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42 algebra, functions, and data analysis in exercises 5 – 8, match the …

Question

42 algebra, functions, and data analysis
in exercises 5 – 8, match the graph with its equation.

  1. $y = 5^x$ 6. $y = 5^{-x}$ 7. $y = 5^{x - 1}$ 8. $y = 5^x - 2$

(a) graph a
(b) graph b
(c) graph c
(d) graph d

  1. applying the skills: you are considering attending college in london. after a little research, you have discovered that the inflation rate in the united kingdom is about 3%, and the first year of college would cost about 10,000 pounds.

(a) determine the growth factor for a 3% inflation rate.
(b) if the yearly inflation rate remains at 3%, what exponential function would you use to determine the cost of a year of college currently at 10,000 pounds after $t$ years?
(c) how much would college tuition be during your fourth year?

Explanation:

Response
For Exercises 5-8 (Matching graphs to equations)
Brief Explanations
  1. For \(y=5^x\) (Exercise 5): This is the basic exponential growth function, which passes through \((0,1)\) and increases rapidly. This matches graph b.
  2. For \(y=5^{-x}\) (Exercise 6): This is an exponential decay function (\(y=(1/5)^x\)), which passes through \((0,1)\) and decreases as \(x\) increases. This matches graph d.
  3. For \(y=5^{x-1}\) (Exercise 7): This is \(y=5^x\) shifted right 1 unit, so it passes through \((1,1)\) and increases. This matches graph c.
  4. For \(y=5^x - 2\) (Exercise 8): This is \(y=5^x\) shifted down 2 units, so it passes through \((0,-1)\) and has a horizontal asymptote at \(y=-2\). This matches graph a.

Step1: Calculate growth factor

The growth factor for a percentage increase is \(1 + \text{inflation rate}\).
\[1 + 0.03 = 1.03\]

Step2: Define exponential function

The general form of exponential growth is \(C(t) = C_0 \times (\text{growth factor})^t\), where \(C_0=10000\).
\[C(t) = 10000 \times (1.03)^t\]

Step3: Compute 4th year cost

Substitute \(t=4\) into the function.
\[C(4) = 10000 \times (1.03)^4\]
First calculate \((1.03)^4 = 1.03 \times 1.03 \times 1.03 \times 1.03 = 1.12550881\)
Then compute \(10000 \times 1.12550881 = 11255.09\)

Answer:

  1. b
  2. d
  3. c
  4. a

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For Question 9