Grade 9 · Algebra & functions

Evaluating functions from percent change practice

Evaluating functions from percent change 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 evaluating functions from percent change problems our math games drill.

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

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Question 1easy

For the function gg, g(0)=100g(0) = 100. For each increase in xx by 11, the value of g(x)g(x) increases by 50%50\%. What is the value of g(2)g(2)?

Each step multiplies by 1+0.50=1.51 + 0.50 = 1.5. So g(2)=100⋅1.52=100⋅2.25=225g(2) = 100 \cdot 1.5^2 = 100 \cdot 2.25 = 225.

Question 2easy

For the function hh, h(0)=64h(0) = 64. For each increase in xx by 11, the value of h(x)h(x) decreases by 25%25\%. What is the value of h(2)h(2)?

Each step multiplies by 1−0.25=0.751 - 0.25 = 0.75. So h(2)=64⋅0.752=64⋅0.5625=36h(2) = 64 \cdot 0.75^2 = 64 \cdot 0.5625 = 36.

Question 3easy

For the function ff, f(0)=125f(0) = 125. For each increase in xx by 11, the value of f(x)f(x) increases by 20%20\%. What is the value of f(2)f(2)?

Each step multiplies by 1.201.20. So f(2)=125⋅1.202=125⋅1.44=180f(2) = 125 \cdot 1.20^2 = 125 \cdot 1.44 = 180.

Question 4easy

For the function pp, p(0)=200p(0) = 200. For each increase in xx by 11, the value of p(x)p(x) decreases by 10%10\%. What is the value of p(2)p(2)?

Each step multiplies by 0.900.90. So p(2)=200⋅0.902=200⋅0.81=162p(2) = 200 \cdot 0.90^2 = 200 \cdot 0.81 = 162.

Question 5easy

For the function ff, f(0)=160f(0) = 160. For each increase in xx by 11, the value of f(x)f(x) decreases by 50%50\%. What is the value of f(3)f(3)?

A decrease of 50%50\% multiplies by 0.500.50 each step. So f(3)=160⋅0.503=160⋅0.125=20f(3) = 160 \cdot 0.50^3 = 160 \cdot 0.125 = 20.

Question 6easy

For the function rr, r(0)=27r(0) = 27. For each increase in xx by 11, the value of r(x)r(x) increases by 100%100\%. What is the value of r(2)r(2)?

Each step multiplies by 22. So r(2)=27⋅22=27⋅4=108r(2) = 27 \cdot 2^2 = 27 \cdot 4 = 108.

Question 7easy

For the function ff, f(0)=250f(0) = 250. For each increase in xx by 11, the value of f(x)f(x) decreases by 40%40\%. What is the value of f(2)f(2)?

Each step multiplies by 1−0.40=0.601 - 0.40 = 0.60. So f(2)=250⋅0.602=250⋅0.36=90f(2) = 250 \cdot 0.60^2 = 250 \cdot 0.36 = 90.

Question 8medium

For the function qq, q(0)=500q(0) = 500. For each increase in xx by 11, the value of q(x)q(x) decreases by 20%20\%. What is the value of q(3)q(3)?

Each step multiplies by 0.800.80. So q(3)=500⋅0.803=500⋅0.512=256q(3) = 500 \cdot 0.80^3 = 500 \cdot 0.512 = 256.

Tests that cover evaluating functions from percent change:
SBACSATACTNC EOG/EOCNY State Math
More Grade 9 skills:
Exponential functions in context (SAT)Evaluating functions at a valueFactoring patternsEquivalent rational expressions (SAT)Factoring quadraticsDiscriminant analysisInconsistent linear systemsStatistics basicsAll Grade 9 practice →

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