Quadratic function

What is it about?

The graph of a quadratic function is a parabola. The simplest one is the standard parabola $y = x^2$ with its vertex at the origin.

In vertex form you can read off the vertex $S(d, e)$ directly:

f(x) = a\,(x - d)^2 + e

$a$ determines the opening and the shape: for $a > 0$ the parabola opens upwards, for $a < 0$ downwards. For $|a| > 1$ it is stretched (narrower), for $|a| < 1$ compressed (wider).
$d$ shifts the parabola horizontally. Watch the sign: $(x - 3)^2$ shifts it 3 to the right.
$e$ shifts the parabola vertically.

Expanding the brackets gives the standard form:

f(x) = a x^2 + b x + c \quad\text{with}\quad b = -2ad,\;\; c = ad^2 + e

The other way round (completing the square): $d = -\frac{b}{2a}$ and $e = c - \frac{b^2}{4a}$ .

Zeros

The discriminant $D = b^2 - 4ac$ tells you whether and where the parabola crosses the x-axis:

Discriminant	Number of zeros
$D > 0$	two zeros
$D = 0$	one (double) zero – the vertex lies on the x-axis
$D < 0$	no zeros

The quadratic formula gives the zeros themselves:

x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Explore

Change only $a$ – also to negative values. Compare with the dashed standard parabola. a = 2 a = −0.5
Set $d = 3$ . Why is there a minus in the equation? Load example
Drag the vertex $S$ up and down. When are there two zeros, one zero or none?
Switch on the stretch triangle: from the vertex you go 1 to the right and $a$ up. Drag the hollow point to the right of the vertex up or down – this changes $a$ . Load example
Switch to standard form at the top. The parabola stays the same – only the representation changes.

Exercises

Exercise 1: Read off the vertex form

Load exercise Read off the vertex and find $a$ with the stretch triangle. Write down the equation and calculate the zeros.

Show solution

Vertex $S(-2, -4.5)$ . Going 1 to the right from the vertex, the graph goes up by 0.5, so $a = 0.5$ :

$f(x) = 0.5\,(x + 2)^2 - 4.5$

Zeros: $0.5\,(x+2)^2 = 4.5 \Rightarrow (x+2)^2 = 9 \Rightarrow x + 2 = \pm 3$ , so $x_1 = -5$ and $x_2 = 1$ .

Exercise 2: Converting

Write $f(x) = 2\,(x - 1)^2 - 8$ in standard form and find the zeros. Check with the simulation.

Show solution

$2\,(x^2 - 2x + 1) - 8 = 2x^2 - 4x - 6$ . Zeros: $x^2 - 2x - 3 = 0 \Rightarrow x_1 = -1,\; x_2 = 3$ . Show in simulation

Exercise 3: How many zeros?

Decide without calculating the zeros how many zeros the functions have:

$f(x) = x^2 + 2x + 3$
$g(x) = -x^2 + 4x - 4$
$h(x) = 0.5x^2 - 2$

Show solution

$D = 4 - 12 = -8 < 0$ : no zeros. Show
$D = 16 - 16 = 0$ : one double zero at $x = 2$ . Show
$D = 0 + 4 = 4 > 0$ : two zeros ( $x = \pm 2$ ). Show

Exercise 4: Throwing a ball (modelling)

The path of a ball is described by $h(x) = -0.2\,(x - 5)^2 + 6$ ( $x$ : distance in m, $h$ : height in m). How high does the ball fly at most, and how far does it fly? Tip: zoom out to check.

Show solution

The vertex $S(5, 6)$ is the highest point: a height of 6 m after 5 m. Landing: $0.2\,(x-5)^2 = 6 \Rightarrow (x-5)^2 = 30 \Rightarrow x = 5 + \sqrt{30} \approx 10.5$ m. Show in simulation

Ideas for the classroom

Use: introducing the vertex form (grade 9), linking the different forms, preparing the quadratic formula.
Common stumbling blocks:
- The sign of $d$ is misread: $(x + 2)^2$ shifts to the left.
- “Stretched” is confused with “wider”. The dashed comparison with the standard parabola helps.
- The coefficient $b$ of the standard form is taken for a shift – dragging the vertex shows that $b$ and $c$ change together.
Regional differences: depending on the region, students work with the $pq$ formula (normal form $x^2 + px + q$ ) or the $abc$ formula. The simulation shows the discriminant $b^2 - 4ac$ ; for $a = 1$ it equals $p^2 - 4q$ .
Rounded values: if zeros or the vertex cannot be shown exactly with two decimal places, the simulation shows “≈” instead of “=” (e.g. $x_2 \approx 10.48$ for the ball). This is a good opportunity to talk about exact values such as $5 + \sqrt{30}$ .
Exercise mode: Share → Hide results hides the equations, the vertex and the zeros. This turns any setup into a reading exercise.
Notation: some curricula and textbooks write the term as $a \cdot (x + d)^2 + e$ . Under Display → Notation of the vertex form you can switch the simulation to this notation; a positive $d$ then shifts to the left.