Question
Which exponential equation can be written with points (-2,0.32) (1,0.625)
Answers
GPT-4o mini
To find the exponential equation that fits the given points \((-2, 0.32)\) and \((1, 0.625)\), we can start by assuming a general form of an exponential function:
\[
y = ab^x
\]
where \(a\) is the initial value (the value of \(y\) when \(x = 0\)), and \(b\) is the growth factor.
### Step 1: Insert the points into the equation.
For the point \((-2, 0.32)\):
\[
0.32 = ab^{-2}
\]
For the point \((1, 0.625)\):
\[
0.625 = ab^1
\]
### Step 2: Express both equations in terms of \(a\) and \(b\).
From the second equation, we can express \(a\) in terms of \(b\):
\[
a = \frac{0.625}{b}
\]
### Step 3: Substitute \(a\) into the first equation.
Substituting \(a\) into the first equation:
\[
0.32 = \left(\frac{0.625}{b}\right)b^{-2}
\]
Simplifying this, we get:
\[
0.32 = 0.625 b^{-1}
\]
Multiplying both sides by \(b\):
\[
0.32b = 0.625
\]
Dividing both sides by \(0.32\):
\[
b = \frac{0.625}{0.32} \approx 1.953125
\]
### Step 4: Find \(a\).
Now that we have the value for \(b\), we can find \(a\) using the equation \(a = \frac{0.625}{b}\):
\[
a = \frac{0.625}{1.953125} \approx 0.32
\]
### Step 5: Final equation.
Now we can write the exponential equation:
\[
y \approx 0.32 (1.953125)^x
\]
### Conclusion:
So the approximate exponential equation that fits the points \((-2, 0.32)\) and \((1, 0.625)\) is:
\[
y \approx 0.32 (1.953125)^x
\]
This is a general method to derive an exponential function from two points. You can plug in different values if more precision or verification is needed.
\[
y = ab^x
\]
where \(a\) is the initial value (the value of \(y\) when \(x = 0\)), and \(b\) is the growth factor.
### Step 1: Insert the points into the equation.
For the point \((-2, 0.32)\):
\[
0.32 = ab^{-2}
\]
For the point \((1, 0.625)\):
\[
0.625 = ab^1
\]
### Step 2: Express both equations in terms of \(a\) and \(b\).
From the second equation, we can express \(a\) in terms of \(b\):
\[
a = \frac{0.625}{b}
\]
### Step 3: Substitute \(a\) into the first equation.
Substituting \(a\) into the first equation:
\[
0.32 = \left(\frac{0.625}{b}\right)b^{-2}
\]
Simplifying this, we get:
\[
0.32 = 0.625 b^{-1}
\]
Multiplying both sides by \(b\):
\[
0.32b = 0.625
\]
Dividing both sides by \(0.32\):
\[
b = \frac{0.625}{0.32} \approx 1.953125
\]
### Step 4: Find \(a\).
Now that we have the value for \(b\), we can find \(a\) using the equation \(a = \frac{0.625}{b}\):
\[
a = \frac{0.625}{1.953125} \approx 0.32
\]
### Step 5: Final equation.
Now we can write the exponential equation:
\[
y \approx 0.32 (1.953125)^x
\]
### Conclusion:
So the approximate exponential equation that fits the points \((-2, 0.32)\) and \((1, 0.625)\) is:
\[
y \approx 0.32 (1.953125)^x
\]
This is a general method to derive an exponential function from two points. You can plug in different values if more precision or verification is needed.