ATLAS Algebra I Functions. Practice it free.

Interprets, compares, and models functions using multiple representations. This Algebra I reporting domain maps to 23 practice skills and 8 representative questions from the playable bank.

Algebra I23 mapped skills
What the test measures

Functions skills

  1. Understand the concept of a function and use function notation

  2. Interpret functions that arise in applications in terms of the context

  3. Analyze functions using different representations

  4. Build a function that models a relationship between two quantities

  5. Construct and compare linear, quadratic, and exponential models and solve problems

Standards basis

Arkansas Academic Standards for Mathematics (CCSS-M–aligned / Arkansas-adopted math standards) — Arkansas

How the ATLAS reports it

Arkansas’s statewide assessment system is ATLAS, which the ADE page says began serving in spring 2024. The official public page does not publish a math blueprint there, so the most defensible coverage map is the Arkansas math standards structure that ATLAS is aligned to.

Blueprint & weighting

ATLAS uses the Independent / state-specific framework (by grade domains structure).

Try it now

8 free questions · 0/0 correct

Practice playeasy

The function ff is defined by f(x)=−3x+22f(x) = -3x + 22. What is the value of f(x)f(x) when x=5x = 5?

Substitute x=5x = 5: f(5)=−3(5)+22=−15+22=7f(5) = -3(5) + 22 = -15 + 22 = 7.

Practice playeasy

The function ff is defined by f(x)=x2+3f(x) = x^2 + 3. What is the value of f(x)f(x) when x=4x = 4?

Substitute x=4x = 4: f(4)=42+3=16+3=19f(4) = 4^2 + 3 = 16 + 3 = 19.

Practice playeasy

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.

Practice playeasy

s=18−n2s = 18 - \dfrac{n}{2}

The equation shown gives the estimated snow depth
s,s\textsf{,} in inches, remaining on a trail nn days after a storm, where 0≤n≤24.0 \le n \le 24\textsf{.} What is the estimated snow depth, in inches, remaining on the trail 1010 days after the storm?

Substitute n=10:n = 10\textsf{:} s=18−102=18−5=13.s = 18 - \dfrac{10}{2} = 18 - 5 = 13\textsf{.} The estimated snow depth remaining is 1313 inches.

Practice playeasy

What is the value of log⁡39?\log_{3}9\textsf{?}

32=9,3^{2} = 9\textsf{,} so log⁡39=2.\log_{3}9 = 2\textsf{.}

Practice playeasy

An account starts with $256\text{\char36}256 and earns 25%25\% interest compounded yearly.

Choose the function model that correctly calculates the amount in the account after
33 years.

The amount after tt years is A(t)=P(1+r)t.A(t) = P(1 + r)^{t}\textsf{.} Here P=256,P = 256\textsf{,} r=0.25,r = 0.25\textsf{,} and t=3,t = 3\textsf{,} so A(3)=256(1.25)3=500.A(3) = 256(1.25)^{3} = 500\textsf{.}

Practice playeasy

The function SS is defined by S(t)=450(0.85)t/3S(t) = 450(0.85)^{t/3}. The function SS models the height, in centimeters, of a snowpack, where tt is the number of days after a winter storm. According to the model, what is the estimated height, in centimeters, of the snowpack 33 days after the storm?

When t=3t = 3, the exponent is 3/3=13/3 = 1. So S(3)=450(0.85)1=382.5S(3) = 450(0.85)^1 = 382.5 centimeters.

Practice playeasy

A substance has a half-life of 11 years. What fraction remains after 22 years?

2÷1=22 \div 1 = 2 half-lives. Remaining: (12)2=14\left(\dfrac{1}{2}\right)^{2} = \dfrac{1}{4}.

Keep practicing

Turn functions into game time.

The ATLAS placement starts with this test's real coverage map and finds the right difficulty.