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Grade 10 Academic Math Chapter 3 – Analyzing and Applying Quadratic Models

Grade 10 Academic Math Chapter 3 – Analyzing and Applying Quadratic Models. Day 1 – Introduction to Quadratic Relations Day 2 - Interpreting Quadratic Graphs and Day 3 - Constructing Quadratic Equations. Agenda – Day 1. Warm-up – interpreting A = w(8-w)/A= -w²+8w

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Grade 10 Academic Math Chapter 3 – Analyzing and Applying Quadratic Models

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  1. Grade 10 Academic Math Chapter 3 – Analyzing and Applying Quadratic Models Day 1 – Introduction to Quadratic Relations Day 2 - Interpreting Quadratic Graphs and Day 3 - Constructing Quadratic Equations

  2. Agenda – Day 1 • Warm-up – interpreting A = w(8-w)/A= -w²+8w 2. Given graph, find vertex, optimal value, equation of the axis of symmetry, zeros of the relationship & sign of 2nd differences 3. Constructing equation of a graph, given zeros and another point

  3. Learning Goal By the end of the lesson… … identify key information from a quadratic graph and interpret, and… … be able to construct a quadratic equation given the graph, or the roots and another point on the graph

  4. Curriculum Expectations • Determine the zeros and the max or min value of a quadratic relation from it graph • Determine, through investigation, and describe the connection between the factors of a quadratic expression and the x-intercepts of the graph of the corresponding quadratic relation expressed in the form y = a(x - S)(s – T) • Ontario Catholic School Graduate Expectations: The graduate is expected to be… a self-directed life long learner who CGE4f applies effective… problem solving… skills

  5. Mathematical Process Expectations • Connecting – make connections among mathematical concepts and procedures; and relate mathematical ideas to situations or phenomena drawn from other contexts

  6. Trinomial Quadratic Forms of the Equation y = 2x² - 8x + 6   y = 2(x – 3)(x – 1) y = ax² + bx - c   y = a(x – S)(x – t) Expanded form Factored form (also called Standard form) (or expand and simplify)

  7. Trinomial Perfect Square Quadratic Forms of the Equation y = 16x² + 16x + 4   y = (4x + 2)(4x + 2) y = ax² + bx + c y = (√16x+ √4) (√16x+ √4) y = (√16x+ √4)² • If the square of ½ of b gives you the product of a x c, you have a perfect square • Ex. 16 ÷ 2 = 8... 8² = 64... 16 x 4 = 64

  8. Trinomial Perfect Square Quadratic Forms of the Equation (Text p.304) y = 16x² + 16x + 4   y = (4x + 2)(4x + 2) y = (4x + 2)² Expanded form Factored form y = a²x² + 2abx +b² y = (ax + b)(ax + b) y = 4²x² + 2(4)(2)x + 2² (gives you the above trinomial)

  9. GCF Binomial Quadratic Forms of the Equation y = 3x² + 27x   y = 3x(x + 9) y = 3x² + (3)(9)x y = ax² + abx  y = ax(x + b)* Expanded form Factored form (also called Standard form) (or expand and simplify) * Where ab (27 in this ex.) is a single number divisible by a (3 in this ex.)

  10. Binomial Difference of Squares Quadratic Form of the Equation y = x² - 9   y = (x + 3)(x – 3) y = a²x² - b²   y = (a*x + b)(a*x – b) Expanded form Factored form (also called Standard form) (or expand and simplify) * a is 1 in this example

  11. Binomial Difference of Squares Quadratic Form of the Equation More complex example where you have to factor out the 3 first y = 3x² - 27 y = 3(x² - 9)   y = 3(x + 3)(x – 3)

  12. Terminology • Vertex: (x, y) of bottom or top of graph • Optimal Value: y value of vertex • Maximum or Minimum: max if graph opens down and min if graph opens up (here min because opens up) • Zeros (or roots or x-intercepts): where the graph crosses the x-axis (0, 1 or 2 places depending on graph – here in 2 places) • Axis of Symmetry: x = # (x of vertex is #) • y-intercept: where graph crosses the y-axis

  13. Graph of y = 2x² - 8x +6 (Standard)y = 2(x – 1)(x – 3) (Factored)

  14. Graph of y = 2x² - 8x +6 (Standard)y = 2(x – 1)(x – 3) (Factored) • Vertex: (x, y) at bottom of graph is (2, -2) • Optimal Value: y value of vertex is -2, eq’n is y = -2 • Maximum or Minimum: minimum of -2 because the graph opens up • Zeros (or roots or x-intercepts): where the graph crosses the x-axis are 1 and 3 (2 zeros) • Axis of Symmetry: x = 2 (x of vertex) • y-intercept: where graph crosses the y-axis is 6

  15. Finding the Equation of the Graph in Factored Form • Start with empty template factored form of the equation • y=a(x – S)(x – t) • Start by substituting zeros and • (x, y) of one other point (other point can be vertex, y-intercept, or any other point other than the zeros) into above

  16. Finding the Equation of the Graph in Factored Form

  17. Finding the Equation of the Graph in Factored Form • y=a(x – S)(x – t) • Let’s use y-intercept of (0, 6) • 6 = a(0 – 1)(0 – 3) • 6 = a(-1)(-3) • 6=3a • 6=3a • --- ---- • 3 3 • a = 2

  18. Finding the Equation of the Graph in Factored Form • Now put “a” (2) value and zeros (1 and 3) into • y=a(x – S)(x – t) leaving x and y as variables • y = 2(x – 1)(x – 3) (You have the factored form of the equation)

  19. Finding the Equation of the Graph in Standard Form • y = 2(x – 1)(x – 3)... expand using FOIL and distributive law • y = 2[x² - 3x – x + 3] • y = 2[x² - 4x + 3] • y = 2x² - 8x + 6 • Note that the 6 is the y-intercept

  20. Finding the Equation of a Quadratic Given Zeros and Another Point • Given zeros of 1 and 3 and point (4, 6) on the graph find the equation of the graph in factored form and standard form

  21. Finding the Equation of the Graph in Factored Form

  22. Finding the Equation of a Quadratic Given Zeros and Another Point • Given zeros of 1 and 3 and point (4, 6) on the graph find the equation of the graph in factored form and standard form • y = a(x – S)(x – t) • Substitute zeros 1 and 3 in for S and t and 4 and 6 in for x and y respectively and solve for a • 6 = a(4 – 1)(4 – 3)... 6 = a(3)(1)... a = 2

  23. Finding the Equation of a Quadratic Given Zeros and Another Point • We have determined that a = 2 • So, now we put 2 in for a & put the 1 and 3 back for S and t • y = 2(x – 1)(x – 3) (factored form)

  24. Finding the Equation of a Quadratic Given Zeros and Another Point • To find the standard or expanded form... • y = 2(x – 1)(x – 3)... expand using FOIL and distributive law • y = 2[x² - 3x – x + 3] • y = 2[x² - 4x + 3] • y = 2x² - 8x + 6 • Note that the 6 is the y-intercept

  25. Finding Zeros, AOS, Vertex and Y-Intercept Given Equation • Given equation y = 2x² - 8x +6, factor and then find the zeros, axis of symmetry (AOS), vertex and y-intercept

  26. Finding the Equation of the Graph in Factored Form

  27. Factor y = 2x² - 8x +6 First Finding the GCF and Then Using Butterfly Method • Take out the GCF of 2 • y = 2(x² - 4x + 3) • When we apply the butterfly method, we see that this factors to y = 2(x – 1)(x – 3)

  28. Finding Zeros of y = 2x² - 8x +6 (now factored to y = 2(x – 1)(x – 3) • So, when y = 2(x – 1)(x – 3), to find the zeros, set y = 0 (because that is the value of y on the x-axis) • 0 = 2(x – 1)(x – 3) • So, x – 1 = 0 and x – 3 = 0 • x = 1 and x = 3 • These are our zeros (or roots or x-intercepts)

  29. Finding AOS of y = 2x² - 8x +6 (now factored to y = 2(x – 1)(x – 3) • To find the AOS, we need the x of the vertex • To find the x of the vertex, we take the average of the zeros 1 and 3 • xv = xzero 1 + xzero 2 --------------------------- 2 • Xv = (1 + 3) --------................ Xv = 2, so the AOS is x = 2 2

  30. Finding y of vertex y = 2x² - 8x +6 (now factored to y = 2(x – 1)(x – 3) • Plug x = 2 of vertex into either factored or standard form of equation and solve for y... • y = 2(2)² - 8(2) + 6 • y = 2(4) -16 + 6 • y = 8 – 16 + 6 • y = -2 • So, the y of the vertex is y = -2 and the vertex coordinates are (2, -2)

  31. Mental Health Break & Then Homework

  32. Homework – Day 1Finding the Equation from the Graph • Page 280, #1abcd • (a) zeros • (b) vertex • (c) Axis of Symmetry • (d) Optimal Value/Max Min • (e) Opens up or down • (f) Value of “a” in y = a (x – S)(x – t) • (g) How “a” affects the steps 1, 3 & 5 • (h) Equation in Factored Form • (i) The equation in Standard Form (use foil)

  33. Homework - Finding Equations Given the Zeros’s and Another Point – Day 1 (Cont’d) • Page 328, #7 • Page 282, #9 (given the y-intercept) • Page 281, #4 (given the y of the vertex) (Hint: Find the x of the vertex by taking the average of the zeros)

  34. Homework - Finding Equations Given the Zeros’s and Another Point – Day 2 • Page 329, #9 • (For all of these above, find the eq’n in both standard an factored form) • Page 281, #5 (here you are given the equations in factored form)

  35. Homework - Finding the Zeros, AOS, Vertex from the Equation (Day 2 – Cont’d) • Page 308, #7acdefghijklmn (change each expression into an equation by putting y = to the left of the expression) • (a) zeros • (b) vertex • (c) Axis of Symmetry • (d) Optimal Value/Max Min • (e) Opens up or down • (f) Value of “a” in y = a (x – S)(x – t) • (g) How “a” affects the steps 1, 3 & 5 • (h) Equation in Factored Form • (i) The equation in Standard Form (use foil)

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