Question 38·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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The scatterplot shows the estimated population of a species of insect at various numbers of days after hatching. Each dot represents one observation. A curve representing an exponential model is shown. The two coordinate labels on the curve indicate exact values of the model.

If $f(d)$ is the population, in thousands, predicted by the model $d$ days after hatching and $f(14)=k\cdot f(4)$ , what is the value of $k$ ?

Your answer

When a fitted exponential curve has exact coordinate labels, use the values on the curve to find a multiplier over a known time interval. An equation of the form $f(b)=k f(a)$ asks for the multiplier from $a$ to $b$ . Raise the known multiplier to the ratio of the two interval lengths, then evaluate it in Desmos. Observed dots need not equal the model’s predictions.

Hints

Hint 1
A dot is an observation, while a label on the fitted curve gives an exact model prediction. Use the two labeled curve values to compare predicted populations, even if other dots are nearby.
Hint 2
An exponential factor tells you what to multiply by over a particular interval. Divide the later labeled population by the earlier one, then count the days between those labels.
Hint 3
In $f(14)=k f(4)$ , $k$ compares the predictions on days $4$ and $14$ . How many days apart are they? Compare that gap with the interval for the factor you found.

Step-by-step

Approach 1: Convert the labeled interval factor

Step 1
Read predictions from the curve
The curve labels $(2,6)$ and $(8,48)$ give $f(2)=6$ and $f(8)=48$ , in thousands. The dots are observations, so the dot at day $4$ need not give $f(4)$ .
Step 2
Find the six-day factor
The labeled days are $8-2=6$ days apart. A factor is the number you multiply by, so divide the later prediction by the earlier one:
$\frac{48}{6}=8.$
The model multiplies by $8$ every $6$ days.
Step 3
Identify the interval for k
The equation says $k$ multiplies the day- $4$ prediction to get the day- $14$ prediction. Divide by $f(4)$ :
$k=\frac{f(14)}{f(4)}.$
Those days are $14-4=10$ days apart, so the six-day factor $8$ stops too early.
Step 4
Convert to a ten-day factor
Exponential growth repeats multiplication. A $10$ -day span contains $10/6$ six-day spans, so raise the six-day factor to that power:
$k=8^{10/6}.$
To change an exponential factor’s interval, use new interval divided by known interval as the exponent.
Step 5
Evaluate the requested multiplier
Type $\left(\frac{48}{6}\right)^{\frac{14-4}{8-2}}$ in Desmos. It prints $32$ . So the model predicts $32$ times as many insects on day $14$ as on day $4$ . Grid in 32.

Approach 2: Count two-day doublings

Step 1
Find a shorter repeating interval
The six-day factor $8$ is $2^3$ . Since six days contain three two-day spans, the model doubles every two days. This shows what the fractional exponent in the first method is counting.
Step 2
Count the doublings
From day $4$ to day $14$ are five two-day spans. Type $2^5$ in Desmos; it prints $32$ . Five doublings make the day- $14$ prediction $32$ times the day- $4$ prediction. Grid in 32.

Question 38·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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