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Mrs. Rivas ISCHS Standard MAFS.912.A-REI.4.11

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Presentation on theme: "Mrs. Rivas ISCHS Standard MAFS.912.A-REI.4.11"β€” Presentation transcript:

1 Mrs. Rivas ISCHS Standard MAFS.912.A-REI.4.11 Explain why the solution of π’š = 𝒇(𝒙) and π’š = π’ˆ(𝒙) is the solutions of the equation 𝒇(𝒙) = π’ˆ(𝒙). Explain the x coordinates of the points where the graphs of the equations intersect are the solutions of the equations. Find the solutions for linear, polynomial, rational, absolute value, exponential and logarithmic

2 MAFS.912.A-REI.4.11 What You Need To Know... Mrs. Rivas Standard
ISCHS Standard MAFS.912.A-REI.4.11 What You Need To Know... 1. The benchmark will be assessed using Equation Response, Multiple Choice Response, Multi-select Response, Table Response, Simulation Response or Open Response. 2. Items may require students to find solutions by using a graph, a table of values, or by successive approximations to a given place value. 3. Items are restricted to exponential with a rational exponent, polynomial of degree greater than 2, rational, absolute value, and logarithmic. 4. Items may require the student to know the role of the x-coordinate and the y-coordinate in the intersection of 𝑓(π‘₯) = 𝑔(π‘₯).

3 MAFS.912.A-REI.4.11 Mrs. Rivas Standard Example 1
ISCHS Standard MAFS.912.A-REI.4.11 Functions f and g are defined below Example 1 𝒇 𝒙 = 𝟏 πŸπ’™ π’ˆ 𝒙 = 𝒙 𝟐 The graph of π’š= 𝒇(𝒙) and π’š= π’ˆ(𝒙) intersect at point 𝑷. Determine the x-coordinate of P. Write your answer with rational exponents. πŸπ’™ 𝟏 βˆ™ 𝟏 πŸπ’™ = 𝒙 𝟐 βˆ™ πŸπ’™ 𝟏 𝟏 𝟐 𝟏 πŸ‘ =𝒙 𝟏= πŸπ’™ πŸ‘ 𝟐 βˆ’πŸ 𝟏 πŸ‘ =𝒙 𝟏 𝟐 = 𝒙 πŸ‘ 𝟐 βˆ’ 𝟏 πŸ‘ =𝒙 𝟏 𝟐 𝟏 πŸ‘ = 𝒙 πŸ‘ 𝟏 πŸ‘

4 MAFS.912.A-REI.4.11 Mrs. Rivas Standard Example 2
ISCHS Standard MAFS.912.A-REI.4.11 Let 𝑓(π‘₯) = π‘Ž π‘₯ 2 where π‘Ž > 0, and let 𝑔(π‘₯) = π‘šπ‘₯ + 𝑏 where π‘š > 0 and 𝑏 < 0. The equation 𝑓(π‘₯) = 𝑔(π‘₯) has n distinct real solution(s). What are all the possible values of 𝑛? Justify your answers. Example 2 Numeric Method: The equation 𝑓(π‘₯)=𝑔(π‘₯) is a quadratic equation. A quadratic equation of this form can has zero real solutions. For example, if 𝑓(π‘₯) = π‘₯ 2 and 𝑔(π‘₯) = π‘₯βˆ’4, the equation has no real solutions because the discriminant of the equation π‘₯ 2 βˆ’π‘₯+4=0 is less than 0. A quadratic equation of this form can has one real solutions. For example, if 𝑓(π‘₯) = π‘₯ 2 and 𝑔(π‘₯) =2π‘₯βˆ’1, the equation has one real solutions because the discriminant of the equation π‘₯ 2 βˆ’2π‘₯+1=0 is equal to 0. A quadratic equation of this form can has two real solutions. For example, if 𝑓(π‘₯) = π‘₯ 2 and 𝑔(π‘₯) =4π‘₯βˆ’1, the equation has one real solutions because the discriminant of the equation π‘₯ 2 βˆ’4π‘₯+1=0 is greater than 0. No other possibilities exist for the number of solutions of a quadratic equation because the number of real solutions of a polynomial of degree 𝑛 is at most 𝑛.

5 MAFS.912.A-REI.4.11 Mrs. Rivas Standard Example 2
ISCHS Standard MAFS.912.A-REI.4.11 Let 𝑓(π‘₯) = π‘Ž π‘₯ 2 where π‘Ž > 0, and let 𝑔(π‘₯) = π‘šπ‘₯ + 𝑏 where π‘š > 0 and 𝑏 < 0. The equation 𝑓(π‘₯) = 𝑔(π‘₯) has n distinct real solution(s). What are all the possible values of 𝑛? Justify your answers. Example 2 Graphing method: There are three graphical examples of where f(x) and g(x) intercept. An example of a parabola and a line that do not intersect (no solution). An example of a parabola and a tangent line (1 no solution). An example of a parabola and a line that intersect at two points (1 no solution). There no other ways to intersect a parabola and a line.

6 MAFS.912.A-REI.4.11 Mrs. Rivas Standard Example 2
ISCHS Standard MAFS.912.A-REI.4.11 Let 𝑓(π‘₯) = π‘Ž π‘₯ 2 where π‘Ž > 0, and let 𝑔(π‘₯) = π‘šπ‘₯ + 𝑏 where π‘š > 0 and 𝑏 < 0. The equation 𝑓(π‘₯) = 𝑔(π‘₯) has n distinct real solution(s). What are all the possible values of 𝑛? Justify your answers. Example 2 Algebraic method: There are three graphical examples of where f(x) and g(x) intercept. Since the polynomial has degree 2 (highest exponent of x0, it either has 0, 1, 2 solutions. We can find the number of solutions by examining the discriminant, which in this case is the quantity (βˆ’π‘š) 2 βˆ’4(π‘Ž)(βˆ’π‘). If the discriminant is negative, there are no real solutions. If the discriminant is equal to zero, there is one real solutions. If the discriminant is positive, there are two distinct real solutions. No other possibilities exist for the number of solutions of a quadratic equation because the number of real solutions of a polynomial of degree 𝑛 is at most 𝑛.

7 Find the answer choices that include the -2 and 1
Mrs. Rivas ISCHS Standard Find the answer choices that include the -2 and 1 MAFS.912.A-REI.4.11 1. Given the functions β„Ž(π‘₯)=|π‘₯βˆ’4|+1 and π‘˜(π‘₯) = π‘₯ 2 +3, which intervals contain a value of π‘₯ for which β„Ž(π‘₯) = π‘˜(π‘₯)? Select ALL that apply. 𝒉(𝒙) = π’Œ(𝒙) |π‘₯βˆ’4|+1= π‘₯ 2 +3 βˆ’4.5<π‘₯<3 βˆ’4.5<π‘₯<βˆ’3 βˆ’1.5<π‘₯<1.5 1.5<π‘₯<3 3<π‘₯<4.5 π‘₯βˆ’4 = π‘₯ 2 +2 π’™βˆ’πŸ’= βˆ’(𝒙 𝟐 +𝟐) π’™βˆ’πŸ’= 𝒙 𝟐 +𝟐 𝒙= 𝒙 𝟐 +πŸ” π’™βˆ’πŸ’= βˆ’π’™ 𝟐 βˆ’πŸ 𝟎= 𝒙 𝟐 βˆ’π’™+πŸ” 𝒙= βˆ’π’™ 𝟐 +𝟐 𝒙= 𝟏± βˆ’πŸπŸ‘ 𝟐 𝟎= βˆ’π’™ 𝟐 βˆ’π’™+𝟐 𝟎= 𝒙 𝟐 +π’™βˆ’πŸ 𝟎=(𝒙+𝟐)(π’™βˆ’πŸ) 𝒙= πŸΒ±π’Š πŸπŸ‘ 𝟐 𝒙=βˆ’πŸ 𝒙=𝟏

8 MAFS.912.A-REI.4.11 Mrs. Rivas Standard 𝒇(𝒙) = π’ˆ(𝒙)
ISCHS Standard MAFS.912.A-REI.4.11 2. Let 𝑓 π‘₯ = 14 π‘₯ π‘₯ 2 βˆ’ 46π‘₯ and 𝑔(π‘₯) = 2π‘₯ + 7. Which is the solution set to the equation 𝑓(π‘₯) = 𝑔(π‘₯)? 𝒇(𝒙) = π’ˆ(𝒙) A. {βˆ’3,0,1} B. {βˆ’3,βˆ’1,2} C. {βˆ’2,1,3} D. {1,5,11} 𝟏 𝟏𝟐 (πŸπŸ’ 𝒙 πŸ‘ + πŸπŸ– 𝒙 𝟐 βˆ’ πŸ’πŸ”π’™) = πŸπ’™ + πŸ• 𝟏 𝟏𝟐 βˆ™πŸ(πŸ• 𝒙 πŸ‘ +πŸπŸ’ 𝒙 𝟐 βˆ’πŸπŸ‘π’™) = πŸπ’™ + πŸ• πŸ” 𝟏 βˆ™ 𝟏 πŸ” (πŸ• 𝒙 πŸ‘ +πŸπŸ’ 𝒙 𝟐 βˆ’πŸπŸ‘π’™) = πŸπ’™ + πŸ• πŸ” πŸ• 𝒙 πŸ‘ +πŸπŸ’ 𝒙 𝟐 βˆ’πŸπŸ‘π’™=πŸπŸπ’™+πŸ’πŸ πŸ• 𝒙 πŸ‘ +πŸπŸ’ 𝒙 𝟐 βˆ’πŸ‘πŸ“π’™βˆ’πŸ’πŸ=𝟎 πŸ•( 𝒙 πŸ‘ +𝟐 𝒙 𝟐 βˆ’πŸ“π’™βˆ’πŸ”)=𝟎

9 MAFS.912.A-REI.4.11 Mrs. Rivas Standard πŸ•( 𝒙 πŸ‘ +𝟐 𝒙 𝟐 βˆ’πŸ“π’™βˆ’πŸ”)=𝟎
ISCHS Standard MAFS.912.A-REI.4.11 2. Let 𝑓 π‘₯ = 14 π‘₯ π‘₯ 2 βˆ’ 46π‘₯ and 𝑔(π‘₯) = 2π‘₯ + 7. Which is the solution set to the equation 𝑓(π‘₯) = 𝑔(π‘₯)? A. {βˆ’3,0,1} B. {βˆ’3,βˆ’1,2} C. {βˆ’2,1,3} D. {1,5,11} πŸ•( 𝒙 πŸ‘ +𝟐 𝒙 𝟐 βˆ’πŸ“π’™βˆ’πŸ”)=𝟎 𝒙 πŸ‘ +𝟐 𝒙 𝟐 βˆ’πŸ“π’™βˆ’πŸ”=𝟎 𝟏 𝟐 βˆ’πŸ“ βˆ’πŸ” 𝟐 + 𝟐 + πŸ– + πŸ” 𝟏 πŸ’ πŸ‘ 𝟎 Possible factor ±𝟏, ±𝟐, Β±πŸ‘, Β±πŸ” (π’™βˆ’πŸ)(𝒙 𝟐 + πŸ’π’™+πŸ‘)=𝟎 Test Possible factor π’™βˆ’πŸ 𝒙+πŸ‘ 𝒙+𝟏 =𝟎 (𝟏) πŸ‘ +𝟐 (𝟏) 𝟐 βˆ’πŸ“(𝟏)βˆ’πŸ”β‰ πŸŽ π’™βˆ’πŸ=𝟎 𝒙+πŸ‘=𝟎 𝒙+𝟏=𝟎 (𝟐) πŸ‘ +𝟐 (𝟐) 𝟐 βˆ’πŸ“ 𝟐 βˆ’πŸ”=𝟎 𝒙=𝟐 𝒙=βˆ’πŸ‘ 𝒙=βˆ’πŸ 𝒙=𝟐 πŸπ’™ πŸ‘ + πŸπ’™ 𝟐 βˆ’πŸ“π’™ βˆ’πŸ”

10 MAFS.912.A-REI.4.11 Mrs. Rivas Standard
ISCHS Standard MAFS.912.A-REI.4.11 3. What is the point of intersection for 𝑓(π‘₯)= 2 π‘₯ and 𝑔(π‘₯)= π‘₯ ? A. (0,1) B. (1,0) C. (1, ) D. (2,4) 𝒇 𝒙 =π’ˆ(𝒙) 𝒇(𝒙)= 𝟐 𝒙 𝟐 𝒙 = 𝟏 𝟐 𝒙 π’š= 𝟐 𝟎 π’š=𝟏 𝟐 𝒙 = 𝟏 𝒙 𝟐 𝒙 π’ˆ(𝒙)= π‘₯ 𝟐 𝒙 = 𝟐 βˆ’π’™ 𝒙=βˆ’π’™ π’š= 𝟎 πŸπ’™=𝟎 𝒙=𝟎 π’š=𝟏

11 MAFS.912.A-REI.4.11 Mrs. Rivas Standard
ISCHS Standard MAFS.912.A-REI.4.11 4. When 𝑔(π‘₯)= 2 π‘₯+2 and β„Ž(π‘₯)=log⁑(π‘₯+1)+3 are graphed on the same set of axes, which coordinates best approximate their point of intersection? A. (βˆ’0.9,1.8) B. (βˆ’0.9,1.9) C. (1.4,3.3) D. (1.4,3.4)

12 MAFS.912.A-REI.4.11 Mrs. Rivas Standard
ISCHS Standard MAFS.912.A-REI.4.11 5. Pedro and Bobby each own an ant farm. Pedro starts with 100 ants and says his farm is growing exponentially at a rate of 15% per month. Bobby starts with 350 ants and says his farm is steadily decreasing by 5 ants per month. Assuming both boys are accurate in describing the population of their ant farms, after how many months will they both have approximately the same number of ants? A.πŸ• B. πŸ– C. 13 D. 36


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