Grade 9 · Algebra & functions

Quadratic function word problems (ACT) practice

Quadratic function word problems (ACT) is a grade 9 math skill aligned to Common Core standard HSF.IF.A.2: use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 quadratic function word problems (act) problems our math games drill.

CCSS HSF.IF.A.210 questions in the bank
Sample questions

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Question 1easy

A football is kicked straight up from a stadium deck. Its height dd, in feet, above the playing field tt seconds after the kick is modeled by d(t)=16t2+64t+48d(t) = -16t^2 + 64t + 48. What is the height of the football, in feet, 33 seconds after it is kicked?

Substitute t=3t = 3: d(3)=16(32)+64(3)+48=144+192+48=96d(3) = -16(3^2) + 64(3) + 48 = -144 + 192 + 48 = 96 feet.

Question 2easy

A ball is thrown straight upward from a platform. Its height hh, in feet, above the ground tt seconds after it is thrown is modeled by h(t)=16t2+32t+20h(t) = -16t^2 + 32t + 20. According to the model, what is the initial height of the ball above the ground, in feet?

The initial height is h(0)h(0). Substituting t=0t = 0 gives h(0)=16(0)2+32(0)+20=20h(0) = -16(0)^2 + 32(0) + 20 = 20 feet.

Question 3easy

A warehouse tracks inventory with I(t)=2t2+6t+85I(t) = -2t^2 + 6t + 85, where I(t)I(t) is the number of units on hand tt weeks after a shipment arrives. According to the model, how many units were on hand when the shipment arrived?

When the shipment arrived, t=0t = 0. So I(0)=2(0)2+6(0)+85=85I(0) = -2(0)^2 + 6(0) + 85 = 85 units.

Question 4easy

A decorative fountain shoots water straight up. The water's height ff, in feet, above the nozzle tt seconds after leaving the nozzle is modeled by f(t)=2t2+12t+20f(t) = -2t^2 + 12t + 20. What is the height of the water, in feet, 33 seconds after it leaves the nozzle?

Substitute t=3t = 3: f(3)=2(32)+12(3)+20=18+36+20=38f(3) = -2(3^2) + 12(3) + 20 = -18 + 36 + 20 = 38 feet.

Question 5medium

A theater's daily ticket revenue RR, in dollars, when each ticket costs pp dollars is modeled by R(p)=3p2+90pR(p) = -3p^2 + 90p. According to the model, what is the daily revenue, in dollars, when each ticket costs 1010?

Substitute p=10p = 10: R(10)=3(102)+90(10)=300+900=600R(10) = -3(10^2) + 90(10) = -300 + 900 = 600 dollars.

Question 6medium

A town's population PP, in thousands, tt years after a census is modeled by P(t)=t2+14t+120P(t) = -t^2 + 14t + 120. According to the model, what is the population, in thousands, 66 years after the census?

Substitute t=6t = 6: P(6)=(62)+14(6)+120=36+84+120=168P(6) = -(6^2) + 14(6) + 120 = -36 + 84 + 120 = 168 thousand.

Question 7medium

A parabolic bridge arch has height hh, in meters, at a horizontal distance xx meters from the left pillar modeled by h(x)=x2+10x+15h(x) = -x^2 + 10x + 15. According to the model, what is the height, in meters, of the arch 88 meters from the left pillar?

Substitute x=8x = 8: h(8)=(82)+10(8)+15=64+80+15=31h(8) = -(8^2) + 10(8) + 15 = -64 + 80 + 15 = 31 meters.

Question 8medium

A company's daily profit PP, in dollars, when nn hundred items are sold is modeled by P(n)=n2+18n+36P(n) = -n^2 + 18n + 36. According to the model, what is the daily profit, in dollars, when 44 hundred items are sold?

Substitute n=4n = 4: P(4)=(42)+18(4)+36=16+72+36=92P(4) = -(4^2) + 18(4) + 36 = -16 + 72 + 36 = 92 dollars.

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