AK STAR Grade 9 Functions. Practice it free.

Measures function concepts, graphing, and building functions in high school. This Grade 9 reporting domain maps to 35 practice skills and 8 representative questions from the playable bank.

Grade 9Functions35 mapped skills
What the test measures

Functions skills

  1. F-IF.1–F-IF.9

  2. F-BF.1–F-BF.4

  3. F-LE.1–F-LE.3, F-LE.5

Standards basis

Alaska Mathematics Standards — Alaska

How the AK STAR reports it

AK STAR is Alaska’s state summative assessment, built around Alaska’s own mathematics standards rather than a national consortium framework. Its math blueprint is organized into grade-level content strands/instructional areas and an adaptive growth component aligned to MAP Growth instructional areas.

Blueprint & weighting

The blueprint gives target item counts and point ranges by grade and strand, rather than fixed percentage weights. Grades 3-5 and 6-9 each include summative item counts plus additional growth items; grade 9 has no summative Geometry strand, with Geometry reported only through growth items.

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