Mathematics
Grade10
Easy

Question

Mia had $ 350 in her bank account at the beginning of the school year. Each week she withdraws $ 50. Two graphs model the situation

Write a function for each graph.

  1. f(x) = -50x + 350 , 0 ≤ x ≤ 7 ; f(x) = 350 – 50 ceiling(x) , 0 ≤ x ≤ 7
  2. f(x) = 50x + 300 , 0 ≤ x ≤ 7 ; f(x) = 300 – 50 ceiling(x) , 0 ≤ x ≤ 7 
  3. f(x) = -50x + 300 , 0 ≤ x ≤ 7 ; f(x) = 300 + 50 floor(x) , 0 ≤ x ≤ 7 
  4. f(x) = 50x + 350 , 0 ≤ x ≤ 7 ; f(x) = 300 – 50 floor(x) , 0 ≤ x ≤ 7

hintHint:

In this question, we have to find the function for each graph. For the first graph we will consider x be the no of the weeks and y be the money in the bank account and now using points (0, 350) and (7, 0) we will find the equation. Now for the second graph, we will find the other function.

The correct answer is: f(x) = -50x + 350 , 0 ≤ x ≤ 7 ; f(x) = 350 – 50 ceiling(x) , 0 ≤ x ≤ 7


    First graph:
    Let x be the no of the weeks and y be the money in the bank account.
    General equation of the line: y = mx +c
    Use points (0, 350) and (7, 0).
    m = fraction numerator left parenthesis 0 minus 350 right parenthesis over denominator left parenthesis 7 minus 0 right parenthesis end fractionfraction numerator negative 350 over denominator 7 end fraction = -50
    Putting (7,0),
    0 = (-50) (7) + c
    c = 350
    First graph: f(x) = -50x + 350, 0 ≤ x ≤ 7
    Second graph:
    It is a ceiling function as can be seen in the graph.
    x = 0, f(x) = 350
    0 < x ≤ 1 , f(x) = 300
    1 < x ≤ 2 , f(x) = 250
    2 < x ≤ 3, f(x) = 200
    3 < x ≤ 4 , f(x) = 150
    4 < x ≤ 5 , f(x) = 100
    5 < x ≤ 6 , f(x) = 50
    6 < x ≤ 7, f(x) = 0
    f(x) = 300 – 50 ceiling(x) + 50 = 350 – 50 ceiling(x)
    Second graph: f(x) = 350 – 50 ceiling(x), 0 ≤ x ≤ 7

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