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Functions & Graphs Christmas Revision

Functions & Graphs Christmas Revision. Put simply, a function is a rule which helps us use one number to produce a different number. Eg . f(x) = 5x + 8 if we input -3 instead of x... f(-3) = 5(-3) + 8 =-15 + 8 f(-3) = -7 When we input -3, we get -7. Ctd.

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Functions & Graphs Christmas Revision

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  1. Functions & GraphsChristmas Revision Put simply, a function is a rule which helps us use one number to produce a different number. Eg. f(x) = 5x + 8 if we input -3 instead of x... f(-3) = 5(-3) + 8 =-15 + 8 f(-3) = -7 When we input -3, we get -7

  2. Ctd... Vica Versa, with the same function f(x) = 3x + 8... If we are told that the number produced is 5 ie. f(x) = 5 we can solve what x was. f(x) = 3x + 8 f(x) = 5 Therefore 3x + 8 = 5 3x = -3 x = -1

  3. Exercisef(x) = 5x + 4 g(x) = -4x + 7 Q1. Find: • f(3) (ii) f(-4) (iii) g(4) (iv) g(-5) Q2. Solve: (i) f(x) = -6 (ii) g(x) = 3 (iii) f(x) = g(x) f(3) = 5(3) + 4 = 15 + 4 f(3) = 19 f(-4) = 5(-4) + 4 = -20 +4 f(-4) = -16 g(4) = -4(4) + 7 = -16 + 7 g(4) = -9 g(-5) = -4(-5) + 7 = 20 + 7 g(-5) = 27 f(x) = 5x + 4 g(x) = -4x + 7 5x + 4 = -4x + 7 5x + 4x = 7 – 4 9x = 3 x = 3/9 = 1/3 f(x) = 5x + 4 5x + 4 = -6 5x = -10 x = -10/5 x = -2 -4x + 7 = 3 -4x = -4 x = -4/-4 x = 1

  4. Problems related to functionsf(x) = 2x + 8 g(x) = -2x + 5 Solve for k if: k[f(5)] = g(k+2) f(5) = -2(5) + 5 = -10 + 5 f(5) = -5 k[f(5)] = -5k g(k+2) = -2(k+2) + 5 = -2k -4 + 5 g(k+2) = -2k +1 -5k = -2k + 1 -3k = 1 k = -⅓ Solve for k when: f(2k + 4) = -4 f(x) = 2x + 8 f(2k +4) = 2(2k + 4)+ 8 = 4k + 16 f(2k + 4) = -4 4k + 16 = -4 4k = -20 k = -5

  5. GraphsIf we graph the function f(x) = 2x + 4 it looks like this. Mathematically Find f(1) f(1) = 2(1) + 4 f(1) = 6 Looking at the Graph f(1) means x =1 When x = 1, y = 6 f(1) = 6 Similarly Solve f(x) = -6 2x + 4 = -6 2x = -10 x = -5 From Graph f(x) = -6 means y = -6 When y = -6, x = -5

  6. Quadratic FunctionsPlot the function f(x) = -x2 + 10x -20 in the domain 1≤ x ≤ 8 Estimate the values for which f(x) = 2 Find f(4.5) and f(1.5) What is the maximum pt of this curve What are the values for x where f(x) ≥ 0 Where does y = 2 hit the graph? x ≈ 3.2 and x ≈ 6.6 Where is x = 4.5? Where is x = 1.5 f(4.5) = 4.5 f(1.5) = -7 Maximum point is the highest pt on the curve Max pt = (5,5) f(x) ≥ 0 above the x axis. Starts when x = 2.6 and finishes where x = 7.2 Domain: 2.6 ≤ x ≤ 7.2

  7. Cubic Functions Cubic functions are treated similarly to the other functions. Some extra characteristics of Cubic f’s: Find the range of value for which f(x) is positive (ii) Find the max point and min point (iii) Find the range of value where f(x) is decreasing. (iv) Find the range of value where f(x) is negative and increasing. Above the x-axis i.e –½ ≤ x ≤ 2.2 Max: (.75, 8.2) Min: (4,-10) Decreasing between max & min i.e. .75 ≤ x ≤ 4 f(x) is negative below the x –axis f(x) is negative and increasing: 4 ≤ x ≤ 5.2

  8. Asymptotes Asymptotes are a special kind of function. They are always in the form f(x) = They have an unusual shaped curve that looks like this. • The curve is mirrored • The curve never touches the x-axis, or where x = 4

  9. Plot the function f(x) = on a graph in the domain

  10. Period RangeConsider the following function • It continues in this fashion forever • The period of this function is 8, as it starts to repeat itself • every 8 steps • The range is [-2 , 2] as the they are the lowest and • highest values for y. Find the following: f(3) f(5) f(9) f(3) = 0 f(5) = 2 f(9) = -2 Hence, find the following f(37) f(25) f(45) f(37) = 2 f(25) = -2 f(45) = 2

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