A website tracks the number of daily views, , that a newly uploaded video receives over several days, . The initial scatterplot of versus shows a curved, decreasing trend. When the data analyst plots on the vertical axis against on the horizontal axis, the points form a nearly perfect straight line with negative slope. Based on these observations, which type of function is most appropriate for modeling as a function of ?

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SAT Two-Variable Data Question 57 | Free SAT Prep | Aniko

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Question 57·Hard·Two-Variable Data: Models and Scatterplots

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A website tracks the number of daily views, $v$ , that a newly uploaded video receives over several days, $d$ . The initial scatterplot of $v$ versus $d$ shows a curved, decreasing trend. When the data analyst plots $\ln v$ on the vertical axis against $d$ on the horizontal axis, the points form a nearly perfect straight line with negative slope.

Based on these observations, which type of function is most appropriate for modeling $v$ as a function of $d$ ?

For SAT questions about choosing a model type from scatterplots and log transformations, memorize the key patterns: if $y$ vs. $x$ is straight, the relationship is linear; if $\ln y$ vs. $x$ is straight, the underlying model is exponential in $x$ ; if $\ln y$ vs. $\ln x$ is straight, it is a power function; and if $y$ vs. $\ln x$ is straight, it is logarithmic. When you see a log applied to one variable in the graph description, quickly write the implied linear equation (like $\ln v = md + b$ ), solve for the original variable, and match the resulting form to the answer choices instead of trying to reason from the scatterplot shape alone.

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Use the information about ln v vs. d

Focus on the second description: $\ln v$ plotted against $d$ is almost a straight line. What kind of algebraic equation connects two quantities when their graph is a straight line?

Write an equation and solve for v

If a straight line relates $\ln v$ and $d$ , write it as $\ln v = md + b$ . How can you undo the natural log on $v$ to get an equation just in terms of $v$ and $d$ ?

Compare your equation with each function type

After you solve for $v$ , write sample equations that fit each answer choice (for instance, $v = 3d + 5$ for one, $v = 3d^2$ for another, $v = 3 + 5\ln d$ for another) and compare their shapes and structure with the equation you found for $v$ . Which option has the same pattern of where $d$ appears?

Desmos Guide

Set up example models for each function type

In Desmos, enter sample functions that represent each answer type using the variable d:

Linear-type example: v(d) = 100 - 5d
Example of the form v(d) = A*B^d
Power-type example: v(d) = 100*d^-1
Logarithmic-type example: v(d) = 50 + 10*ln(d) (Desmos will create sliders for constants like A, B if you want.)

Graph ln v versus d for each model

For each function you defined, create a new expression for the transformed output, such as g(d) = ln(v(d)). Plot each g(d) on the same axes as a function of d. Visually compare which transformed graph looks like a straight line, since the problem states that $\ln v$ versus $d$ forms an almost perfect straight line.

Match the straight-line transformation to an answer choice

Notice which type of original model (from your four examples) produces a graph of $\ln v$ versus $d$ that appears linear in Desmos. That is the same function type that best matches the situation described in the question.

Step-by-step Explanation

Translate the straight line in the transformed plot into an equation

The problem says that when the analyst plots $\ln v$ on the vertical axis against $d$ on the horizontal axis, the points form a nearly perfect straight line with negative slope.

A straight line in these variables means $\ln v$ and $d$ are related linearly, so we can write

\ln v = md + b,

where $m<0$ is the (negative) slope and $b$ is the intercept.

Rewrite the equation in terms of v instead of ln v

To solve for $v$ , undo the natural log by exponentiating both sides with base $e$ :

\ln v = md + b \implies e^{\ln v} = e^{md + b}.

Since $e^{\ln v} = v$ , we get

v = e^{md + b}.

Now separate the terms in the exponent:

v = e^{md + b} = e^b \cdot e^{md}.

Let $A = e^b$ and $k = m$ , so

v = A \cdot e^{kd}.

Match the form of v(d) to a function type

The model we found is

v = A \cdot e^{kd},

where $A$ and $k$ are constants and the variable $d$ appears in the exponent.

Among the answer choices, a function where the input variable is in the exponent (with a constant base) is called an exponential function, so the most appropriate model is B) Exponential.

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