Graphs Of Quadratic Functions Quizzes
Introduction
Graphs of quadratic functions are an essential topic in GCSE Maths. A quadratic function is a function of the form:
$$y = ax^2 + bx + c$$
where a, b, and c are constants and a ≠ 0. The graph of a quadratic function is called a parabola. Understanding the shape, key features, and transformations of quadratic graphs is essential for solving equations, analysing data, and modelling real-life scenarios such as projectile motion, area problems, and financial modelling.
For example, the quadratic function $$y = x^2 - 4x + 3$$ produces a parabola that opens upwards, has a vertex, and intersects the x-axis at two points. Graphing quadratic functions allows students to visualise solutions, interpret maximum or minimum values, and solve algebraic problems efficiently.
Core Concepts
Standard Form of a Quadratic Function
A quadratic function can be written as:
- Standard form: $$y = ax^2 + bx + c$$
- Vertex form: $$y = a(x - h)^2 + k$$, where (h, k) is the vertex
Shape of a Quadratic Graph (Parabola)
- If a > 0 → parabola opens upwards → minimum point at vertex
- If a < 0 → parabola opens downwards → maximum point at vertex
- The vertex represents the highest or lowest point of the graph
Vertex of a Quadratic Function
The vertex can be found using:
$$x = -\frac{b}{2a}$$
Then substitute x into the function to find y-coordinate:
$$y = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c$$
Example:
- Function: $$y = 2x^2 - 4x + 1$$
- Vertex x-coordinate: $$x = -(-4)/(2×2) = 4/4 = 1$$
- Vertex y-coordinate: $$y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1$$
- Vertex: (1, -1)
Y-Intercept
The y-intercept occurs when x = 0:
$$y = c$$
Example:
- Function: $$y = x^2 - 3x + 2$$ → y-intercept = 2 → point (0,2)
X-Intercepts (Roots / Zeros)
The x-intercepts occur when y = 0:
$$ax^2 + bx + c = 0$$
- Use factorisation if possible
- Use quadratic formula: $$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$$
Axis of Symmetry
Vertical line passing through the vertex:
$$x = -\frac{b}{2a}$$
The parabola is symmetric about this line.
Direction and Stretch
- If |a| > 1 → parabola is narrower (stretched vertically)
- If 0 < |a| < 1 → parabola is wider (compressed vertically)
- If a < 0 → opens downwards
Plotting Quadratic Graphs
- Find y-intercept (c)
- Find vertex using $$x = -b/(2a)$$ and corresponding y
- Find x-intercepts (if possible) using factorisation or formula
- Plot additional points if necessary
- Draw smooth parabola through points
Real-Life Applications
- Projectile motion: height = f(time)
- Area problems: maximising area with fixed perimeter
- Revenue or profit functions in finance
- Physics: parabolic paths
- Engineering: design curves
Worked Examples
Example 1 (Foundation): Simple parabola
Function: $$y = x^2 - 4x + 3$$
- Vertex x-coordinate: $$x = -(-4)/(2×1) = 2$$
- Vertex y-coordinate: $$y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$$ → Vertex: (2, -1)
- Y-intercept: c = 3 → point (0,3)
- X-intercepts: factorise: $$x^2 - 4x + 3 = (x - 1)(x - 3) = 0 → x = 1, 3$$
Example 2 (Foundation): Parabola opening downwards
Function: $$y = -x^2 + 6x - 5$$
- Vertex x-coordinate: $$x = -6/(2 × -1) = 3$$
- Vertex y-coordinate: $$y = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4$$ → Vertex: (3, 4)
- Y-intercept: c = -5 → point (0, -5)
- X-intercepts: $$-x^2 + 6x - 5 = 0 → x^2 - 6x + 5 = 0 → (x-1)(x-5) = 0 → x = 1, 5$$
Example 3 (Higher): Using quadratic formula
Function: $$y = 2x^2 - 3x - 5$$ → find x-intercepts
- $$x = [-(-3) ± \sqrt{(-3)^2 - 4(2)(-5)}]/(2 × 2)$$
- $$x = [3 ± \sqrt{9 + 40}]/4 = [3 ± \sqrt{49}]/4$$
- $$x = (3 ± 7)/4 → x = 10/4 = 2.5, x = -4/4 = -1$$
Example 4 (Higher): Parabola transformations
Function: $$y = 3(x - 2)^2 + 1$$
- Vertex: (h, k) = (2,1)
- Opens upwards (a = 3 > 0)
- Steeper than y = x^2 (a > 1)
Example 5 (Higher): Real-life application
Projectile: height h = -5t^2 + 20t + 15, t in seconds
- Vertex: t = -b/(2a) = -20/(2×-5) = 2 → maximum height
- Height at t=2: h = -5(2)^2 + 20×2 + 15 = -20 + 40 + 15 = 35 m
- Maximum height = 35 m at t = 2 s
Example 6 (Higher): Finding x for given y
Function: $$y = x^2 - 6x + 8$$, find x when y = 0
- $$x^2 - 6x + 8 = 0 → (x-2)(x-4) = 0 → x = 2, 4$$
Example 7 (Higher): Intersection of two quadratics
y = x^2 + x and y = 2x + 3
- Set equal: $$x^2 + x = 2x + 3 → x^2 - x - 3 = 0$$
- Quadratic formula: $$x = [1 ± \sqrt{1 + 12}]/2 = [1 ± \sqrt{13}]/2$$
Example 8 (Higher): Axis of symmetry
Function: $$y = 4x^2 - 16x + 11$$
- Axis: $$x = -(-16)/(2×4) = 16/8 = 2$$
- Vertex y-coordinate: $$y = 4(2)^2 -16×2 +11 = 16 - 32 + 11 = -5$$ → Vertex: (2, -5)
Example 9 (Higher): Maximum/Minimum
Function: $$y = -3x^2 + 12x - 7$$
- Opens downwards (a < 0) → maximum
- Vertex x-coordinate: $$x = -b/(2a) = -12/(2×-3) = 2$$
- Maximum y: $$y = -3(2)^2 + 12×2 -7 = -12 + 24 -7 = 5$$
Example 10 (Higher): Y-intercept
Function: $$y = 2x^2 - 3x + 4$$ → y-intercept: c = 4 → point (0,4)
---Common Mistakes
- Confusing vertex and y-intercept
- Errors in quadratic formula calculation
- Not accounting for negative coefficient a (direction of parabola)
- Incorrect factorisation leading to wrong x-intercepts
- Plotting points inaccurately on the graph
Tips to avoid errors:
- Identify a, b, c clearly in standard form
- Calculate vertex using $$x = -b/(2a)$$
- Check discriminant for number of x-intercepts
- Use quadratic formula when factorisation is difficult
- Plot multiple points to ensure accuracy of parabola
Applications
- Physics: projectile motion, maximum height
- Economics: profit functions, cost/revenue curves
- Geometry: area problems, parabolic shapes
- Engineering: design of arches, bridges, and curves