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Educora
Intermediate22 min9 / 14

Advanced Math II: functions, polynomials, radicals and rational expressions

Function notation and composition, graph transformations, polynomial zeros and the remainder theorem, rational exponents, and extraneous solutions in radical and rational equations.

Check yourself
In this lesson you will learn
  • Evaluate expressions like f(g(x)) and recognize graph transformations
  • Use the link between a polynomial's zeros and its factors
  • Solve radical and rational equations and reject extraneous solutions

The second half of Advanced Math is questions that look scary but follow a few rules: f(g(x)), x³ − 4x² + x + 6, x^(3/2), √(x + 7) = x − 5. Most of them take 3–4 lines — if you know the rule. After each topic we will also show a fast way to check your answer.

  1. 1
    Rewrite what is given

    Copy f(x), g(x) or the polynomial onto scratch paper so it is in front of you.

  2. 2
    Note restrictions

    A denominator cannot be zero and an expression under a square root cannot be negative — write this down before solving.

  3. 3
    Inside out

    For compositions and transformations, work out the inside of the brackets first.

  4. 4
    Check

    Check every root in the original equation; plug in a number to test equivalence.

Function notation and transformations

f(3) means “put 3 in place of x”. f(g(x)) is worked from the inside out: first g, then f. For example, if f(x) = 2x + 1 and g(x) = x², then f(g(2)) = f(4) = 9, but g(f(2)) = g(5) = 25. Order matters!

y = a · f(x − h) + k
where:
  • hthe graph shifts h units right (left if h < 0)
  • kthe graph shifts k units up (down if k < 0)
  • avertical stretch; if a < 0 the graph is reflected over the x-axis

f(−x) reflects the graph over the y-axis.

Interactive
Loading simulation…
The first graph is y = |x|; the second is its transformation. With h = 3 and k = 1 the corner moves from (0, 0) to (3, 1). Make a negative and watch the graph flip.
Example 1: move a point

The graph of y = f(x) passes through (2, 5). Which point lies on the graph of y = f(x − 3) + 1?

Show solution
x − 3 → 3 units right: 2 + 3 = 5.
+1 → 1 unit up: 5 + 1 = 6.
Answer: (5, 6).
Check: at x = 5, f(5 − 3) + 1 = f(2) + 1 = 5 + 1 = 6.

Polynomials: zeros, factors and remainders

p(a) = 0 ⇔ (x − a) is a factor of p(x) p(x) ÷ (x − a) ⇒ remainder = p(a)
where:
  • p(x)a polynomial
  • aa zero of the polynomial: the graph crosses the x-axis at x = a

Remainder theorem: when p(x) is divided by (x − a), the remainder equals p(a).

Example 2: from one zero to all zeros

p(x) = x³ − 4x² + x + 6. Find p(2) and all the zeros of the polynomial.

Show solution
p(2) = 8 − 16 + 2 + 6 = 0 ⇒ (x − 2) is a factor.
Divide: x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3).
x² − 2x − 3 = (x − 3)(x + 1).
So p(x) = (x − 2)(x − 3)(x + 1), and the zeros are −1, 2, 3.
Check: the zeros add up to 4, the opposite of the x² coefficient.

Rational exponents, radical and rational equations

x^(m/n) = ⁿ√(xᵐ) = (ⁿ√x)ᵐ x⁻ⁿ = 1/xⁿx^(m/n) = ⁿ√(xᵐ) = (ⁿ√x)ᵐ x⁻ⁿ = 1/xⁿ
where:
  • nthe index of the root (the denominator)
  • mthe power (the numerator)

For example: 8^(2/3) = (∛8)² = 2² = 4.

Example 3: extraneous solutions

a) Solve √(x + 7) = x − 5.
b) Solve x/(x − 2) = 2/(x − 2) + 3.

Show solution
a) Square both sides: x + 7 = x² − 10x + 25 ⇒ x² − 11x + 18 = 0 ⇒ (x − 9)(x − 2) = 0.
Check: x = 9: √16 = 4 = 9 − 5 ✓; x = 2: √9 = 3, but 2 − 5 = −3 ✗.
Answer: only x = 9 (x = 2 is extraneous).
b) Multiply both sides by (x − 2): x = 2 + 3(x − 2) ⇒ x = 3x − 4 ⇒ x = 2.
But x = 2 makes the denominator zero! So there is no solution.
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Key points

  • f(g(x)) is evaluated from the inside; f(g(x)) and g(f(x)) are usually different.
  • f(x − h) + k: h units right, k units up.
  • p(a) = 0 ⇔ (x − a) is a factor; the remainder on division by (x − a) is p(a).
  • x^(m/n) = ⁿ√(xᵐ).
  • In radical and rational equations, check every root in the original equation.

Check yourself

10 questions. Every correct answer earns XP.

1 / 10
If f(x) = 3x − 2 and g(x) = x² + 1, what is the value of f(g(2))?