Linear, Quadratic, and Exponential Models: Constructing Linear and Exponential Functions (CCSS.F-LE.2)
Common Core High School Functions · Learn by Concept
Help Questions
Common Core High School Functions › Linear, Quadratic, and Exponential Models: Constructing Linear and Exponential Functions (CCSS.F-LE.2)
A savings account earns 5 percent interest compounded yearly. The balance is 1000 dollars initially and about 1157.625 dollars after 3 years. This situation has a constant growth ratio, so it is exponential.
Which function best fits the data?
$A=1000\cdot 1.5^t$
$A=1000\cdot 1.05^t$
$A=1050\cdot 1.05^t$
$A=1000+50t$
Explanation
A 5% yearly growth gives ratio \$1.05$ and initial value 1000, so $A=1000\cdot 1.05^t$. Choice C has the right ratio but the wrong starting value; D is a linear model that ignores compounding; A uses the wrong base.
A student saves the same amount each week, starting with 120 already in the account. Balance (in dollars) by week: Week 0: 120, Week 1: 170, Week 2: 220, Week 3: 270.
Which function best fits the data?
$B=120+50t$
$B=120+60t$
$B=100+50t$
$B=120+45t$
Explanation
The balance increases by a constant difference of $+50$ each week, with initial value 120, so $B=120+50t$. Choice C has the correct slope but wrong starting amount; B uses the wrong slope; D reflects an arithmetic slip (miscomputing the weekly increase as 45).
A car's value decreases by 15 percent each year. Its value when new is 24000 dollars; after one year it's 20400 dollars and after two years it's about 17340 dollars. This is an exponential decay model with a constant ratio.
Which model represents this situation?
$V=24000(0.15)^t$
$V=20000(0.85)^t$
$V=24000(0.85)^t$
$V=24000-3600t$
Explanation
A 15% yearly decrease means multiply by \$0.85$ each year: $V=24000(0.85)^t$. Choice D is a tempting linear model subtracting a fixed 3600 each year, but the decreases should get smaller over time; A uses the wrong base; B has the correct ratio but wrong starting value.
A used car is worth 25,000 at purchase and then loses 20% of its value each year. The values over the first years are: Year 0: 25,000; Year 1: 20,000; Year 2: 16,000; Year 3: 12,800. Assume the same percent change each year.
Which function best fits the data?
$V=25000-5000t$
$V=20000\cdot 0.8^t$
$V=25000\cdot 0.75^t$
$V=25000\cdot 0.8^t$
Explanation
Successive ratios are $\frac{20000}{25000}=0.8$, so the exponential decay model is $V=25000\cdot 0.8^t$. Choice B has the correct ratio but wrong starting value; C is an arithmetic slip using \$0.75$; A is linear with a constant difference, not a constant percent.
At noon a water tank holds 900 liters. By 2 pm it holds 720 liters, and the tank drains at a constant rate (same number of liters each hour).
Which model represents this situation?
$y=-80x+900$
$y=-90x+900$
$y=-90x+720$
$y=900\cdot 0.9^x$
Explanation
Let $x$ be hours after noon. Using $(0,900)$ and $(2,720)$, the slope is $m=\frac{720-900}{2-0}=-\frac{180}{2}=-90$, so $y=-90x+900$. Choice C keeps the slope but uses the wrong intercept (plugging in a data point as $b$). A has the wrong slope, and D is the wrong function type (exponential).
A colony of bacteria starts with 300 cells and doubles every hour. After 1 hour there are 600 cells and after 3 hours there are 2400 cells, showing multiplicative growth.
Which function best fits the data?
$y=300\cdot 2^t$
$y=600\cdot 2^t$
$y=300\cdot 3^t$
$y=300+300t$
Explanation
The growth factor is a constant ratio $\frac{600}{300}=2$, with initial value 300, so $y=300\cdot 2^t$. Choice B uses the wrong starting value, C uses the wrong ratio, and D is linear (constant difference) not exponential.
A plumber charges a fixed trip fee plus a constant hourly rate. A 2-hour job cost 170, and a 5-hour job cost 365. Assume the cost increases by the same amount each additional hour.
Which model represents this situation?
$y=60x+40$
$y=65x+45$
$y=65x+40$
$y=40\cdot 1.65^x$
Explanation
Using $(2,170)$ and $(5,365)$, the slope is $m=\frac{365-170}{5-2}=\frac{195}{3}=65$. Solve $170=65\cdot 2+b$ to get $b=40$, so $y=65x+40$. Choice B has the correct slope but the wrong intercept, and D is the wrong function type (exponential).
A food truck charges a fixed fee plus a cost per mile to travel to events. For a 5-mile trip, the total charge is 35 dollars, and for an 11-mile trip, the total is 59 dollars. Assume the relationship is linear with a constant per-mile rate.
Which model represents this situation?
$y=6x+5$
$y=4x+15$
$y=4x+20$
$y=15\cdot 4^x$
Explanation
Slope $m=\frac{59-35}{11-5}=\frac{24}{6}=4$, and using $(5,35)$ gives $35=4\cdot 5+b\Rightarrow b=15$, so $y=4x+15$. Choice C has the correct slope but the wrong intercept; D is an exponential model (wrong type) though the numbers may look related.