Question

The curve begins at approximately left parenthesis negative 10 comma negative 1 right parenthesis and passes through the plotted point left parenthesis negative 9 comma negative 1 right parenthesis. It continues almost horizontally until it begins to curve upward at approximately left parenthesis negative 5 comma negative 1 right parenthesis. The curves passes through approximately left parenthesis negative 2 comma 0 right parenthesis, left parenthesis 0 comma 3 right parenthesis, and left parenthesis 1 comma 7 right parenthesis before passing through the plotted point left parenthesis 2 comma 15 right parenthesis. The curve ends at approximately left parenthesis 2.4 comma 18 right parenthesis.

Compare the estimated average rates of change of the exponential function pictured above and the quadratic function f(x)=x2−20
over the interval [−9,2]
and identify which function has a greater rate of change than the other.

(1 point)

The estimated average rate of change of the exponential function over the interval [−9,2]
is less than the magnitude of the estimated average rate of change of f(x)
over the interval [−9,2]
.

The estimated average rate of change of the exponential function over the interval [−9,2]
is greater than the magnitude of the estimated average rate of change of f(x)
over the interval [−9,2]
.

The estimated average rate of change of both the exponential function and f(x)
over the interval [−9,2]
is approximately 1.5
.

The estimated average rate of change of the exponential function over the interval [−9,2]
is negative, while the estimated average rate of change of f(x)
over the interval [−9,2]
is positive.

Answer 1

Compute averages on [−9, 2]:

- Exponential: y(−9) ≈ −1, y(2) = 15, so average rate = (15 − (−1)) / (2 − (−9)) = 16/11 ≈ 1.45.
- Quadratic f(x)=x^2−20: f(−9)=81−20=61, f(2)=4−20=−16, so average rate = (−16−61)/11 = −77/11 = −7 (magnitude 7).

Thus the exponential’s average rate (~1.45) is less than the magnitude of the quadratic’s average rate (7). Select the first option.