Quadratic Functions: Graphs, Vertex, Intercepts, and Axis of Symmetry

Quadratic Functions and Their Graphs

Understanding Quadratic Functions

A quadratic function is a polynomial function of degree two, typically written in the form:

• Standard Form: $f(x) = ax^2 + bx + c$, where $a \neq 0$.

The graph of a quadratic function is a parabola, which opens upward if $a > 0$ and downward if $a < 0$.

Vertex of a Parabola

The vertex is the highest or lowest point on the parabola, depending on the direction it opens.

• Vertex Formula: The vertex $(h, k)$ can be found using:

$h = -\frac{b}{2a}$ $k = f(h) = a(h)^2 + b(h) + c$

• The vertex is a key point for sketching the graph.

Intercepts

• Y-intercept: The point where the graph crosses the y-axis. Found by evaluating $f(0)$.

• X-intercepts (Roots): The points where the graph crosses the x-axis. Found by solving $f(x) = 0$.

Quadratic Formula: The x-intercepts are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two mirror images.

• Equation: $x = h = -\frac{b}{2a}$

Domain and Range

• Domain: For all quadratic functions, the domain is all real numbers: $(-\infty, \infty)$.

• Range: Depends on the direction the parabola opens:

  • If $a > 0$ (opens upward): $[k, \infty)$

  • If $a < 0$ (opens downward): $(-\infty, k]$

• $k$ is the y-coordinate of the vertex.

Example

Given the quadratic function $f(x) = (x + 2)^2$:

• Standard Form: $f(x) = x^2 + 4x + 4$

• Vertex: $h = -2$, $k = 0$ (since $f(-2) = 0$)

• Axis of Symmetry: $x = -2$

• Y-intercept: $f(0) = (0 + 2)^2 = 4$

• X-intercept: Set $f(x) = 0$:

$(x + 2)^2 = 0$ $x + 2 = 0$ $x = -2$

• Domain: $(-\infty, \infty)$

• Range: $[0, \infty)$

Graphing Quadratic Functions

• Plot the vertex and intercepts.

• Draw the axis of symmetry.

• Sketch the parabola, ensuring it passes through the intercepts and is symmetric about the axis.

Summary Table: Key Features of a Parabola

Feature	How to Find	Example ($f(x) = (x+2)^2$)
Vertex	$(-\frac{b}{2a}, f(-\frac{b}{2a}))$	$(-2, 0)$
Axis of Symmetry	$x = -\frac{b}{2a}$	$x = -2$
Y-intercept	$f(0)$	$4$
X-intercept(s)	Solve $f(x) = 0$	$x = -2$
Domain	All real numbers	$(-\infty, \infty)$
Range	Depends on $a$ and $k$	$[0, \infty)$

Additional info: The original file contained fragmented instructions and references to graphing tools. Academic context and a full example were added for completeness.