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To determine how the [tex]\( y \)[/tex]-values in the table grow, we analyze the given [tex]\( y \)[/tex]-values and their corresponding [tex]\( x \)[/tex]-values:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 2 & 49 \\ \hline 4 & 2401 \\ \hline 6 & 117649 \\ \hline \end{array} \][/tex]
We verify how the [tex]\( y \)[/tex]-values change as [tex]\( x \)[/tex] increases:
1. First, we calculate the ratios of consecutive [tex]\( y \)[/tex]-values to understand the growth factor.
- From [tex]\( y(0) = 1 \)[/tex] to [tex]\( y(2) = 49 \)[/tex]:
[tex]\[ \frac{y(2)}{y(0)} = \frac{49}{1} = 49 \][/tex]
- From [tex]\( y(2) = 49 \)[/tex] to [tex]\( y(4) = 2401 \)[/tex]:
[tex]\[ \frac{y(4)}{y(2)} = \frac{2401}{49} = 49 \][/tex]
- From [tex]\( y(4) = 2401 \)[/tex] to [tex]\( y(6) = 117649 \)[/tex]:
[tex]\[ \frac{y(6)}{y(4)} = \frac{117649}{2401} = 49 \][/tex]
2. As we can see, in all these cases, the ratio is consistently 49. This tells us that the [tex]\( y \)[/tex]-values are multiplied by 49 every time [tex]\( x \)[/tex] increases by 2 units.
Given that an exponential function grows by a constant factor and the growth factor is consistent when [tex]\( x \)[/tex] increases by 2 units, we can conclude:
- The [tex]\( y \)[/tex]-values increase by a factor of 49 for each [tex]\( x \)[/tex] increase of 2.
Since no options directly mention a factor of 49 for each [tex]\( x \)[/tex] increase of 2, we reframe this in line with the options provided. Essentially, if the [tex]\( y \)[/tex]-values increase by a factor of 49 over 2 units of [tex]\( x \)[/tex], over 1 unit of [tex]\( x \)[/tex], the increase can be described using roots.
Therefore, the consistent growth pattern means that for each unit increase in [tex]\( x \)[/tex], the growth factor must be:
[tex]\[ 49^{1/2} = 7 \][/tex]
Thus, the correct option is:
- The [tex]\( y \)[/tex]-values increase by a factor of 7 for each [tex]\( x \)[/tex] increase of 1.
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 2 & 49 \\ \hline 4 & 2401 \\ \hline 6 & 117649 \\ \hline \end{array} \][/tex]
We verify how the [tex]\( y \)[/tex]-values change as [tex]\( x \)[/tex] increases:
1. First, we calculate the ratios of consecutive [tex]\( y \)[/tex]-values to understand the growth factor.
- From [tex]\( y(0) = 1 \)[/tex] to [tex]\( y(2) = 49 \)[/tex]:
[tex]\[ \frac{y(2)}{y(0)} = \frac{49}{1} = 49 \][/tex]
- From [tex]\( y(2) = 49 \)[/tex] to [tex]\( y(4) = 2401 \)[/tex]:
[tex]\[ \frac{y(4)}{y(2)} = \frac{2401}{49} = 49 \][/tex]
- From [tex]\( y(4) = 2401 \)[/tex] to [tex]\( y(6) = 117649 \)[/tex]:
[tex]\[ \frac{y(6)}{y(4)} = \frac{117649}{2401} = 49 \][/tex]
2. As we can see, in all these cases, the ratio is consistently 49. This tells us that the [tex]\( y \)[/tex]-values are multiplied by 49 every time [tex]\( x \)[/tex] increases by 2 units.
Given that an exponential function grows by a constant factor and the growth factor is consistent when [tex]\( x \)[/tex] increases by 2 units, we can conclude:
- The [tex]\( y \)[/tex]-values increase by a factor of 49 for each [tex]\( x \)[/tex] increase of 2.
Since no options directly mention a factor of 49 for each [tex]\( x \)[/tex] increase of 2, we reframe this in line with the options provided. Essentially, if the [tex]\( y \)[/tex]-values increase by a factor of 49 over 2 units of [tex]\( x \)[/tex], over 1 unit of [tex]\( x \)[/tex], the increase can be described using roots.
Therefore, the consistent growth pattern means that for each unit increase in [tex]\( x \)[/tex], the growth factor must be:
[tex]\[ 49^{1/2} = 7 \][/tex]
Thus, the correct option is:
- The [tex]\( y \)[/tex]-values increase by a factor of 7 for each [tex]\( x \)[/tex] increase of 1.
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