Question

Use the table to answer the question.

This table displays the exponential growth of a function over equal intervals.
x
f(x)=a⋅bx

x0
a⋅bx0

x0+Δx
a⋅b(x0+Δx)

x0+2Δx
a⋅b(x0+2Δx)

Ramona wants to use the table to prove something about how exponential functions grow. What does she need to do next and what will she be able to prove?

(1 point)
Responses

Ramona needs to add another column to the table and find the growth factors between the consecutive values of f(x)
. She will be able to prove that exponential functions grow by equal factors over equal intervals.
Ramona needs to add another column to the table and find the growth factors between the consecutive values of f left parenthesis x right parenthesis . She will be able to prove that exponential functions grow by equal factors over equal intervals.

Ramona needs to subtract the x
-values from the f(x)
-values. She will be able to prove that exponential functions grow by equal factors over equal intervals.
Ramona needs to subtract the x -values from the f left parenthesis x right parenthesis -values. She will be able to prove that exponential functions grow by equal factors over equal intervals.

Ramona needs to subtract the x
-values from the f(x)
-values. She will be able to prove that exponential functions grow by equal differences over equal intervals.
Ramona needs to subtract the x -values from the f left parenthesis x right parenthesis -values. She will be able to prove that exponential functions grow by equal differences over equal intervals.

Ramona needs to add another column to the table and find the change in f(x)
-values. She will be able to prove that exponential functions grow by equal differences over equal intervals.

Answer 1

Ramona should add a column with the growth factors (ratios) between consecutive f(x) values. Then compute

f(x0+Δx)/f(x0) = a b^{x0+Δx} / (a b^{x0}) = b^{Δx},
f(x0+2Δx)/f(x0+Δx) = a b^{x0+2Δx} / (a b^{x0+Δx}) = b^{Δx}.

Thus the ratios are equal, so exponential functions grow by equal factors over equal intervals.