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 RelationsDay 2 - Interpreting Quadratic Graphs and Day 3 - Constructing Quadratic Equations

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

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

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

Mathematical Process Expectations

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

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)

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

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)

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.)

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

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)

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

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

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

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


•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


•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)

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

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


• 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


• 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)


• 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

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

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)

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)

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

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)

Mental Health Break & Then Homework

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)

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)

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)

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)

grade 10 academic math chapter 3 – analyzing and applying quadratic models

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