 18. Example 2 - For the quadratic equation y = −x 2 + 2x + 3 . Find:(a) its axes intercepts(b) the equation of the axis of symmetry(c) the coordinates of the vertex(d) sketch the function, labeling the maximum or minimum turning point
19. 19. Example 2 - For the quadratic equation y = −x 2 + 2x + 3 . Find:(a) its axes intercepts(b) the equation of the axis of symmetry(c) the coordinates of the vertex(d) sketch the function, labeling the maximum or minimum turning pointa) Firstly, to find the x-intercepts we must factorise!But watch out for the negative, take it out first.y = −x 2 + 2x + 3= − ⎡ x 2 − 2x − 3⎤ ⎣ ⎦= − ⎡( x − 3) ( x + 1) ⎤ ⎣ ⎦hence the x-intercepts are 3 and -1.The y-intercept we can just read from the eqn as 3.
20. 20. Example 2 - For the quadratic equation y = −x 2 + 2x + 3 . Find:(a) its axes intercepts(b) the equation of the axis of symmetry(c) the coordinates of the vertex(d) sketch the function, labeling the maximum or minimum turning pointa) Firstly, to find the x-intercepts we must factorise!But watch out for the negative, take it out first.y = −x 2 + 2x + 3= − ⎡ x 2 − 2x − 3⎤ ⎣ ⎦= − ⎡( x − 3) ( x + 1) ⎤ ⎣ ⎦hence the x-intercepts are 3 and -1.The y-intercept we can just read from the eqn as 3.b) The axis of symmetry comes from the coefficientsnow a = −1,b = 2, c = 3 −b − ( 2 )x= = =1 2a 2 ( −1)hence the axis of symmetry is x = 1.
21. 21. Example 2 - For the quadratic equation y = −x 2 + 2x + 3 . Find:(a) its axes intercepts(b) the equation of the axis of symmetry(c) the coordinates of the vertex(d) sketch the function, labeling the maximum or minimum turning pointa) Firstly, to find the x-intercepts we must factorise!But watch out for the negative, take it out first.y = −x 2 + 2x + 3= − ⎡ x 2 − 2x − 3⎤ ⎣ ⎦= − ⎡( x − 3) ( x + 1) ⎤ ⎣ ⎦hence the x-intercepts are 3 and -1.The y-intercept we can just read from the eqn as 3.b) The axis of symmetry comes from the coefficientsnow a = −1,b = 2, c = 3 −b − ( 2 )x= = =1 2a 2 ( −1)hence the axis of symmetry is x = 1.c) The vertex comes from substituting x = 1, 2y = − (1) + 2 (1) + 3 = 4hence the coordinates of the vertex is (1,4).
22. 22. Example 2 - For the quadratic equation y = −x 2 + 2x + 3 . Find:(a) its axes intercepts(b) the equation of the axis of symmetry(c) the coordinates of the vertex(d) sketch the function, labeling the maximum or minimum turning pointa) Firstly, to find the x-intercepts we must factorise!But watch out for the negative, take it out first.y = −x 2 + 2x + 3= − ⎡ x 2 − 2x − 3⎤ ⎣ ⎦= − ⎡( x − 3) ( x + 1) ⎤ ⎣ ⎦hence the x-intercepts are 3 and -1.The y-intercept we can just read from the eqn as 3.b) The axis of symmetry comes from the coefficientsnow a = −1,b = 2, c = 3 −b − ( 2 )x= = =1 2a 2 ( −1)hence the axis of symmetry is x = 1.c) The vertex comes from substituting x = 1, 2y = − (1) + 2 (1) + 3 = 4hence the coordinates of the vertex is (1,4).
23. 23. Now complete these in your books: You only have to do the ﬁrst column here! Click here when you’ve completed them and marked them all!
24. 24. Last ones for today...You only haveto do the ﬁrst column here! Click here when you’ve completed them and marked them all!
25. 25. Well done! I’m impressed!