Characteristics of Quadratic Functions

Shape of Quadratic Function Graphs

The graph of a quadratic function always forms a parabola. This parabola can open upward or downward, depending on the value of the coefficient $a$.

Influence of Coefficient a on Graph Shape

When a > 0

If $a > 0$, the graph of the quadratic function will open upward. This means the graph has a minimum point.

Examples of functions with $a > 0$:

• $f(x) = x^2$ (the simplest function with $a = 1$ )
• $f(x) = 2x^2 + 1$ (example with $a = 2$ )

When a < 0

If $a < 0$, the graph of the quadratic function will open downward. This means the graph has a maximum point.

Examples of functions with $a < 0$:

• $f(x) = -x^2$ with $a = -1$
• $f(x) = -3x^2 - 12x - 15$ with $a = -3$

Why Can't a = 0?

When $a = 0$, the function form becomes $f(x) = bx + c$. This is no longer a quadratic function, but a linear function. A quadratic function must have $a \neq 0$ so that the highest power of the variable $x$ is 2.

Important Characteristics of Quadratic Functions

Vertex

The vertex is the highest point (if $a < 0$) or the lowest point (if $a > 0$) on the graph. The coordinates of the vertex are expressed as $(-\frac{b}{2a}, f(-\frac{b}{2a}))$.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two symmetrical parts. The equation of the axis of symmetry is $x = -\frac{b}{2a}$.

Y-Intercept

The y-intercept is obtained when $x = 0$. Its value is $f(0) = c$.

X-Intercepts

The x-intercepts are obtained when $f(x) = 0$, i.e., when $ax^2 + bx + c = 0$. The solutions can be found using the formula:

Steps to Graph a Quadratic Function

1. Determine whether the parabola opens upward ($a > 0$) or downward ($a < 0$).
2. Calculate the coordinates of the vertex $(-\frac{b}{2a}, f(-\frac{b}{2a}))$.
3. Calculate the y-intercept: $(0, c)$.
4. Calculate the x-intercepts (if any).
5. Choose several other x-values and calculate their corresponding y-values.
6. Plot all points in the coordinate system.
7. Connect the points with a parabolic curve.

Drawing Quadratic Function Graphs

f(x) = x² - 2x - 3

Let's graph the function $f(x) = x^2 - 2x - 3$:

1. Coefficient $a = 1 > 0$, so the parabola opens upward.

2. Vertex:

   $$x = -\frac{b}{2a} = -\frac{-2}{2 \cdot 1} = 1$$

   $$f(1) = 1^2 - 2 \cdot 1 - 3 = 1 - 2 - 3 = -4$$

   So the vertex is at (1, -4).

3. Y-intercept:

   $$f(0) = 0^2 - 2 \cdot 0 - 3 = -3$$

   So the y-intercept is at (0, -3).

4. X-intercepts: $f(x) = 0$ or $x^2 - 2x - 3 = 0$

   Using the quadratic formula:

   $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}$$

   $$x = \frac{2 + 4}{2} = 3 \text{ or } x = \frac{2 - 4}{2} = -1$$

   So the x-intercepts are at (-1, 0) and (3, 0).

5. Let's calculate some additional points:

   $$f(-2) = (-2)^2 - 2 \cdot (-2) - 3 = 4 + 4 - 3 = 5$$

   $$f(2) = 2^2 - 2 \cdot 2 - 3 = 4 - 4 - 3 = -3$$

f(x) = -x²

Let's graph the function $f(x) = -x^2$:

1. Coefficient $a = -1 < 0$, so the parabola opens downward.

2. Vertex:

   $$x = -\frac{b}{2a} = -\frac{0}{2 \cdot (-1)} = 0$$

   $$f(0) = -(0)^2 = 0$$

   So the vertex is at (0, 0).

3. Y-intercept:

   $$f(0) = 0$$

   So the y-intercept is at (0, 0).

4. X-intercepts: $f(x) = 0$ or $-x^2 = 0$

   $$x^2 = 0$$

   $$x = 0$$

   So the x-intercept is at (0, 0).

5. Let's calculate some additional points:

   $$f(-2) = -((-2)^2) = -4$$

   $$f(-1) = -((-1)^2) = -1$$

   $$f(1) = -(1^2) = -1$$

   $$f(2) = -(2^2) = -4$$

Table of Quadratic Function Graph Shapes