PRECALCULUS | CHAPTER STUDY MATERIAL
Polynomial and
Rational Functions
A Complete Study Guide with Worked Examples,
Tricky Practice Questions & Answer Keys
Topics Covered
Quadratic Functions · Power & Polynomial Functions · Graphs of Polynomials
Dividing Polynomials · Zeros of Polynomial Functions · Rational Functions
Inverse & Radical Functions · Modeling Using Variation
Compiled Study Notes
,Polynomial & Rational Functions Study Guide
Contents
1
, Polynomial & Rational Functions Study Guide
1. Quadratic Functions
A quadratic function is a polynomial function of degree 2. Its graph is a parabola — a U-shaped curve
that is symmetric about a vertical line called the axis of symmetry.
Forms of a Quadratic Function
• Standard (general) form: f (x) = ax2 + bx + c, a ̸= 0
• Vertex form: f (x) = a(x − h)2 + k, where (h, k) is the vertex
• Conversion: complete the square on ax2 + bx + c to get
b b2
h=− , k = f (h) = c −
2a 4a
KEY IDEA
• If a > 0 the parabola opens upward and the vertex is a minimum.
• If a < 0 the parabola opens downward and the vertex is a maximum.
b
• The axis of symmetry is the vertical line x = h = − .
2a
• y-intercept: set x = 0 ⇒ (0, c).
• x-intercepts (zeros): solve ax2 + bx + c = 0 using factoring or the quadratic formula
√
−b ± b2 − 4ac
x=
2a
• Discriminant D = b2 − 4ac tells the nature of the roots: D > 0 (two real roots), D = 0 (one repeated
real root), D < 0 (two complex roots).
• Domain: all real numbers, (−∞, ∞).
• Range: [k, ∞) if a > 0; (−∞, k] if a < 0.
Worked Example
Find the vertex, axis of symmetry, and intercepts of f (x) = 2x2 − 6x + 4, then sketch its behaviour.
Solution. Here a = 2, b = −6, c = 4.
b −6 3 (3) ( 3 )2 (3)
h=− =− = , k=f 2 =2 2 −6 2 + 4 = − 12
2a 4 2
( )
3 1 3
Vertex: , − , axis of symmetry x = , opens upward (min). y-intercept: (0, 4). For x-intercepts
2 2 2
solve 2x2 − 6x + 4 = 0 ⇒ x2 − 3x + 2 = 0 ⇒ (x − 1)(x − 2) = 0 ⇒ x = 1, 2.
Application: Maximizing Area
A farmer has 80 m of fencing to enclose a rectangular field bordered on one side by a river (no fence needed
there). Find the dimensions that maximize the area.
Solution. Let x = width (perpendicular to river), so length along river = 80 − 2x.
A(x) = x(80 − 2x) = −2x2 + 80x
2
Polynomial and
Rational Functions
A Complete Study Guide with Worked Examples,
Tricky Practice Questions & Answer Keys
Topics Covered
Quadratic Functions · Power & Polynomial Functions · Graphs of Polynomials
Dividing Polynomials · Zeros of Polynomial Functions · Rational Functions
Inverse & Radical Functions · Modeling Using Variation
Compiled Study Notes
,Polynomial & Rational Functions Study Guide
Contents
1
, Polynomial & Rational Functions Study Guide
1. Quadratic Functions
A quadratic function is a polynomial function of degree 2. Its graph is a parabola — a U-shaped curve
that is symmetric about a vertical line called the axis of symmetry.
Forms of a Quadratic Function
• Standard (general) form: f (x) = ax2 + bx + c, a ̸= 0
• Vertex form: f (x) = a(x − h)2 + k, where (h, k) is the vertex
• Conversion: complete the square on ax2 + bx + c to get
b b2
h=− , k = f (h) = c −
2a 4a
KEY IDEA
• If a > 0 the parabola opens upward and the vertex is a minimum.
• If a < 0 the parabola opens downward and the vertex is a maximum.
b
• The axis of symmetry is the vertical line x = h = − .
2a
• y-intercept: set x = 0 ⇒ (0, c).
• x-intercepts (zeros): solve ax2 + bx + c = 0 using factoring or the quadratic formula
√
−b ± b2 − 4ac
x=
2a
• Discriminant D = b2 − 4ac tells the nature of the roots: D > 0 (two real roots), D = 0 (one repeated
real root), D < 0 (two complex roots).
• Domain: all real numbers, (−∞, ∞).
• Range: [k, ∞) if a > 0; (−∞, k] if a < 0.
Worked Example
Find the vertex, axis of symmetry, and intercepts of f (x) = 2x2 − 6x + 4, then sketch its behaviour.
Solution. Here a = 2, b = −6, c = 4.
b −6 3 (3) ( 3 )2 (3)
h=− =− = , k=f 2 =2 2 −6 2 + 4 = − 12
2a 4 2
( )
3 1 3
Vertex: , − , axis of symmetry x = , opens upward (min). y-intercept: (0, 4). For x-intercepts
2 2 2
solve 2x2 − 6x + 4 = 0 ⇒ x2 − 3x + 2 = 0 ⇒ (x − 1)(x − 2) = 0 ⇒ x = 1, 2.
Application: Maximizing Area
A farmer has 80 m of fencing to enclose a rectangular field bordered on one side by a river (no fence needed
there). Find the dimensions that maximize the area.
Solution. Let x = width (perpendicular to river), so length along river = 80 − 2x.
A(x) = x(80 − 2x) = −2x2 + 80x
2
