Question

Use the table to answer the question.

Which exponential equation contains the points shown in the input-output table?

x y
−2
−
2
0.32
0.32
1 0.625
0.625
(1 point)
Responses

y=−0.5⋅(1.25)x
𝑦
=
−
0.5
⋅
(
1.25
)
𝑥
y is equal to negative 0 point 5 times 1 point 2 5 to the x th power

y=(0.625)x
𝑦
=
(
0.625
)
𝑥
y is equal to 0 point 6 2 5 to the x th power

y=0.05⋅−(1.25)x
𝑦
=
0.05
⋅
−
(
1.25
)
𝑥
y is equal to 0 point 0 5 times negative 1 point 2 5 to the x th power

y=0.5⋅(1.25)x

Answer 1

To determine which exponential equation contains the given points in the input-output table, we can analyze the values provided in the table and the options given.

From the input-output table, we see:

- When \( x = -2 \), \( y = 0.32 \)
- When \( x = 1 \), \( y = 0.625 \)

We need to see which of the provided options satisfies these points.

Let's check the equations:

1. **Option 1:** \( y = -0.5 \cdot (1.25)^x \)
- For \( x = -2 \):
\[ y = -0.5 \cdot (1.25)^{-2} = -0.5 \cdot \frac{1}{(1.25^2)} \approx -0.5 \cdot 0.64 \approx -0.32 \] (not valid since \( y \) cannot be negative)
- This option is not valid.

2. **Option 2:** \( y = (0.625)^x \)
- For \( x = -2 \):
\[ y = (0.625)^{-2} = \frac{1}{(0.625^2)} \approx \frac{1}{0.390625} \approx 2.56 \] (not valid)
- This option is not valid.

3. **Option 3:** \( y = 0.05 \cdot -(1.25)^x \)
- For \( x = -2 \):
\[ y = 0.05 \cdot -(1.25)^{-2} = 0.05 \cdot -\frac{1}{1.5625} \approx 0.05 \cdot -0.64 \approx -0.032 \] (not valid)
- This option is also not valid.

4. **Option 4:** \( y = 0.5 \cdot (1.25)^x \)
- For \( x = -2 \):
\[ y = 0.5 \cdot (1.25)^{-2} = 0.5 \cdot \frac{1}{1.5625} \approx 0.5 \cdot 0.64 = 0.32 \] (valid for \( x = -2 \))
- For \( x = 1 \):
\[ y = 0.5 \cdot (1.25)^1 = 0.5 \cdot 1.25 = 0.625 \] (valid for \( x = 1 \))

The correct exponential equation that contains the points shown in the input-output table is:

**\( y = 0.5 \cdot (1.25)^x \)**.