← Back to the Interpreting Functions worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: F-IF.B.4, F-IF.B.5, F-IF.B.6, F-IF.C.8a, F-IF.C.9 · Level: Grade 11 Algebra 2
Problem 1 Domain in context
Set h(t) = 0: −16t(t − 4) = 0, so t = 0 or 4. The ball is in the air from 0 to 4 seconds. Lands at t = 4; domain 0 ≤ t ≤ 4.
Problem 2 Average rate of change from an equation
s(1) = 3 and s(4) = 72. Rate = (72 − 3)/(4 − 1) = 69/3. 23 feet per second.
Problem 3 Interpret a linear model
640 is the starting volume in gallons, and −16 means the tank loses 16 gallons every minute. Solve 640 − 16t = 200: 16t = 440, so t = 27.5. Starts at 640 gal; drains 16 gal/min; 200 gal remain at 27.5 min.
Problem 4 Average rate of change from a table
From 2 to 5: (26 − 11)/(5 − 2) = 5. From 5 to 9: (50 − 26)/(9 − 5) = 24/4 = 6. 5 miles per minute, then 6 miles per minute; the second interval is greater.
Problem 5 Average rate of change from a graph
The points are (1, 1) and (9, 3). Rate = (3 − 1)/(9 − 1) = 2/8. 1/4.
Problem 6 Complete the square
Half of 6 is 3, so x2 + 6x + 9 − 9 + 5 = (x + 3)2 − 4. (x + 3)2 − 4; vertex (−3, −4).
Problem 7 Zeros, axis and minimum
(x − 2)(x − 5), so zeros 2 and 5. The axis is midway: x = 3.5. f(3.5) = 12.25 − 24.5 + 10 = −2.25. Zeros 2 and 5; axis x = 3.5; minimum −2.25.
Problem 8 Maximum height
−16(t2 − 3t) + 6 = −16(t2 − 3t + 2.25) + 36 + 6 = −16(t − 1.5)2 + 42. Maximum 42 ft at t = 1.5 s.
Problem 9 Revenue model
Factor: −6p(p − 28), zeros 0 and 28. The vertex is at p = 14, and R(14) = −1176 + 2352 = 1176. Revenue is nonnegative for 0 ≤ p ≤ 28. Price $14; maximum $1,176; domain 0 ≤ p ≤ 28.
Problem 10 Exponential decay features
f(0) = 250. The factor 0.8 means a 20% decrease. The asymptote is y = 0. f(3) = 250(0.512) = 128, so the rate is (128 − 250)/3 = -40.67. 250; 20% decrease; y = 0; ≈ −40.67.
Problem 11 Logarithmic features
The argument must be positive: x + 3 > 0. The asymptote is where it equals 0. For the intercept, x + 3 = 20 = 1. Domain x > −3; asymptote x = −3; intercept (−2, 0).
Problem 12 Tide features
Period = 2π ÷ (π/6) = 12 hours. Maximum 14 ft and minimum 6 ft. d(0) = 10 and d(3) = 14, so the rate is 4/3. 12 h; 14 and 6 ft; ≈ 1.33 ft/h.
Problem 13 Average cost with an asymptote
Items are counted, so x > 0. Since C(x) = 500/x + 20, the output approaches 20 for large x: y = 20 is the horizontal asymptote, the per-item material cost once the fixed cost of 500 is spread thin. C(25) = (500 + 500)/25 = 40. x > 0; y = 20; C(25) = $40.
Problem 14 Features of a cubic
Zeros: 0, 2 and 4. Testing sign: p(1) = 3 > 0, p(3) = −3 < 0, p(5) = 15 > 0, p(−1) = −15 < 0. Zeros 0, 2, 4; positive on (0, 2) and (4, ∞); as x → ∞, p → ∞ and as x → −∞, p → −∞.
Problem 15 Compare a line and a table
f(0) = 8 and g(0) = 5. The rate of f is 3; for g it is (9 − 5)/1 = 4 every step. At x = 10: f = 38 and g = 5 + 40 = 45. f starts higher, but g grows faster and is larger at x = 10 (45 versus 38).
Problem 16 Compare two quadratics
f has maximum 7 at x = 2. For g, the equal outputs 4 at x = 2 and x = 3 put the axis midway at x = 2.5, so its maximum is a little above 4 (it is 4.25). f has the greater maximum (7 versus about 4.25); g has its axis farther right (2.5 versus 2).
Problem 17 Population change from a table
2010–2015: (13,750 − 12,400)/5 = 1350/5 = 270 people per year. 2015–2020: (15,900 − 13,750)/5 = 2150/5 = 430 people per year. 270 then 430 people per year; growth is speeding up.
Problem 18 Completing the square with a leading coefficient
3(x2 − 6x + 9) − 27 + 10 = 3(x − 3)2 − 17. Minimum −17 at x = 3. Zeros: (x − 3)2 = 17/3, so x = 3 ± 2.3805. 3(x − 3)2 − 17; min −17; zeros ≈ 0.62 and 5.38.
Problem 19 Exponential versus linear savings
At 10: A ≈ $1,628.89, B = $2,000.00. At 20: A ≈ $2,653.30, B = $2,800.00. At 30: A ≈ $4,321.94, B = $3,600.00. B is ahead at 10 and 20; A is ahead at 30, so the crossing is between 20 and 30, near t ≈ 23. A overtakes B after about 23 years.
Problem 20 Average rate of change of sine
From 0 to π/2: (1 − 0)/(π/2) = 2/π ≈ 0.6366. From π/2 to π: (0 − 1)/(π/2) = −2/π ≈ -0.6366. ≈ 0.64 and ≈ −0.64: sine is increasing on the first interval and decreasing on the second.