grade 10 academic math chapter 3 – analyzing and applying quadratic models day 1 – introduction...
TRANSCRIPT
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Grade 10 Academic Math Chapter 3 – Analyzing and Applying
Quadratic Models
Day 1 – Introduction to Quadratic RelationsDay 2 - Interpreting Quadratic Graphs and Day 3 - Constructing Quadratic Equations
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Agenda – Day 1
1. Warm-up – interpreting A = w(8-w)/A= -w²+8w2. Given graph, find vertex, optimal value, equation of the axis of symmetry, zeros of the relationship & sign of 2nd differences3. Constructing equation of a graph, given zeros and another point
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Learning GoalBy 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 + 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
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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•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
x² 3
x -1
x -3
•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)
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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 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) -16 + 6• y = 8 – 16 + 6• 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 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)
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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)