ISASP High School Functions. Practice it free.

Students understand and use functions, function notation, and functional relationships. This High School reporting domain maps to 30 practice skills and 8 representative questions from the playable bank.

High School30 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

  6. Interpret expressions for functions in terms of the situation they model

Standards basis

Iowa Academic Standards / Iowa Core Mathematics — Iowa

How the ISASP reports it

ISASP is Iowa’s statewide assessment program administered by the state Department of Education and aligned to Iowa Academic Standards / Iowa Core rather than a multi-state blueprint like SBAC or ACT. The public materials accessible here establish the tested grades and that mathematics results are reported statewide, but I could not locate an official public math blueprint PDF on the Iowa DOE site in the available sources.

Blueprint & weighting

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

Try it now

8 free questions · 0/0 correct

Practice playmedium

The first five terms of an arithmetic sequence are 8,14,20,26,328, 14, 20, 26, 32. What is the 1818th term?

The common difference is d=6d = 6. With a1=8a_1 = 8, the 1818th term is a18=8+17(6)=110a_{18} = 8 + 17(6) = 110.

Practice playeasy

The function ff is defined by f(x)=7x−4f(x) = 7x - 4. What is the value of f(x)f(x) when x=3x = 3?

Substitute x=3x = 3: f(3)=7(3)−4=21−4=17f(3) = 7(3) - 4 = 21 - 4 = 17.

Practice playeasy

The function ff is defined by f(x)=2x+1f(x) = \sqrt{2x + 1}. What is the value of f(x)f(x) when x=4x = 4?

Substitute x=4x = 4: f(4)=2(4)+1=9=3f(4) = \sqrt{2(4) + 1} = \sqrt{9} = 3.

Practice playeasy

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.

Practice playeasy

g=9−m4g = 9 - \dfrac{m}{4}

The equation shown gives the estimated amount of paint
g,g\textsf{,} in gallons, remaining after covering mm square meters of wall, where 0≤m≤36.0 \le m \le 36\textsf{.} What is the estimated amount of paint, in gallons, remaining after covering 1212 square meters of wall?

Substitute m=12:m = 12\textsf{:} g=9−124=9−3=6.g = 9 - \dfrac{12}{4} = 9 - 3 = 6\textsf{.} The estimated amount remaining is 66 gallons.

Practice playeasy

What is the value of log⁡416?\log_{4}16\textsf{?}

42=16,4^{2} = 16\textsf{,} so log⁡416=2.\log_{4}16 = 2\textsf{.}

Practice playmedium

A charity drive starts with 8080 donors. Each day after the first day, a model estimates that the number of donors increases by 25%25\% of the number of donors the previous day. Which equation defines this model, where d(n)d(n) is the estimated number of donors nn days after the drive begins?

The starting number is 8080, and a 25%25\% daily increase means multiplying by 1.251.25 each day, so d(n)=80(1.25)nd(n) = 80(1.25)^n.

Practice playeasy

A 3232-milligram sample of a radioactive isotope has a half-life of 11 year.

Choose the function model that correctly calculates the amount remaining after
44 years.

The remaining amount after tt years is A(t)=32(12)th,A(t) = 32\left(\dfrac{1}{2}\right)^{\textstyle\frac{t}{h}}\textsf{,} where hh is the half-life. Here t=4t = 4 and h=1,h = 1\textsf{,} so A(4)=32(12)41=32(12)4=2.A(4) = 32\left(\dfrac{1}{2}\right)^{\textstyle\frac{4}{1}} = 32\left(\dfrac{1}{2}\right)^{4} = 2\textsf{.}

Keep practicing

Turn functions into game time.

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