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ponents & radicals revisiting exponents & their functions back to intro…

Question

ponents & radicals revisiting exponents & their functions back to intro page
the curve begins in quadrant 2 then continues through the plotted points left parenthesis 0 comma 8 right parenthesis, left parenthesis 1 comma 4 right parenthesis, and left parenthesis 2 comma 2 right parenthesis. the curve continues past the last plotted point to run rightward on the x - axis beginning at left parenthesis 8 comma 0 right parenthesis.
use the information given in the graph to find the exponential function that represents the graph, then find f(5). round the answer to two decimal places.
(2 points)
f(x) = squarecdotsquare^x
f(5) = square

Explanation:

Step1: Identify the exponential function form

The general form of an exponential function is \( f(x) = a \cdot b^x \). We have points \((0, 8)\), \((1, 4)\), and \((2, 2)\). When \( x = 0 \), \( f(0) = a \cdot b^0 = a \cdot 1 = a \). From the point \((0, 8)\), we get \( a = 8 \).

Step2: Find the base \( b \)

Using the point \((1, 4)\) and \( a = 8 \) in \( f(x) = 8 \cdot b^x \), we substitute \( x = 1 \) and \( f(1) = 4 \): \( 4 = 8 \cdot b^1 \). Solving for \( b \), we divide both sides by 8: \( b = \frac{4}{8} = 0.5 \). So the function is \( f(x) = 8 \cdot (0.5)^x \).

Step3: Calculate \( f(5) \)

Substitute \( x = 5 \) into the function: \( f(5) = 8 \cdot (0.5)^5 \). First, calculate \( (0.5)^5 = 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 = \frac{1}{32} = 0.03125 \). Then multiply by 8: \( 8 \times 0.03125 = 0.25 \). Wait, let's check again. Wait, \( (0.5)^5 = 1/32 = 0.03125 \), 8 0.03125 = 0.25? Wait, no, wait: 8(0.5)^5 = 8(1/32) = 8/32 = 1/4 = 0.25? Wait, but let's check with the points. Wait, at x=2, f(2)=8(0.5)^2=8(0.25)=2, which matches the point (2,2). At x=3, f(3)=8(0.5)^3=8(0.125)=1, which matches the graph (at x=3, y=1). At x=4, f(4)=8(0.5)^4=8(0.0625)=0.5. At x=5, f(5)=8(0.5)^5=8*(0.03125)=0.25. So that's correct.

Answer:

The exponential function is \( f(x) = 8 \cdot (0.5)^x \) and \( f(5) = 0.25 \).