IB Mathematics AA SL The quadratic function Study Notes
IB Mathematics AA SL The quadratic function Study Notes Offer a clear explanation of The quadratic function , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic The quadratic function
The Quadratic Function: \( f(x) = ax^2 + bx + c \)
A quadratic function is a polynomial function of degree 2. It has the general form:
\( f(x) = ax^2 + bx + c \), where \( a \ne 0 \)
Graph of a Quadratic Function
The graph of a quadratic function is a parabola:
- Opens upward if \( a > 0 \)
- Opens downward if \( a < 0 \)
The shape is symmetrical and curved
Y-Intercept
When \( x = 0 \), we find the y-intercept:
\( f(0) = a(0)^2 + b(0) + c = c \)
Thus, the y-intercept is at the point \( (0, c) \).
Axis of Symmetry
The graph of a quadratic is symmetric about a vertical line called the axis of symmetry.
The formula to find the axis of symmetry is:
\( x = -\frac{b}{2a} \)
This is the x-value where the parabola reaches its maximum or minimum (its vertex).
Example: Consider the quadratic function:
\( f(x) = 2x^2 – 4x + 1 \)
▶️Answer/Explanation
Y-intercept: Set \( x = 0 \):
\( f(0) = 2(0)^2 – 4(0) + 1 = 1 \)
→ Y-intercept = (0, 1)
Axis of symmetry: Use the formula \( x = -\frac{b}{2a} \):
\( x = -\frac{-4}{2(2)} = \frac{4}{4} = 1 \)
→ Axis of symmetry: \( x = 1 \)
Vertex: Use \( x = 1 \), find \( f(1) \):
\( f(1) = 2(1)^2 – 4(1) + 1 = 2 – 4 + 1 = -1 \)
→ Vertex = (1, -1)
Shape of graph: Since \( a = 2 > 0 \), the parabola opens upward.
Different Forms of a Quadratic Function
Factored Form of a Quadratic Function
The quadratic function can be written in the factored form:
\( f(x) = a(x – p)(x – q) \)
- \( a \): determines the direction and width of the parabola (if \( a > 0 \), it opens upward; if \( a < 0 \), downward)
- \( p \) and \( q \): are the x-intercepts (or roots) of the function
- Vertex: lies halfway between \( p \) and \( q \):
\( x = \frac{p + q}{2} \)
Vertex Form of a Quadratic Function
The quadratic function can be expressed in vertex form as:
\( f(x) = a(x – h)^2 + k \)
- \( a \): controls the direction and width of the parabola.
- \( h \) and \( k \): represent the vertex of the parabola.
- The axis of symmetry is the vertical line \( x = h \).
- The graph opens:
- Upward if \( a > 0 \)
- Downward if \( a < 0 \)
Example: Consider the quadratic function:
\( f(x) = 2(x – 1)(x – 3) \)
▶️Answer/Explanation
X-intercepts (roots):
Set each factor to 0:
\( x – 1 = 0 \Rightarrow x = 1 \)
\( x – 3 = 0 \Rightarrow x = 3 \)
→ X-intercepts: (1, 0) and (3, 0)
Vertex:
\( x = \frac{1 + 3}{2} = 2 \)
\( f(2) = 2(2 – 1)(2 – 3) = 2(1)(-1) = -2 \)
→ Vertex = (2, -2)
Y-intercept:
Set \( x = 0 \): \( f(0) = 2(-1)(-3) = 6 \)
→ Y-intercept = (0, 6)
Shape:
Since \( a = 2 > 0 \), the parabola opens upward.
Example: Consider the function:
\( f(x) = 3(x – 2)^2 – 5 \)
▶️Answer/Explanation
Vertex:
The form \( f(x) = a(x – h)^2 + k \) gives vertex = \( (h, k) \)
→ \( h = 2, \quad k = -5 \)
→ Vertex = (2, -5)
Y-intercept:
Set \( x = 0 \):
\( f(0) = 3(0 – 2)^2 – 5 = 3(4) – 5 = 12 – 5 = 7 \)
→ Y-intercept = (0, 7)
Axis of Symmetry:
\( x = h = 2 \)
Shape:
Since \( a = 3 > 0 \), the parabola opens upward.
Example: Convert the quadratic function \( f(x) = x^2 – 4x – 5 \) into:
- Factorised form
- Vertex (completed square) form
▶️Answer/Explanation
Factorised Form
Factor the quadratic:
\( f(x) = x^2 – 4x – 5 = (x – 5)(x + 1) \)
Factorised form: \( f(x) = (x – 5)(x + 1) \)
Vertex Form (Complete the Square)
Start from the standard form:
\( f(x) = x^2 – 4x – 5 \)
Group and complete the square:
\( f(x) = (x^2 – 4x + 4) – 4 – 5 = (x – 2)^2 – 9 \)
→ Vertex form: \( f(x) = (x – 2)^2 – 9 \)
→ Vertex: \( (2, -9) \)
Connections Between Quadratic Functions and Physics
1. Kinematics:
The displacement of an object under constant acceleration follows a quadratic relationship:
\( s(t) = ut + \frac{1}{2}at^2 \)
- \( s(t) \): displacement as a function of time
- \( u \): initial velocity
- \( a \): constant acceleration
2. Projectile Motion:
The vertical position of a projectile is modeled by a quadratic function:
\( y(t) = v_0 \sin(\theta)t – \frac{1}{2}gt^2 \)
- \( v_0 \): initial speed
- \( \theta \): angle of projection
- \( g \): acceleration due to gravity
This results in a parabolic path for the projectile.
3. Simple Harmonic Motion (SHM):
The position of an object in SHM is modeled by:
\( x(t) = A \cos(\omega t + \phi) \) or \( x(t) = A \sin(\omega t + \phi) \)
- \( A \): amplitude
- \( \omega \): angular frequency
- \( \phi \): phase shift
Although SHM is not quadratic in form, the energy relationships (e.g., potential energy vs. displacement) involve quadratic expressions:
\( E_{PE} = \frac{1}{2}kx^2 \)