Quadratic Equation and Graphs
Quadratic Functions and Their Graphs
A quadratic function is a polynomial function of degree 2, typically written in the form:
$$f(x) = ax^2 + bx + c$$
where a, b, and c are constants and a ≠ 0.
Important Properties of Quadratic Function Graphs
- Y-Intercept: The point where the parabola crosses the y-axis (0, c)
- X-Intercepts: Also known as zeros, roots, or solutions - points where the parabola crosses the x-axis
- Vertex: The highest or lowest point of the parabola
- Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two symmetrical halves
Forms of Quadratic Functions
1. Standard Form
$$f(x) = ax^2 + bx + c$$
To find the axis of symmetry: $$x = -\frac{b}{2a}$$
2. Vertex Form
$$f(x) = a(x - h)^2 + k$$
Where (h, k) is the vertex of the parabola
Characteristics of Parabolas
- If a > 0, the parabola opens upward (U-shaped)
- If a < 0, the parabola opens downward (∩-shaped)
- The larger the absolute value of a, the narrower the parabola
Finding Key Points
Vertex
For f(x) = ax^2 + bx + c:
$$x = -\frac{b}{2a}$$
$$y = f(-\frac{b}{2a})$$
Y-Intercept
Always at (0, c)
X-Intercepts
Solve the equation: ax^2 + bx + c = 0
Using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Graphical Representation
Parent Function
The simplest quadratic function is f(x) = x^2. Its characteristics include:
- Vertex at (0, 0)
- Axis of symmetry at x = 0
- Opens upward
- Y-intercept at (0, 0)
- No x-intercepts other than (0, 0)
Practice Problems: Quadratic Functions
Problem 1
Given the quadratic function f(x) = 2x² - 8x + 6, find:
- The vertex
- The axis of symmetry
- The y-intercept
- The x-intercepts (if any)
Solution:
1. To find the vertex:
Using the formula x = -b/(2a), where a = 2 and b = -8
x = -(-8)/(2(2)) = 8/4 = 2
y = 2(2)² - 8(2) + 6 = 8 - 16 + 6 = -2
Therefore, the vertex is (2, -2)
2. The axis of symmetry is x = 2
3. The y-intercept is found by setting x = 0:
f(0) = 2(0)² - 8(0) + 6 = 6
The y-intercept is (0, 6)
4. To find x-intercepts, set f(x) = 0:
2x² - 8x + 6 = 0
Using the quadratic formula: x = [-(-8) ± √((-8)² - 4(2)(6))] / (2(2))
x = (8 ± √(64 - 48)) / 4 = (8 ± √16) / 4 = (8 ± 4) / 4
x = 3 or x = 1
The x-intercepts are (1, 0) and (3, 0)
Problem 2
Sketch the graph of the quadratic function f(x) = -(x - 3)² + 4
Solution:
1. This function is in vertex form: f(x) = a(x - h)² + k
Where a = -1, h = 3, and k = 4
2. The vertex is (3, 4)
3. The axis of symmetry is x = 3
4. The parabola opens downward because a is negative
5. To find y-intercept, set x = 0:
f(0) = -(0 - 3)² + 4 = -9 + 4 = -5
The y-intercept is (0, -5)
6. To find x-intercepts, set f(x) = 0:
-(x - 3)² + 4 = 0
(x - 3)² = 4
x - 3 = ±2
x = 3 ± 2
x = 1 or x = 5
The x-intercepts are (1, 0) and (5, 0)
Here's a sketch of the graph:
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