Question 32·SAT Desmos Regression Practice·Algebra

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A town’s yearly water demand is modeled by a linear function $W(t)$ , where $W(t)$ is the demand (in millions of gallons) $t$ years after 2015.

In 2019, the demand was 12% greater than the demand in 2017.

In 2023, the demand was 5 million gallons less than twice the demand in 2018.

Assuming the demand changed at a constant rate, which choice gives $W(t)$ ?

A linear model has a constant change each year, even when the facts compare demands by percentages. Convert each calendar year to years after the starting year, then turn the two comparisons into equations for the same line. A matched-list regression in Desmos can find the line’s rate and starting value. The key trap is treating a percent increase as a fixed number of gallons.

Hints

Hint 1
The input is years after 2015, not the calendar year itself. Subtract 2015 from each year before putting it into $W(t)$ .
Hint 2
For a 12% increase, the later demand is 112% of the earlier demand. Which demand is the earlier one, and what multiplier represents 112%?
Hint 3
A linear function can be written $W(t)=at+b$ , where $a$ is the yearly change and $b$ is the demand at $t=0$ . Turn each comparison into a condition on the same $a$ and $b$ .

Step-by-step

Translate both facts and fit the line
Step 1
Convert the years to inputs
The input $t$ counts years after 2015, so subtract 2015 from each year. The inputs for 2017, 2019, 2018, and 2023 are $2$ , $4$ , $3$ , and $8$ , respectively. Using calendar years as inputs would shift the line.
Step 2
Write the line in terms of its unknowns
A linear function changes by the same amount each year, so write $W(t)=at+b$ . Here $a$ is the yearly change in millions of gallons, and $b=W(0)$ is the demand in 2015. This gives $W(2)=2a+b$ and $W(4)=4a+b$ for the first comparison.
Step 3
Translate the percent comparison
A percent increase multiplies the earlier demand; it isn't a fixed increase in gallons. Since $100\%+12\%=112\%=1.12$ , the 2019 demand is $1.12$ times the 2017 demand:
$4a+b=1.12(2a+b).$
Subtract the right side to make this a condition equal to $0$ :
$4a+b-1.12(2a+b)=0.$
Step 4
Translate the second comparison
The demand in 2023 is 5 million gallons less than twice the 2018 demand, not twice the 2023 demand. Using $W(8)=8a+b$ and $W(3)=3a+b$ gives
$8a+b=2(3a+b)-5.$
Subtract twice the 2018 demand so this condition equals $-5$ :
$8a+b-2(3a+b)=-5.$
Step 5
Fit both conditions together in Desmos
Type $[0,-5]\sim[4a+b-1.12(2a+b),8a+b-2(3a+b)]$ . The lists pair the $0$ with the percent condition and the $-5$ with the 2023 condition; $\sim$ asks Desmos to find one $a$ and one $b$ that fit both. Under PARAMETERS, it shows $a\approx0.394736842$ and $b\approx5.789473684$ .
Step 6
Read the exact coefficients and choose the rule
Add $a$ and $b$ on separate Desmos lines and use their fraction buttons. They give $a=\frac{15}{38}$ and $b=\frac{110}{19}$ . Put those values into $W(t)=at+b$ : the town’s modeled demand, in millions of gallons, is $W(t)=\frac{15}{38}t+\frac{110}{19}$ . Choice D.

Question 32·SAT Desmos Regression Practice·Algebra

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