← Module 1 Visual Lab Algebra II Syllabus
Mathematical Architects · Algebra II · Foundations

Get Ready: Module 1 — Extending Linear Relationships

Everyone starts somewhere. Before we raise the building, we'll pour the floor. This page rebuilds the handful of skills Module 1 leans on — clearly, one step at a time, with nothing assumed. Go at your own pace. There's no rush and no wrong place to begin.

5 Foundation Skills Worked Examples Self-Paced

Module 1 takes the straight lines you already know and gives them a bend (absolute value) and a job (systems of inequalities). If lines, slopes, or \(f(x)\) feel fuzzy right now, that's completely okay — we'll firm them up together below, then you'll be ready for the lab.

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Skills to build first

Five short skills. Read the idea, follow the worked example, then try one yourself — the answer is one click away whenever you want it.

1 Skill

Graphing & writing linear equations (slope–intercept)

What it is: a line written as \(y = mx + b\), where \(m\) is the slope (how steep, and which way it tilts) and \(b\) is the y-intercept (where the line crosses the y-axis). Why you need it: the whole module is "lines, extended." The V-shape of an absolute value is built from two lines, and every inequality fence is just a line you've shaded one side of. If you can graph and write a line, you can do this module.

Worked example — graph \(y = 2x - 1\)
  1. Start at the y-intercept. Here \(b = -1\), so plot the point \((0, -1)\).
  2. Use the slope to step. Slope \(m = 2 = \tfrac{2}{1}\) means "rise \(2\), run \(1\)": from \((0,-1)\) go up \(2\) and right \(1\) to \((1, 1)\).
  3. Repeat once more for a third point: up \(2\), right \(1\) lands on \((2, 3)\).
  4. Connect the points with a straight line. Done — that's \(y = 2x - 1\).

To write a line instead: read the slope (rise over run) and the y-intercept off the graph, then drop them into \(y = mx + b\).

Try it 1 — Graph (in your head or on paper): \(y = -\tfrac{1}{2}x + 3\). What is the slope and the y-intercept?
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Slope \(m = -\tfrac{1}{2}\) (down \(1\), right \(2\)); y-intercept \(b = 3\), so the line crosses the y-axis at \((0, 3)\).

From \((0,3)\): go right \(2\), down \(1\) to \((2, 2)\); again to \((4, 1)\). Connect — a gentle line falling to the right.

Try it 2 — Write the equation of a line with slope \(3\) that passes through \((0, -4)\).
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The point \((0,-4)\) is on the y-axis, so it is the y-intercept: \(b = -4\). With \(m = 3\), the equation is \(\;y = 3x - 4\).

Practice more (free) Khan Academy → IXL →
2 Skill

Solving linear equations & inequalities

What it is: getting the variable by itself by doing the same thing to both sides (add, subtract, multiply, divide). An inequality works the same way, with one extra rule. Why you need it: you'll solve for where lines cross (the corners of a feasible region) and work with inequalities to describe the shaded regions in the systems part of the module.

Worked example — solve \(3x - 5 = 7\)
  1. Undo the −5 by adding \(5\) to both sides: \(3x = 12\).
  2. Undo the ×3 by dividing both sides by \(3\): \(x = 4\).
  3. Check: \(3(4) - 5 = 12 - 5 = 7\). ✓

The one inequality rule: when you multiply or divide both sides by a negative number, flip the inequality sign. Example: \(-2x < 6 \;\Rightarrow\; x > -3\) (sign flipped).

Try it 1 — Solve \(4x + 3 = 19\).
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Subtract \(3\): \(4x = 16\). Divide by \(4\): \(\;x = 4\). Check: \(4(4)+3 = 19\). ✓

Try it 2 — Solve the inequality \(-3x + 1 \ge 10\).
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Subtract \(1\): \(-3x \ge 9\). Divide by \(-3\) and flip the sign: \(\;x \le -3\).

Practice more (free) Khan Academy → IXL →
3 Skill

Absolute value as distance from 0

What it is: \(|x|\) means "how far \(x\) is from \(0\) on the number line" — and distance is never negative. So \(|5| = 5\) and \(|-5| = 5\): both are \(5\) steps from zero. Why you need it: the headline function of this module is \(y = a|x - h| + k\). Once you see absolute value as distance, the V-shape (and why it bounces back up) stops being a mystery.

Worked example — evaluate absolute values
  1. \(|7| = 7\) — seven is \(7\) units from \(0\).
  2. \(|-4| = 4\) — negative four is also \(4\) units from \(0\); the sign drops away.
  3. \(|3 - 10| = |-7| = 7\) — do the inside first, then take the distance.
  4. \(-|6| = -6\) — the bars give \(6\), and the minus in front stays. (The minus outside is not protected by the bars.)
Try it 1 — Find \(|-12|\) and \(|0|\).
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\(|-12| = 12\) (twelve units from zero). \(|0| = 0\) (zero is no distance from itself).

Try it 2 — Simplify \(|2 - 9|\) and then \(-|2 - 9|\).
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Inside first: \(2 - 9 = -7\), so \(|2-9| = |-7| = 7\). Then the outside minus gives \(-|2-9| = -7\).

Practice more (free) Khan Academy → IXL →
4 Skill

The coordinate plane

What it is: the grid where every point has an address \((x, y)\) — \(x\) tells you left/right, \(y\) tells you up/down. They meet at the origin \((0,0)\). Why you need it: everything in this module is drawn on this plane. The vertex of a V, the corners of a region, where two lines cross — they're all points you'll read and plot.

Worked example — plot \((3, -2)\)
  1. Start at the origin \((0,0)\), the center of the grid.
  2. Read \(x\) first (always): \(x = 3\) means move right \(3\) units.
  3. Then read \(y\): \(y = -2\) means move down \(2\) units.
  4. Mark the point. You've landed in the lower-right area (Quadrant IV).

Quick map: right + up = Quadrant I, left + up = II, left + down = III, right + down = IV.

Try it 1 — In which quadrant is the point \((-4, 5)\)?
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\(x = -4\) (left) and \(y = 5\) (up) → upper-left → Quadrant II.

Try it 2 — Give the coordinates of a point that is \(2\) left and \(3\) down from the origin.
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Left \(2\) makes \(x = -2\); down \(3\) makes \(y = -3\). The point is \((-2, -3)\) (Quadrant III).

Practice more (free) Khan Academy → IXL →
5 Skill

Function notation \(f(x)\)

What it is: \(f(x)\) is just a fancier name for \(y\). It says "the function named \(f\), fed an input \(x\)." So \(f(3)\) means "put \(3\) in for \(x\) and see what comes out." Why you need it: Algebra II talks about whole families of functions, and \(f(x)\) notation is how we name them and ask "what's the output at this input?" — including for absolute-value functions.

Worked example — if \(f(x) = 2x + 1\), find \(f(3)\)
  1. The \(3\) replaces every \(x\): \(f(3) = 2(3) + 1\).
  2. Multiply, then add: \(2(3) = 6\), then \(6 + 1 = 7\).
  3. So \(f(3) = 7\). In plain terms: input \(3\), output \(7\) — the point \((3, 7)\) on the graph.
Try it 1 — If \(f(x) = -3x + 4\), find \(f(2)\).
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\(f(2) = -3(2) + 4 = -6 + 4 = -2\). So \(f(2) = -2\), i.e. the point \((2, -2)\).

Try it 2 — If \(g(x) = |x| + 1\), find \(g(-5)\). (Hint: this is exactly the kind of function you'll meet in the lab.)
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\(g(-5) = |-5| + 1\). The bars give \(|-5| = 5\), so \(g(-5) = 5 + 1 = 6\). Point \((-5, 6)\).

Practice more (free) Khan Academy → IXL →
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Quick Readiness Check

Six short questions spanning all five skills. Try each one first, then reveal the answer to check yourself.

  1. What are the slope and y-intercept of \(y = -4x + 7\)?
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    Slope \(m = -4\); y-intercept \(b = 7\) (crosses at \((0,7)\)).

  2. Solve \(2x + 6 = 14\).
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    Subtract \(6\): \(2x = 8\). Divide by \(2\): \(x = 4\).

  3. Solve the inequality \(-5x > 20\).
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    Divide by \(-5\) and flip the sign: \(x < -4\).

  4. Evaluate \(|4 - 11|\).
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    Inside first: \(4 - 11 = -7\), so \(|-7| = 7\).

  5. In which quadrant is the point \((5, -3)\)?
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    Right and down → Quadrant IV.

  6. If \(f(x) = |x - 2| + 3\), find \(f(2)\).
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    \(f(2) = |2-2| + 3 = |0| + 3 = 0 + 3 = 3\). (Notice: that's the vertex of the V!)

If these feel comfortable, you're ready for the module. If one or two felt shaky, that's your cue — scroll back up to that skill, re-read the worked example, and try a couple from the free practice links. No score to hit, no clock running. Come back when it clicks.

Floor poured. When the skills above feel steady, step into the module — the Visual Lab lets you see every one of these ideas move at once.

Module 1 Overview & Visual Lab Jump to the Visual Lab Course Syllabus