Answer :
To determine which functions could be exponential, we will analyze the given set of values for [tex]\( f(x) \)[/tex], [tex]\( g(x) \)[/tex], [tex]\( h(x) \)[/tex], and [tex]\( k(x) \)[/tex] to see if they follow the common pattern of exponential functions.
An exponential function has the form [tex]\( y = a \cdot b^x \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are constants, and for any two consecutive points, the ratio of [tex]\( y \)[/tex]-values should be constant.
Let’s analyze each function step by step:
### Function [tex]\( f(x) \)[/tex]:
[tex]\[ \begin{aligned} &x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ &f(x) & 1 & 4 & 9 & 16 & 25 & 36 & 49 \\ \end{aligned} \][/tex]
Let’s calculate the ratios of consecutive [tex]\( y \)[/tex]-values:
[tex]\[ \frac{4}{1} = 4, \quad \frac{9}{4} = 2.25, \quad \frac{16}{9} \approx 1.78, \quad \frac{25}{16} \approx 1.56, \quad \frac{36}{25} = 1.44, \quad \frac{49}{36} \approx 1.36 \][/tex]
The ratios are not constant. Thus, [tex]\( f(x) \)[/tex] is not an exponential function.
### Function [tex]\( g(x) \)[/tex]:
[tex]\[ \begin{aligned} &x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ &g(x) & 0.125 & 0.25 & 0.5 & 1 & 2 & 4 & 8 \\ \end{aligned} \][/tex]
Let’s calculate the ratios of consecutive [tex]\( y \)[/tex]-values:
[tex]\[ \frac{0.25}{0.125} = 2, \quad \frac{0.5}{0.25} = 2, \quad \frac{1}{0.5} = 2, \quad \frac{2}{1} = 2, \quad \frac{4}{2} = 2, \quad \frac{8}{4} = 2 \][/tex]
The ratios are constant. Thus, [tex]\( g(x) \)[/tex] could be an exponential function.
### Function [tex]\( h(x) \)[/tex]:
[tex]\[ \begin{aligned} &x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ &h(x) & 0.25 & 0.5 & 0.75 & 0 & 1.25 & 1.5 & 1.75 \\ \end{aligned} \][/tex]
Let’s calculate the ratios of consecutive [tex]\( y \)[/tex]-values:
[tex]\[ \frac{0.5}{0.25} = 2, \quad \frac{0.75}{0.5} = 1.5, \quad \frac{0}{0.75} = 0, \quad \frac{1.25}{0} \ (\text{undefined}), \quad \frac{1.5}{1.25} = 1.2, \quad \frac{1.75}{1.5} \approx 1.17 \][/tex]
The ratios are not constant. Thus, [tex]\( h(x) \)[/tex] is not an exponential function.
### Function [tex]\( k(x) \)[/tex]:
[tex]\[ \begin{aligned} &x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ &k(x) & 64 & 16 & 4 & 1 & 0.25 & 0.0625 & 0.0156 \\ \end{aligned} \][/tex]
Let’s calculate the ratios of consecutive [tex]\( y \)[/tex]-values:
[tex]\[ \frac{16}{64} = 0.25, \quad \frac{4}{16} = 0.25, \quad \frac{1}{4} = 0.25, \quad \frac{0.25}{1} = 0.25, \quad \frac{0.0625}{0.25} = 0.25, \quad \frac{0.0156}{0.0625} = 0.25 \][/tex]
The ratios are constant. Thus, [tex]\( k(x) \)[/tex] could be an exponential function.
### Conclusion:
The functions [tex]\( g(x) \)[/tex] and [tex]\( k(x) \)[/tex] give values that could come from exponential functions.
So, the correct answers are:
- A. [tex]\( g(x) \)[/tex]
- D. [tex]\( k(x) \)[/tex]
An exponential function has the form [tex]\( y = a \cdot b^x \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are constants, and for any two consecutive points, the ratio of [tex]\( y \)[/tex]-values should be constant.
Let’s analyze each function step by step:
### Function [tex]\( f(x) \)[/tex]:
[tex]\[ \begin{aligned} &x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ &f(x) & 1 & 4 & 9 & 16 & 25 & 36 & 49 \\ \end{aligned} \][/tex]
Let’s calculate the ratios of consecutive [tex]\( y \)[/tex]-values:
[tex]\[ \frac{4}{1} = 4, \quad \frac{9}{4} = 2.25, \quad \frac{16}{9} \approx 1.78, \quad \frac{25}{16} \approx 1.56, \quad \frac{36}{25} = 1.44, \quad \frac{49}{36} \approx 1.36 \][/tex]
The ratios are not constant. Thus, [tex]\( f(x) \)[/tex] is not an exponential function.
### Function [tex]\( g(x) \)[/tex]:
[tex]\[ \begin{aligned} &x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ &g(x) & 0.125 & 0.25 & 0.5 & 1 & 2 & 4 & 8 \\ \end{aligned} \][/tex]
Let’s calculate the ratios of consecutive [tex]\( y \)[/tex]-values:
[tex]\[ \frac{0.25}{0.125} = 2, \quad \frac{0.5}{0.25} = 2, \quad \frac{1}{0.5} = 2, \quad \frac{2}{1} = 2, \quad \frac{4}{2} = 2, \quad \frac{8}{4} = 2 \][/tex]
The ratios are constant. Thus, [tex]\( g(x) \)[/tex] could be an exponential function.
### Function [tex]\( h(x) \)[/tex]:
[tex]\[ \begin{aligned} &x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ &h(x) & 0.25 & 0.5 & 0.75 & 0 & 1.25 & 1.5 & 1.75 \\ \end{aligned} \][/tex]
Let’s calculate the ratios of consecutive [tex]\( y \)[/tex]-values:
[tex]\[ \frac{0.5}{0.25} = 2, \quad \frac{0.75}{0.5} = 1.5, \quad \frac{0}{0.75} = 0, \quad \frac{1.25}{0} \ (\text{undefined}), \quad \frac{1.5}{1.25} = 1.2, \quad \frac{1.75}{1.5} \approx 1.17 \][/tex]
The ratios are not constant. Thus, [tex]\( h(x) \)[/tex] is not an exponential function.
### Function [tex]\( k(x) \)[/tex]:
[tex]\[ \begin{aligned} &x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ &k(x) & 64 & 16 & 4 & 1 & 0.25 & 0.0625 & 0.0156 \\ \end{aligned} \][/tex]
Let’s calculate the ratios of consecutive [tex]\( y \)[/tex]-values:
[tex]\[ \frac{16}{64} = 0.25, \quad \frac{4}{16} = 0.25, \quad \frac{1}{4} = 0.25, \quad \frac{0.25}{1} = 0.25, \quad \frac{0.0625}{0.25} = 0.25, \quad \frac{0.0156}{0.0625} = 0.25 \][/tex]
The ratios are constant. Thus, [tex]\( k(x) \)[/tex] could be an exponential function.
### Conclusion:
The functions [tex]\( g(x) \)[/tex] and [tex]\( k(x) \)[/tex] give values that could come from exponential functions.
So, the correct answers are:
- A. [tex]\( g(x) \)[/tex]
- D. [tex]\( k(x) \)[/tex]
