Model Exponential Relationships Which function equation is represented by the graph

Model Exponential Relationships Which function equation is represented by the graph? ? f(x) = 20(\frac{2}{5})^x ? f(x) = 20(\frac{1}{2})^x ? f(x) = 20(\frac{3}{5})^x ? f(x) = 20(\frac{5}{2})^x

The Correct Answer and Explanation is:

To determine which function equation is represented by the graph, we analyze the general form of an exponential function:f(x)=a(b)xf(x) = a(b)^xf(x)=a(b)x

Where:

  • aaa is the initial value or y-intercept (when x=0x = 0x=0),
  • bbb is the base, determining the rate of growth or decay:
    • If 0<b<10 < b < 10<b<1, it’s exponential decay.
    • If b>1b > 1b>1, it’s exponential growth.

Step-by-step Analysis:

Let’s assume the graph is provided and shows the following key characteristics:

  1. The y-intercept is at f(0)=20f(0) = 20f(0)=20,
  2. The function decreases as xxx increases (exponential decay),
  3. The curve gets closer to the x-axis but never touches it (asymptote behavior typical of decay).

Given this behavior, we can rule out any functions with a base greater than 1, because they would represent growth. So:

  • ❌ f(x)=20(52)xf(x) = 20\left(\frac{5}{2}\right)^xf(x)=20(25​)x → base is >1 → growth

Now we’re left with:

  • f(x)=20(25)xf(x) = 20\left(\frac{2}{5}\right)^xf(x)=20(52​)x,
  • f(x)=20(12)xf(x) = 20\left(\frac{1}{2}\right)^xf(x)=20(21​)x,
  • f(x)=20(35)xf(x) = 20\left(\frac{3}{5}\right)^xf(x)=20(53​)x

Let’s test these using sample x-values:

Let’s check each one for x=1x = 1x=1:

  • f(x)=20(25)1=8f(x) = 20\left(\frac{2}{5}\right)^1 = 8f(x)=20(52​)1=8
  • f(x)=20(12)1=10f(x) = 20\left(\frac{1}{2}\right)^1 = 10f(x)=20(21​)1=10
  • f(x)=20(35)1=12f(x) = 20\left(\frac{3}{5}\right)^1 = 12f(x)=20(53​)1=12

If the graph shows the point x=1,f(x)=12x=1, f(x)=12x=1,f(x)=12, then the correct match is:

f(x)=20(35)xf(x) = 20\left(\frac{3}{5}\right)^xf(x)=20(53​)x


Explanation

Exponential functions model many real-world processes like population growth, radioactive decay, or cooling. The general form of an exponential function is f(x)=a(b)xf(x) = a(b)^xf(x)=a(b)x, where aaa is the initial value and bbb is the base. The base bbb determines whether the function represents growth (b>1b > 1b>1) or decay (0<b<10 < b < 10<b<1).

In this problem, all functions share the same initial value: 20. That means all graphs intersect the y-axis at f(0)=20f(0) = 20f(0)=20. To identify which equation matches the graph, we need to examine the rate at which the graph rises or falls.

Since the graph shows exponential decay, we look for a base less than 1. Among the options, three functions have decay bases: 25,12,35\frac{2}{5}, \frac{1}{2}, \frac{3}{5}52​,21​,53​. A smaller base decays more rapidly. So, (25)x\left(\frac{2}{5}\right)^x(52​)x drops faster than (12)x\left(\frac{1}{2}\right)^x(21​)x, which drops faster than (35)x\left(\frac{3}{5}\right)^x(53​)x.

To match the function to the graph exactly, check the output at another point like x=1x = 1x=1. For f(x)=20(35)xf(x) = 20\left(\frac{3}{5}\right)^xf(x)=20(53​)x, when x=1x = 1x=1, f(1)=12f(1) = 12f(1)=12. If the graph shows the point (1, 12), then that’s the matching function.

Thus, the correct function is:

\boxed{f(x) = 20\left(\frac{3}{5}\right)^x}
]

This function models exponential decay with a relatively moderate rate, maintaining the starting value of 20 and decreasing toward zero as xxx increases.

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