Question

part 3 of 4
some values of functions f, g, h, and k are provided in the table below. find a possible equation of
each function. verify your results with a graphing calculator table.

x	0	1	2	3	4
g(x)	60	12	\\(\frac{12}{5}\\)	\\(\frac{12}{25}\\)	\\(\frac{12}{125}\\)
h(x)	6	60	600	6000	60,000
k(x)	150	50	\\(\frac{50}{3}\\)	\\(\frac{50}{9}\\)	\\(\frac{50}{27}\\)

f(x) = \\(4\left(2^x\
ight)\\)
(use integers or fractions for any numbers in the expression.)

g(x) = \\(60\left(\frac{1}{5}\
ight)^x\\)
(use integers or fractions for any numbers in the expression.)

h(x) = \\(\square\\)
(use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Analyze h(x) values

For \( x = 0 \), \( h(0)=6 \); \( x = 1 \), \( h(1)=60 \); \( x = 2 \), \( h(2)=600 \); \( x = 3 \), \( h(3)=6000 \); \( x = 4 \), \( h(4)=60000 \). Notice the pattern: each time \( x \) increases by 1, \( h(x) \) is multiplied by 10. So it's an exponential function with base 10 and initial value 6 when \( x = 0 \)? Wait, no, when \( x = 0 \), \( h(0)=6 \), \( x = 1 \): \( 6\times10 = 60 \), \( x = 2 \): \( 60\times10 = 600 \), etc. So the general form of an exponential function is \( h(x)=a\times b^x \). When \( x = 0 \), \( h(0)=a\times b^0=a = 6 \). Then \( b \): from \( x = 0 \) to \( x = 1 \), \( 6\times b=60\Rightarrow b = 10 \). So \( h(x)=6\times10^x \). Let's check: \( x = 0 \): \( 6\times10^0 = 6 \), correct. \( x = 1 \): \( 6\times10^1 = 60 \), correct. \( x = 2 \): \( 6\times10^2 = 600 \), correct. \( x = 3 \): \( 6\times10^3 = 6000 \), correct. \( x = 4 \): \( 6\times10^4 = 60000 \), correct.

Step2: Analyze k(x) values

For \( x = 0 \), \( k(0)=150 \); \( x = 1 \), \( k(1)=50 \); \( x = 2 \), \( k(2)=\frac{50}{3} \); \( x = 3 \), \( k(3)=\frac{50}{9} \); \( x = 4 \), \( k(4)=\frac{50}{27} \). The pattern: each time \( x \) increases by 1, \( k(x) \) is multiplied by \( \frac{1}{3} \). So exponential function \( k(x)=a\times b^x \). When \( x = 0 \), \( k(0)=a\times b^0=a = 150 \)? Wait, no, \( x = 0 \): \( k(0)=150 \), \( x = 1 \): \( 150\times b = 50\Rightarrow b=\frac{50}{150}=\frac{1}{3} \). Then check \( x = 2 \): \( 150\times(\frac{1}{3})^2=150\times\frac{1}{9}=\frac{50}{3} \), correct. \( x = 3 \): \( 150\times(\frac{1}{3})^3=150\times\frac{1}{27}=\frac{50}{9} \), correct. \( x = 4 \): \( 150\times(\frac{1}{3})^4=150\times\frac{1}{81}=\frac{50}{27} \), correct. So \( k(x)=150\times(\frac{1}{3})^x \) or \( k(x)=150\times3^{-x} \).

(Assuming the question is to find h(x) or k(x) equation, let's focus on h(x) as per the partial solution shown. Wait, the user's image shows f(x) is given as \( 4(2^x) \), g(x) as \( 60(\frac{1}{5})^x \), and h(x) and k(x) to find. For h(x), as above, the pattern is multiplying by 10 each time, so \( h(x)=6\times10^x \).)

Answer:

For \( h(x) \), a possible equation is \( \boldsymbol{h(x) = 6 \times 10^x} \) (or verifying the pattern, since each term is 10 times the previous, starting from 6 when \( x = 0 \), the exponential function fits \( h(x)=6(10)^x \)).