Free Study Materials for Grades 10 to 12
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Quadratic Functions
Learning Outcomes
- Identify and understand the standard form of quadratic functions
- Find roots using factorization, quadratic formula, and completing the square
- Determine axis of symmetry, vertex, and y-intercept
- Sketch graphs of quadratic functions accurately
- Apply quadratic functions to solve real-world problems
Introduction to Quadratic Functions
A quadratic function is a polynomial function of degree two. Its general form is given by:
f(x) = ax² + bx + c
where:
- a, b, c are constants
- a ≠ 0 (if a = 0, it becomes a linear function)
- The graph of a quadratic function is called a parabola
Standard Form and Coefficients
Students should be able to identify the coefficients a, b, and c in a given quadratic function and understand how they affect the graph.
Identifying Coefficients
Given the quadratic function f(x) = 2x² - 5x + 3, identify the coefficients a, b, and c, and determine the direction the parabola opens.
- Identify coefficients:
a = 2, b = -5, c = 3
- Determine direction:
Since a = 2 > 0, the parabola opens upwards.
- Conclusion:
The quadratic function has a minimum point (vertex) because it opens upwards.
Finding the Roots (x-intercepts)
The roots of a quadratic function are the values of x for which f(x) = 0. These are also known as the x-intercepts of the parabola.
Factorization Method
Express the quadratic as a product of two linear factors.
Find roots of f(x) = x² - 4x + 3
- Factorize: x² - 4x + 3 = (x - 1)(x - 3)
- Set factors to zero: x - 1 = 0 or x - 3 = 0
- Solve: x = 1 or x = 3
- Roots: x = 1, x = 3
Quadratic Formula
x = (-b ± √(b² - 4ac)) / 2a
Find roots of f(x) = 2x² + 3x - 2
- Identify: a = 2, b = 3, c = -2
- Apply formula:
x = (-3 ± √(3² - 4(2)(-2))) / (2(2))
- Simplify:
x = (-3 ± √(9 + 16)) / 4 = (-3 ± √25) / 4 = (-3 ± 5) / 4
- Solve:
x = (-3 + 5)/4 = 1/2x = (-3 - 5)/4 = -2
Completing the Square
Rewrite in the form a(x - h)² + k
Find roots of f(x) = x² + 6x + 5
- Complete square:
x² + 6x + 5 = (x² + 6x + 9) - 9 + 5 = (x + 3)² - 4
- Set to zero:
(x + 3)² - 4 = 0
- Solve:
(x + 3)² = 4x + 3 = ±2x = -1orx = -5
Axis of Symmetry and Vertex
The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two symmetrical halves.
x = -b / 2a
(-b / 2a, f(-b / 2a))
Finding Axis of Symmetry and Vertex
Find the axis of symmetry and vertex of the quadratic function f(x) = x² - 2x + 4.
- Identify coefficients:
a = 1, b = -2, c = 4
- Find axis of symmetry:
x = -b / 2a = -(-2) / (2 × 1) = 2 / 2 = 1
Axis of symmetry: x = 1
- Find vertex:
x-coordinate = 1y-coordinate = f(1) = (1)² - 2(1) + 4 = 1 - 2 + 4 = 3
Vertex: (1, 3)
y-intercept
The y-intercept is where the parabola intersects the y-axis (when x = 0).
Finding y-intercept
Find the y-intercept of the quadratic function f(x) = 3x² + x - 2.
- Set x = 0:
f(0) = 3(0)² + (0) - 2 = -2
- Identify y-intercept:
The y-intercept is the point (0, c) where c = -2
- Conclusion:
y-intercept: (0, -2)
Sketching Quadratic Graphs
Step-by-Step Method:
- Find roots (x-intercepts)
- Find y-intercept (0, c)
- Find axis of symmetry x = -b/2a
- Find vertex (-b/2a, f(-b/2a))
- Determine shape (upward if a > 0, downward if a < 0)
- Plot points and draw smooth parabola
Sketching a Quadratic Function
Sketch the graph of the quadratic function f(x) = x² - 6x + 8.
- Find roots:
x² - 6x + 8 = (x - 2)(x - 4)
Roots: x = 2 and x = 4
- Find y-intercept:
f(0) = (0)² - 6(0) + 8 = 8
y-intercept: (0, 8)
- Find axis of symmetry:
x = -(-6) / (2 × 1) = 6 / 2 = 3
- Find vertex:
f(3) = (3)² - 6(3) + 8 = 9 - 18 + 8 = -1
Vertex: (3, -1)
- Determine shape:
a = 1 > 0, so parabola opens upwards
- Sketch:
Plot points: (2,0), (4,0), (0,8), (3,-1)
Draw smooth curve through points
Applications of Quadratic Functions
Real-World Problem Solving:
📈 Projectile Motion
A ball is thrown vertically upwards from the ground with an initial velocity of 20 m/s. The height h(t) of the ball after t seconds is given by h(t) = -5t² + 20t. Find:
- The maximum height reached by the ball
- The time when the ball hits the ground
- Maximum height (vertex):
t = -b / 2a = -20 / (2 × -5) = -20 / -10 = 2 secondsh(2) = -5(2)² + 20(2) = -20 + 40 = 20 meters
Maximum height: 20 meters
- Time when ball hits ground (roots):
-5t² + 20t = 0-5t(t - 4) = 0t = 0(start) ort = 4seconds
Ball hits ground after 4 seconds
📊 Optimization Problem
A farmer has 100 meters of fencing to create a rectangular garden. Find the dimensions that maximize the area of the garden.
- Define variables:
Let length = x meters
Then width = (100 - 2x)/2 = 50 - x meters
- Area function:
A(x) = x(50 - x) = -x² + 50x
- Find maximum area (vertex):
x = -b / 2a = -50 / (2 × -1) = 25 metersWidth = 50 - 25 = 25 meters
- Maximum area:
A(25) = -25² + 50(25) = 625 m²
Maximum area: 625 m² with dimensions 25m × 25m
Assessment Guidelines
Assessment of quadratic functions in Grade 10 should include:
Tests and Examinations
- Identifying coefficients and parabola properties
- Finding roots using different methods
- Sketching graphs accurately
- Solving application problems
Assignments and Projects
- Problem-solving tasks and investigations
- Real-world application projects
- Mathematical modeling exercises
- Group investigations of quadratic patterns
Download Notes
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Download Quadratic Functions Notes (PDF)