SAT Math: Scatterplots and Lines of Best Fit (Drill 2)

Question 1. Which type of function best models the relationship between x and y in the data?

• A) Linear, because y increases by a constant amount as x increases by 1.
• B) Quadratic, because the data forms a U-shaped pattern.
• C) Exponential, because y approximately doubles as x increases by 1. ✓
• D) Linear, because all the data points lie exactly on a line.

Explanation: Check the ratios of consecutive y-values: 6/3 = 2, 11/6 ≈ 1.8, 25/11 ≈ 2.3, 49/25 ≈ 2.0, 98/49 = 2.0. The y-values approximately double each time x increases by 1, which is the hallmark of exponential growth. A linear model would require constant differences (not ratios), and the differences here are 3, 5, 14, 24, 49, clearly not constant. Choice A is incorrect because the increase is not constant.

Question 2. A scatterplot shows the amount of a medication, in milligrams, remaining in a patient's bloodstream over time. An equation for the exponential model can be written as y = a · b^x, where a and b are positive constants, x is the number of hours, and y is the amount remaining. The model shows a decreasing trend. Which of the following is closest to the value of b?

• A) 0.72 ✓
• B) 1.00
• C) 1.28
• D) 2.50

Explanation: For an exponential model y = a · b^x where a and b are positive constants: if 0 < b < 1, the model is decreasing (decay); if b > 1, the model is increasing (growth). Since the model shows a decreasing trend, b must be between 0 and 1. Of the given choices, only 0.72 satisfies 0 < b < 1. Choice B (1.00) would mean no change. Choices C and D are greater than 1 and would indicate growth.

Question 3. Of the following, which equation best models the data?

• A) y = −5x² + 30x + 2 ✓
• B) y = 5x² + 30x + 2
• C) y = −5x² − 30x + 2
• D) y = −5x + 47

Explanation: The data shows a parabolic shape, height increases, reaches a maximum near x = 3, then decreases symmetrically. This indicates a downward-opening parabola (negative leading coefficient). Check Choice A at x = 0: y = −5(0) + 30(0) + 2 = 2 ✓. At x = 3: y = −5(9) + 30(3) + 2 = −45 + 90 + 2 = 47 ✓. At x = 6: y = −5(36) + 30(6) + 2 = −180 + 180 + 2 = 2 ✓. Choice B has a positive leading coefficient, which would open upward. Choice D is linear and cannot model the curved pattern.

Question 4. A line of best fit for a data set has the equation y = 1.5x + 8. A new data set is created by multiplying the y-coordinate of each data point by 4. Which of the following could be an equation of a line of best fit for the new data set?

• A) y = 6x + 32 ✓
• B) y = 1.5x + 32
• C) y = 6x + 8
• D) y = 5.5x + 12

Explanation: When every y-value is multiplied by 4, the entire equation is scaled by 4: y_new = 4(1.5x + 8) = 6x + 32. The slope is multiplied by 4 (1.5 × 4 = 6) and the y-intercept is multiplied by 4 (8 × 4 = 32). Choice B only multiplies the intercept. Choice C only multiplies the slope. Choice D adds 4 instead of multiplying.

Question 5. According to the line of best fit, what is the predicted price of a 3-year-old car, and how does it compare to the actual data point?

• A) Predicted: $22,100; the actual price is $700 higher than predicted. ✓
• B) Predicted: $22,100; the actual price is $700 lower than predicted.
• C) Predicted: $22,800; the actual price equals the predicted price.
• D) Predicted: $25,200; the actual price is $2,400 lower than predicted.

Explanation: Substitute x = 3 into the equation: y = −3,100(3) + 31,400 = −9,300 + 31,400 = 22,100. The actual price at age 3 is $22,800. The difference is 22,800 − 22,100 = 700. The actual price is $700 higher than the predicted price. This difference (called the residual) is positive, meaning the data point lies above the line of best fit.