Answer :
To determine the base of the exponential function, which follows a pattern of decay through the points [tex]\((-2, 25)\)[/tex], [tex]\((-1, 5)\)[/tex], and [tex]\((0, 1)\)[/tex], we start by recognizing that the exponential function is of the form [tex]\( y = a \cdot b^x \)[/tex], where [tex]\(a\)[/tex] is a constant and [tex]\(b\)[/tex] is the base we need to find.
Using the given points, we can set up equations based on the form [tex]\( y = a \cdot b^x \)[/tex]:
1. For the point [tex]\((-2, 25)\)[/tex]:
[tex]\[ 25 = a \cdot b^{-2} \][/tex]
2. For the point [tex]\((-1, 5)\)[/tex]:
[tex]\[ 5 = a \cdot b^{-1} \][/tex]
3. For the point [tex]\((0, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^0 \][/tex]
Next, we simplify these equations to find the base [tex]\(b\)[/tex]:
First, from the point [tex]\((0, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^0 \][/tex]
Since any number raised to the power of 0 is 1, [tex]\(b^0 = 1\)[/tex]. Thus, we have:
[tex]\[ 1 = a \cdot 1 \][/tex]
This simplifies to:
[tex]\[ a = 1 \][/tex]
Now, substitute [tex]\(a = 1\)[/tex] into the equations derived from the other points.
For the point [tex]\((-1, 5)\)[/tex]:
[tex]\[ 5 = 1 \cdot b^{-1} \][/tex]
[tex]\[ 5 = \frac{1}{b} \][/tex]
Solving for [tex]\(b\)[/tex]:
[tex]\[ b = \frac{1}{5} \][/tex]
To confirm, we can use the remaining point [tex]\((-2, 25)\)[/tex]:
[tex]\[ 25 = 1 \cdot b^{-2} \][/tex]
[tex]\[ 25 = \left(\frac{1}{5}\right)^{-2} \][/tex]
Since [tex]\(\left(\frac{1}{5}\right)^{-2} = 5^2 = 25\)[/tex], the equation holds true.
Therefore, the base of the exponential function is:
[tex]\[ b = \frac{1}{5} \][/tex]
Using the given points, we can set up equations based on the form [tex]\( y = a \cdot b^x \)[/tex]:
1. For the point [tex]\((-2, 25)\)[/tex]:
[tex]\[ 25 = a \cdot b^{-2} \][/tex]
2. For the point [tex]\((-1, 5)\)[/tex]:
[tex]\[ 5 = a \cdot b^{-1} \][/tex]
3. For the point [tex]\((0, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^0 \][/tex]
Next, we simplify these equations to find the base [tex]\(b\)[/tex]:
First, from the point [tex]\((0, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^0 \][/tex]
Since any number raised to the power of 0 is 1, [tex]\(b^0 = 1\)[/tex]. Thus, we have:
[tex]\[ 1 = a \cdot 1 \][/tex]
This simplifies to:
[tex]\[ a = 1 \][/tex]
Now, substitute [tex]\(a = 1\)[/tex] into the equations derived from the other points.
For the point [tex]\((-1, 5)\)[/tex]:
[tex]\[ 5 = 1 \cdot b^{-1} \][/tex]
[tex]\[ 5 = \frac{1}{b} \][/tex]
Solving for [tex]\(b\)[/tex]:
[tex]\[ b = \frac{1}{5} \][/tex]
To confirm, we can use the remaining point [tex]\((-2, 25)\)[/tex]:
[tex]\[ 25 = 1 \cdot b^{-2} \][/tex]
[tex]\[ 25 = \left(\frac{1}{5}\right)^{-2} \][/tex]
Since [tex]\(\left(\frac{1}{5}\right)^{-2} = 5^2 = 25\)[/tex], the equation holds true.
Therefore, the base of the exponential function is:
[tex]\[ b = \frac{1}{5} \][/tex]
