Question

What is equation shown in the graph? (-2, 1.28) (1, 2.5)
(1 point) Responses y=−2⋅(1.25)x y is equal to negative 2 times 1 point 2 5 to the x th power y=2⋅−(1.25)x y is equal to 2 times negative 1 point 2 5 to the x th power y=1.25⋅(2)x y is equal to 1 point 2 5 times 2 to the x th power y=2⋅(1.25)x

Answer 1

To determine the equation represented by the graph, we can analyze the given points: (-2, 1.28) and (1, 2.5).

1. **Identifying the Base and Growth Factor**: For an exponential function in the form \( y = a \cdot b^x \) where:
- \( a \) is the initial value,
- \( b \) is the base (growth factor),
- \( x \) is the exponent.

2. **Finding the Values**: Using point (1, 2.5):
\[
y = a \cdot b^1 = a \cdot b
\]
Thus,
\[
a \cdot b = 2.5
\]

Using the other point (-2, 1.28):
\[
y = a \cdot b^{-2} = \frac{a}{b^2}
\]
Thus,
\[
\frac{a}{b^2} = 1.28 \rightarrow a = 1.28 \cdot b^2
\]

3. **Solving the Equations**:
From \( a \cdot b = 2.5 \):
- Substituting \( a \) from the second equation:
\[
(1.28 \cdot b^2) \cdot b = 2.5 \rightarrow 1.28 \cdot b^3 = 2.5
\]
- Thus, solving for \( b \):
\[
b^3 = \frac{2.5}{1.28} \approx 1.953125
\]
- Taking the cube root gives us:
\[
b \approx 1.25
\]

4. **Finding 'a'**:
- Substitute \( b \) back into \( a \cdot b = 2.5 \):
\[
a \cdot 1.25 = 2.5 \rightarrow a = \frac{2.5}{1.25} = 2
\]

5. **Determining the Exponential Function**:
Thus, we find:
\[
y = 2 \cdot (1.25)^x
\]

**Final Equation**:
The equation that matches the graph is
\[
y = 2 \cdot (1.25)^x
\]

This corresponds to the option \( y = 2 \cdot (1.25)^x \).