← Module 2 Visual Lab Pre-Calculus
Mathematical Architects · Pre-Calculus · Foundations

Get Ready: Module 2 — Analyzing Function Families

Everyone starts somewhere. Before we analyze whole families of curves, we'll firm up the four skills they lean 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.

4 Foundation Skills Worked Examples Self-Paced

Module 2 takes the parent curves you've met and transforms them — shifting, stretching, and reflecting whole families at once. If the shapes of \(2^x\), \(\log x\), or \(\tfrac1x\), the rules of exponents and logs, factoring, or interval notation feel fuzzy right now, that's completely okay — we'll rebuild them together below, then you'll be ready for the Studio.

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Skills to build first

Four 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

The parent functions & their shapes

What it is: the "starter" graph for each family before any transformation — the line \(y = x\), the parabola \(y = x^2\), the reciprocal \(y = \tfrac1x\), the exponential \(y = 2^x\), and the logarithm \(y = \log_2 x\). Why you need it: the whole module is "parent, then transform." If you can sketch the parent's basic shape and recall its key features (where it starts, which way it bends, any asymptote), every transformation is just that shape moved.

Worked example — the four families at a glance
  1. Quadratic \(y = x^2\): a U-shaped parabola with its vertex at the origin, opening up, symmetric about the y-axis.
  2. Exponential \(y = 2^x\): hugs the x-axis on the left (horizontal asymptote \(y = 0\)), passes through \((0,1)\), then shoots up fast on the right.
  3. Logarithm \(y = \log_2 x\): the mirror of the exponential across \(y = x\). It has a vertical asymptote at \(x = 0\), passes through \((1,0)\), and rises slowly.
  4. Rational \(y = \tfrac1x\): two branches with a vertical asymptote at \(x = 0\) and a horizontal asymptote at \(y = 0\).

Key idea: each parent has a signature feature (a vertex, or an asymptote). Transformations move that feature — they never erase it.

Try it 1 — Which parent function has a vertical asymptote at \(x = 0\) and passes through \((1, 0)\)?
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The logarithm \(y = \log_2 x\) (or \(\log\) of any base): it never reaches \(x = 0\) and crosses the x-axis at \((1,0)\), since \(\log_2 1 = 0\).

Try it 2 — The exponential \(y = 2^x\) passes through which point on the y-axis, and what is its horizontal asymptote?
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It passes through \((0, 1)\) because \(2^0 = 1\). Its horizontal asymptote is \(y = 0\) (the curve approaches the x-axis as \(x \to -\infty\) but never touches it).

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

Exponent & logarithm rules

What it is: the algebra that moves exponents and logs around — \(x^a x^b = x^{a+b}\), \(\tfrac{x^a}{x^b} = x^{a-b}\), \((x^a)^b = x^{ab}\), \(x^0 = 1\), \(x^{-a} = \tfrac{1}{x^a}\) — and the fact that a logarithm is just "the exponent": \(\log_b y = x\) means \(b^x = y\). Why you need it: exponentials and logs are two of the four families, and you'll solve log and exponential equations later in the module. Fluent rules make those one-liners instead of mysteries.

Worked example — simplify and convert
  1. Combine like bases: \(2^3 \cdot 2^4 = 2^{3+4} = 2^7 = 128\) — add the exponents.
  2. Negative exponent: \(5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}\) — the minus flips it to the denominator.
  3. Zero exponent: \(7^0 = 1\) (anything nonzero to the zero power is \(1\)).
  4. Log ↔ exponent: \(\log_2 8 = 3\) because \(2^3 = 8\). Read a log as "what power do I raise the base to?"

The log rules mirror the exponent rules: \(\log_b(MN) = \log_b M + \log_b N\) and \(\log_b(M^p) = p\,\log_b M\).

Try it 1 — Simplify \(\dfrac{3^5}{3^2}\) and evaluate \(4^{-1}\).
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\(\dfrac{3^5}{3^2} = 3^{5-2} = 3^3 = 27\) (subtract exponents). And \(4^{-1} = \dfrac{1}{4}\).

Try it 2 — Find \(\log_3 81\). (Hint: \(81\) is a power of \(3\).)
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Ask "\(3\) to what power is \(81\)?" Since \(3^4 = 81\), we get \(\log_3 81 = 4\).

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

Factoring polynomials

What it is: rewriting a polynomial as a product — pulling out a common factor, undoing FOIL on a trinomial, or recognizing a difference of squares \(a^2 - b^2 = (a-b)(a+b)\). Why you need it: factoring finds zeros (where a graph crosses the x-axis), exposes the holes vs. asymptotes of a rational function (does a factor cancel?), and is the engine behind solving polynomial equations and inequalities — all core Module 2 moves.

Worked example — factor and find the zeros of \(x^2 - x - 6\)
  1. Find two numbers that multiply to \(-6\) and add to \(-1\): they are \(-3\) and \(+2\).
  2. Write the factors: \(x^2 - x - 6 = (x - 3)(x + 2)\).
  3. Set each factor to zero (zero-product rule): \(x - 3 = 0\) or \(x + 2 = 0\).
  4. Solve: \(x = 3\) or \(x = -2\) — the zeros (x-intercepts) of the parabola.

Difference of squares is even faster: \(x^2 - 9 = (x-3)(x+3)\), so its zeros are \(x = \pm 3\).

Try it 1 — Factor \(x^2 + 5x + 6\).
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Two numbers multiplying to \(6\), adding to \(5\): \(2\) and \(3\). So \(x^2 + 5x + 6 = (x + 2)(x + 3)\).

Try it 2 — Factor \(x^2 - 16\) and state its zeros.
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Difference of squares: \(x^2 - 16 = (x - 4)(x + 4)\). Zeros at \(x = 4\) and \(x = -4\).

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

Interval notation

What it is: a compact way to write a set of numbers. A bracket \([\ ]\) includes the endpoint; a parenthesis \((\ )\) excludes it; and \(\infty\) always gets a parenthesis (it's a direction, not a reachable number). Why you need it: every domain, range, and inequality solution in this module is written this way — "\(x > 3\)" becomes \((3, \infty)\), and a feasible answer to a polynomial inequality is a union of intervals.

Worked example — translate between notations
  1. \(x \ge -1\) includes \(-1\) and everything above: \([-1, \infty)\) — bracket on \(-1\), parenthesis on \(\infty\).
  2. \(2 < x \le 7\) excludes \(2\), includes \(7\): \((2, 7]\).
  3. "All reals except \(x = 0\)" splits into two pieces: \((-\infty, 0) \cup (0, \infty)\) — the \(\cup\) means "or / union."
  4. Read it back: \([-4, 3)\) means "\(-4 \le x < 3\)" — start included, end excluded.

Rule of thumb: solid dot / "or equal to" \(\Rightarrow\) bracket; open dot / strict inequality \(\Rightarrow\) parenthesis.

Try it 1 — Write \(x \le 5\) in interval notation.
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Everything from \(-\infty\) up to and including \(5\): \((-\infty, 5]\). Parenthesis on \(-\infty\), bracket on \(5\).

Try it 2 — A rational function's domain is "all reals except \(x = 2\)." Write it in interval notation.
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Remove the single point \(x = 2\): \((-\infty, 2) \cup (2, \infty)\). (This is exactly how you'll state a domain around a vertical asymptote.)

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

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

  1. Which parent function has a horizontal asymptote at \(y = 0\) and passes through \((0, 1)\)?
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    The exponential \(y = 2^x\) (any base \(b > 1\)).

  2. Simplify \(2^3 \cdot 2^5\).
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    Add exponents: \(2^{3+5} = 2^8 = 256\).

  3. Evaluate \(\log_2 16\).
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    \(2^4 = 16\), so \(\log_2 16 = 4\).

  4. Factor \(x^2 - 9\).
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    Difference of squares: \((x - 3)(x + 3)\); zeros at \(x = \pm 3\).

  5. Write \(-2 \le x < 4\) in interval notation.
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    \([-2, 4)\) — bracket on \(-2\) (included), parenthesis on \(4\) (excluded).

  6. \(5^{-2}\) equals what fraction?
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    \(5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}\). (Negative exponent → reciprocal.)

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.

Foundation poured. When the skills above feel steady, step into the module — the Transformation Studio lets you see every one of these families move at once.

Module 2 Overview & Visual Lab Jump to the Transformation Studio Pre-Calculus Course