Quadratic Function: Complete Lesson with Interactive Parabola

The general form

A quadratic polynomial function is written:

$f(x) = ax^2 + bx + c \quad \text{with } a \neq 0$

• a is the leading coefficient (sets the parabola’s opening direction)
• b is the linear coefficient
• c is the y-intercept (= $f(0)$)

The parabola in action

Move the sliders below to see the parabola, discriminant and roots update in real time:

a =1b =-2c =-3

Discriminant Δ

—

Racines

—

Sommet

—

Things to try:

• Flip the sign of a → the parabola turns upside down
• Make a very small → the parabola flattens
• Change b and c → roots slide or vanish

The discriminant: $\Delta = b^2 - 4ac$

The magic number that decides whether $ax^2 + bx + c = 0$ has solutions.

Case	Number of solutions	Formulas
$\Delta > 0$	2 distinct real roots	$x_1 = \dfrac{-b - \sqrt{\Delta}}{2a}$, $x_2 = \dfrac{-b + \sqrt{\Delta}}{2a}$
$\Delta = 0$	1 double root	$x_0 = \dfrac{-b}{2a}$
$\Delta < 0$	0 real roots	(complex roots)

The vertex

The vertex of the parabola is at:

$S\left(-\dfrac{b}{2a} \; ; \; f\left(-\dfrac{b}{2a}\right)\right)$

It appears in green on the interactive parabola above.

• If $a > 0$: parabola opens upward, vertex is a minimum.
• If $a < 0$: parabola opens downward, vertex is a maximum.

Vertex form

Completing the square:

$f(x) = a(x - \alpha)^2 + \beta$

where $\alpha = -\dfrac{b}{2a}$ and $\beta = f(\alpha)$. The point $(\alpha, \beta)$ is exactly the vertex.

Example

$f(x) = 2x^2 - 8x + 5$

• $\alpha = -\dfrac{-8}{4} = 2$
• $\beta = 2 \times 4 - 16 + 5 = -3$
• Vertex form: $f(x) = 2(x - 2)^2 - 3$

Three classic examples

Example 1: $\Delta > 0$ (two roots)

Solve $x^2 - 5x + 6 = 0$.

• $\Delta = 25 - 24 = 1 > 0$
• $x_1 = 2$, $x_2 = 3$.

Example 2: $\Delta = 0$ (double root)

Solve $x^2 - 6x + 9 = 0$.

• $\Delta = 0$, double root $x_0 = 3$.
• Factors as $(x - 3)^2$.

Example 3: $\Delta < 0$ (no real root)

Solve $x^2 + x + 1 = 0$.

• $\Delta = -3 < 0$
• No real solution. The parabola never touches the x-axis.

Sign of the trinomial

One-line rule: $f(x)$ is the same sign as $a$ outside the roots, and opposite to $a$ between them.

For $a > 0$ with $x_1 < x_2$:

$x$	$-\infty$		$x_1$		$x_2$		$+\infty$
$f(x)$		$+$	0	$-$	0	$+$

Factored form

When $\Delta \geq 0$:

$f(x) = a(x - x_1)(x - x_2)$

(or $a(x - x_0)^2$ for a double root).

Example: $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Step-by-step method

For $ax^2 + bx + c = 0$:

1. Identify $a$, $b$, $c$.
2. Compute $\Delta = b^2 - 4ac$.
3. Check the sign of $\Delta$.
4. Apply the matching formula.
5. Verify by plugging back in.

Common mistakes

1. Forgetting $2a$ in the root formula — not just $2$.
2. Sign of $b$: it’s $-b$ in the formula; if $b = -5$, write $-(-5) = +5$.
3. Squaring $b$ without parentheses: $(-3)^2 = 9$, not $-9$.
4. Mixing equation and function.

Solved exercises

Exercise 1

Solve $2x^2 + 3x - 2 = 0$.

Show solution

$\Delta = 9 + 16 = 25$ $x_1 = -2$, $x_2 = 1/2$.

Exercise 2

Find the vertex of $f(x) = -x^2 + 4x + 1$.

Show solution

$\alpha = 2$, $\beta = 5$. Vertex $S(2, 5)$. Since $a < 0$, it’s a maximum.

Exercise 3

For which $m$ does $x^2 - 2x + m = 0$ have a double root?

Show solution

Need $\Delta = 0$: $4 - 4m = 0$, so $m = 1$.

Takeaways

• $\Delta = b^2 - 4ac$ decides the root count.
• Vertex: $x_S = -\dfrac{b}{2a}$.
• Sign: same as $a$ outside roots.
• Parabola opens up if $a > 0$, down if $a < 0$.
• Vertex form: $a(x - \alpha)^2 + \beta$.

Going further

• Related tips: Discriminant in 5 seconds, Factoring from roots.
• Application: optimization (max/min problems).
• Next: biquadratic equations $ax^4 + bx^2 + c = 0$ reduce to quadratic via $X = x^2$.