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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
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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
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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
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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
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Mathematical Process Expectations
Connecting – make connections among mathematical concepts and procedures; and relate mathematical ideas to situations or phenomena drawn from other contexts
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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)
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Trinomial Perfect Square Quadratic Forms of the Equation
y = 16x² + 16x 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 = ² = x 4 = 64
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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)
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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.)
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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
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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)
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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
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Graph of y = 2x² - 8x +6 (Standard) y = 2(x – 1)(x – 3) (Factored)
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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
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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
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Finding the Equation of the Graph in Factored Form
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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 3 3 a = 2
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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)
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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
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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
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Finding the Equation of the Graph in Factored Form
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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
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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)
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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
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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
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Finding the Equation of the Graph in Factored Form
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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 x² 3 x -1 -3 y = 2(x – 1)(x – 3)
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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)
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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
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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) y = 8 – y = -2 So, the y of the vertex is y = -2 and the vertex coordinates are (2, -2)
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Mental Health Break & Then Homework
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Homework – Day 1 Finding 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)
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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)
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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)
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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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