Question

Use the graph to answer the question.

What is equation shown in the graph?

A coordinate plane going from negative 5 to 5 in increments of 1 labeled every 5 units on the x-axis and from negative 1 to 10 in increments of 1 labeled every 5 units on the y-axis.

the two marked points are
-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 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

Answer 1

Use the graph to answer the question.

What is the equation shown in the graph?

A coordinate plane going from negative 10 to 10 in increments of 1 labeled every 5 units on the x-axis and from negative 1 to 15 in increments of 1 labeled every 5 units on the y-axis.

the makred points are
0,10 and 1, 10.5

(1 point)
Responses

y=1.05⋅(10)x
y is equal to 1 point 0 5 times 10 to the x th power

y=−10⋅(1.05)x
y is equal to negative 10 times 1 point 0 5 to the x th power

y=10⋅−(1.05)x
y is equal to 10 times negative 1 point 0 5 to the x th power

y=10⋅(1.05)x

Answer 2

To determine the correct equation from the provided graph points (-2, 1.28) and (1, 2.5), we can analyze the points. The relationship likely indicates an exponential function, typically in the form of \( y = a \cdot b^x \).

1. From point (-2, 1.28):
\[
1.28 = a \cdot b^{-2}
\]
This implies:
\[
a \cdot \frac{1}{b^2} = 1.28 \quad \text{(1)}
\]

2. From point (1, 2.5):
\[
2.5 = a \cdot b^{1}
\]
This implies:
\[
a \cdot b = 2.5 \quad \text{(2)}
\]

Next, we can solve these equations simultaneously to find \( a \) and \( b \).
From equation (2):
\[
a = \frac{2.5}{b}
\]

Substituting \( a \) into equation (1):
\[
\frac{2.5}{b} \cdot \frac{1}{b^2} = 1.28
\]
\[
\frac{2.5}{b^3} = 1.28
\]
\[
2.5 = 1.28b^3
\]
\[
b^3 = \frac{2.5}{1.28}
\]
\[
b^3 \approx 1.953125
\]
Taking the cube root:
\[
b \approx 1.26
\]

Now substituting this \( b \) back to find \( a \):
\[
a \cdot 1.26 = 2.5
\]
\[
a \approx \frac{2.5}{1.26} \approx 1.984
\]

Now that we have values for \( a \) and \( b \), we can conclude that one of the equations seems to be approaching:

If we round those numbers, we find:
1. \( a \approx 2 \)
2. \( b \approx 1.25 \)

Resulting in the equation being approximately:
\[
y = 2 \cdot (1.25)^x
\]

Among your choices, the correct response would be:
**y = 2⋅(1.25)x** (y is equal to 2 times 1 point 2 5 to the x th power).

Answer 3

To determine the correct equation based on the marked points \( (0, 10) \) and \( (1, 10.5) \), we can assume a form similar to \( y = a \cdot b^x \).

1. From point \( (0, 10) \):
\[
10 = a \cdot b^0
\]
This simplifies to:
\[
a = 10
\]

2. From point \( (1, 10.5) \):
\[
10.5 = 10 \cdot b^1
\]
Therefore:
\[
b = \frac{10.5}{10} = 1.05
\]

Putting these values back into the equation gives us:
\[
y = 10 \cdot (1.05)^x
\]

Among the provided options, the correct equation is:
**y = 10⋅(1.05)x** (y is equal to 10 times 1 point 0 5 to the x th power).