Sure! Let's break down the problem and find the exponential function that passes through the points [tex]\( (-2, 3) \)[/tex] and [tex]\( (-1, 9) \)[/tex].
### Step-by-Step Solution:
1. Identify the Form of the Exponential Function:
We are looking for an exponential function of the form:
[tex]\[ y = a \cdot b^x \][/tex]
2. Substitute the Given Points into the Equation:
Let's substitute the points [tex]\( (-2, 3) \)[/tex] and [tex]\( (-1, 9) \)[/tex] into the exponential function and create two equations.
For the point [tex]\( (-2, 3) \)[/tex]:
[tex]\[ 3 = a \cdot b^{-2} \][/tex]
For the point [tex]\( (-1, 9) \)[/tex]:
[tex]\[ 9 = a \cdot b^{-1} \][/tex]
3. Divide the Equations to Eliminate [tex]\( a \)[/tex]:
We can divide the second equation by the first to solve for [tex]\( b \)[/tex]:
[tex]\[ \frac{9}{3} = \frac{a \cdot b^{-1}}{a \cdot b^{-2}} \][/tex]
[tex]\[ 3 = \frac{b^{-1}}{b^{-2}} \][/tex]
[tex]\[ 3 = b \][/tex]
Hence, we have [tex]\( b = 3 \)[/tex].
4. Substitute [tex]\( b \)[/tex] Back to Find [tex]\( a \)[/tex]:
Now that we know [tex]\( b = 3 \)[/tex], we can substitute [tex]\( b \)[/tex] back into one of the original equations to solve for [tex]\( a \)[/tex].
Using the equation [tex]\( 3 = a \cdot 3^{-2} \)[/tex]:
[tex]\[ 3 = a \cdot \frac{1}{9} \][/tex]
[tex]\[ 3 = \frac{a}{9} \][/tex]
[tex]\[ a = 3 \cdot 9 \][/tex]
[tex]\[ a = 27 \][/tex]
So, the constants are [tex]\( a = 27 \)[/tex] and [tex]\( b = 3 \)[/tex].
5. Write the Exponential Function:
The exponential function that passes through the points [tex]\( (-2, 3) \)[/tex] and [tex]\( (-1, 9) \)[/tex] is:
[tex]\[ y = 27 \cdot 3^x \][/tex]
This is the required exponential function.
