Mathematical Architects · Algebra I · Foundations

Get Ready: Module 5 — Maximizing & Minimizing

Everyone starts somewhere. Before we work with parabolas, let’s build the floor under the building — one small skill at a time, at your own pace.

Foundations · Build These First
Module 5 asks you to find the highest or lowest point of a curve like \(y = ax^2 + bx + c\). That sounds big, but it’s really just six familiar moves stacked together. This page warms up each one with a plain explanation, a fully worked example, and a couple of practice problems you can check yourself.

No rush, no shame. If a skill already feels easy, skim it and move on. If one feels shaky, slow down and do the “Try it” problems — the answers are right there when you’re ready to peek. You don’t need to be fast. You just need to be willing.
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Skills to Build First

Six prerequisite skills. Work top to bottom — each one shows up inside Module 5.

1

Distributing & Multiplying Binomials (FOIL)

What it is & why you need it To “multiply out” means turning something like \((x+3)(x+2)\) into \(x^2 + 5x + 6\). In Module 5 you’ll go backward and forward between a factored parabola and its standard form, so you need to be comfortable multiplying two binomials. FOIL just means multiply the First, Outer, Inner, and Last pairs, then add them up.

Worked example — multiply \((x+3)(x+2)\)
  1. First: \(x \cdot x = x^2\)
  2. Outer: \(x \cdot 2 = 2x\)
  3. Inner: \(3 \cdot x = 3x\)
  4. Last: \(3 \cdot 2 = 6\)
  5. Add them: \(x^2 + 2x + 3x + 6\), then combine the middle: \(2x + 3x = 5x\).

Answer: \((x+3)(x+2) = x^2 + 5x + 6\)

Try it 1. Multiply \((x+4)(x+5)\).

Show answer
F: \(x^2\); O: \(5x\); I: \(4x\); L: \(20\). Combine the middle: \(5x + 4x = 9x\).
Answer: \(x^2 + 9x + 20\).

Try it 2. Multiply \((x-2)(x+6)\). (Watch the negative sign!)

Show answer
F: \(x^2\); O: \(+6x\); I: \(-2x\); L: \(-12\). Combine the middle: \(6x - 2x = 4x\).
Answer: \(x^2 + 4x - 12\).
2

Factoring Out a GCF

What it is & why you need it The GCF (greatest common factor) is the biggest thing every term shares. Factoring it out is like un-distributing: you pull the common piece to the front and put what’s left in parentheses. In Module 5 this is usually your first move when factoring a quadratic — pulling out a common number or \(x\) makes everything after it simpler.

Worked example — factor \(6x^2 + 9x\)
  1. Numbers: the GCF of \(6\) and \(9\) is \(3\).
  2. Variables: both terms have at least one \(x\), so pull out \(x\). Together the GCF is \(3x\).
  3. Divide each term by \(3x\): \(\;6x^2 \div 3x = 2x\;\) and \(\;9x \div 3x = 3\).
  4. Write the GCF outside, the leftovers inside.

Answer: \(6x^2 + 9x = 3x(2x + 3)\). Check by distributing back: \(3x\cdot 2x = 6x^2\), \(3x\cdot 3 = 9x\). ✓

Try it 1. Factor out the GCF of \(10x^2 + 15x\).

Show answer
GCF of \(10\) and \(15\) is \(5\); both have an \(x\), so GCF \(= 5x\).
Answer: \(5x(2x + 3)\).

Try it 2. Factor out the GCF of \(4x^2 - 8x + 12\).

Show answer
GCF of \(4, 8, 12\) is \(4\). Not every term has an \(x\) (the \(12\) doesn’t), so we only pull out \(4\).
Answer: \(4(x^2 - 2x + 3)\).
3

Integer Operations (Positives & Negatives)

What it is & why you need it Quadratics are full of negative numbers — coefficients, roots, the value of \(-\tfrac{b}{2a}\), and more. The two rules that save you: same signs → the answer is positive, and different signs → the answer is negative (for multiplying and dividing). For adding, think of a number line: subtracting is just adding the opposite.

Worked example — evaluate \(-3 \cdot (-4)\) and \(-7 + 2\)
  1. \(-3 \cdot (-4)\): the signs are the same (both negative), so the answer is positive: \(12\).
  2. \(-7 + 2\): start at \(-7\), move \(2\) steps in the positive direction → you land on \(-5\).

Answers: \(-3 \cdot (-4) = 12\) and \(-7 + 2 = -5\).

Try it 1. Compute \(-5 \cdot 6\) and \(8 + (-3)\).

Show answer
\(-5 \cdot 6\): different signs → negative → \(-30\).
\(8 + (-3)\): start at \(8\), move \(3\) back → \(5\).

Try it 2. Compute \(-4 - (-9)\).

Show answer
Subtracting a negative is adding its opposite: \(-4 - (-9) = -4 + 9\). Start at \(-4\), move \(9\) forward.
Answer: \(5\).
4

Exponents (Especially Squaring)

What it is & why you need it The little raised \(2\) in \(x^2\) means “multiply the base by itself”: \(x^2 = x \cdot x\). The whole module lives on the \(x^2\) term, and computing the discriminant \(b^2 - 4ac\) means squaring \(b\). The trap to remember: a negative number squared is positive, because two negatives multiply to a positive.

Worked example — evaluate \((-5)^2\) and \(3^2 - 4 \cdot 1 \cdot 2\)
  1. \((-5)^2 = (-5)\cdot(-5)\). Same signs → positive → \(25\).
  2. \(3^2 = 9\). Then \(4 \cdot 1 \cdot 2 = 8\). So \(3^2 - 4\cdot 1\cdot 2 = 9 - 8 = 1\).

Answers: \((-5)^2 = 25\) and \(3^2 - 4\cdot1\cdot2 = 1\). (That second one is exactly a discriminant!)

Try it 1. Evaluate \((-6)^2\).

Show answer
\((-6)\cdot(-6)\), same signs → positive. Answer: \(36\).

Try it 2. Evaluate the discriminant \(b^2 - 4ac\) when \(a = 1,\ b = 4,\ c = 3\).

Show answer
\(b^2 = 4^2 = 16\). \(4ac = 4\cdot 1\cdot 3 = 12\). So \(16 - 12 = \mathbf{4}\).
(Positive → this parabola would have two real roots.)
5

Solving Linear Equations

What it is & why you need it Solving means getting \(x\) alone by doing the same thing to both sides until it’s by itself. In Module 5 you’ll set each factor equal to zero (like \(x - 3 = 0\)) to find the roots, and you’ll rearrange formulas. If you can undo addition with subtraction and undo multiplication with division, you’re set.

Worked example — solve \(2x + 6 = 0\)
  1. Undo the \(+6\) by subtracting \(6\) from both sides: \(2x = -6\).
  2. Undo the \(\times 2\) by dividing both sides by \(2\): \(x = -3\).
  3. Check: \(2(-3) + 6 = -6 + 6 = 0\). ✓

Answer: \(x = -3\).

Try it 1. Solve \(x - 5 = 0\).

Show answer
Add \(5\) to both sides. Answer: \(x = 5\). (This is exactly how you read a root off the factor \((x-5)\).)

Try it 2. Solve \(3x - 12 = 0\).

Show answer
Add \(12\): \(3x = 12\). Divide by \(3\): \(x = 4\).
6

Plotting on the Coordinate Plane

What it is & why you need it A point is written \((x, y)\): the first number is how far left/right, the second is how far up/down. Always go across first, then up or down. Module 5 is a graphing module — you’ll plot the vertex, the y-intercept, and the roots, then connect them into the U-shaped parabola.

Worked example — plot \((3, -2)\)
  1. Start at the origin \((0,0)\) — the center where the axes cross.
  2. The \(x\)-value is \(3\): move 3 units right.
  3. The \(y\)-value is \(-2\): move 2 units down.
  4. Mark the point. You’re in the lower-right region (Quadrant IV).

Answer: \((3, -2)\) sits 3 right and 2 down from the origin.

Try it 1. Where does \((-2, 5)\) land — left or right? up or down?

Show answer
\(x = -2\) → 2 units left. \(y = 5\) → 5 units up. (Upper-left, Quadrant II.)

Try it 2. A parabola’s vertex is \((1, -4)\). From the origin, which way do you move to plot it?

Show answer
1 unit right, then 4 units down. That’s the lowest point of the curve when it opens up.
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Quick Readiness Check

Five short questions spanning the prerequisites. Try them on paper, then peek.

CHECK 1 · FOIL

Multiply \((x+1)(x+7)\).

Show answer
\(x^2 + 8x + 7\).
CHECK 2 · GCF

Factor out the GCF of \(8x^2 + 12x\).

Show answer
\(4x(2x + 3)\).
CHECK 3 · INTEGERS

Compute \(-6 \cdot (-2)\) and \(-9 + 4\).

Show answer
\(12\) and \(-5\).
CHECK 4 · EXPONENTS

Evaluate \(b^2 - 4ac\) for \(a=1,\ b=5,\ c=6\).

Show answer
\(25 - 24 = \) \(1\).
CHECK 5 · SOLVING

Solve \(x + 2 = 0\) and \(2x - 8 = 0\).

Show answer
\(x = -2\) and \(x = 4\).
CHECK 6 · PLOTTING

From the origin, how do you reach the point \((-3, 2)\)?

Show answer
3 units left, then 2 units up.

If these feel comfortable, you’re ready for the module. If a couple tripped you up, scroll back to that skill card — that’s exactly where to spend a few more minutes.

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Practice More (Free)

Want to drill a skill until it’s automatic? These free sites open in a new tab.

1 · Distributing & FOIL Khan Academy ↗ IXL ↗
2 · Factoring out a GCF Khan Academy ↗ IXL ↗
3 · Integer operations Khan Academy ↗ IXL ↗
4 · Exponents & squaring Khan Academy ↗ IXL ↗
5 · Solving linear equations Khan Academy ↗ IXL ↗
6 · Plotting on the coordinate plane Khan Academy ↗ IXL ↗
When You’re Ready

You’ve built the floor. Time to build up.

Head into the module whenever you feel steady. You can always come back here.

Module 5 Overview → Open the Visual Lab Read the Syllabus