Graphing quadratic functions — vertex, axis, direction, roots from equation

Question

Given a quadratic equation $y = ax^2 + bx + c$ , how do we determine the vertex, axis of symmetry, direction of opening, and roots to sketch the graph?

Solution — Step by Step

The coefficient $a$ tells you everything about direction:

$a > 0$ : parabola opens upward (U-shaped, happy face)
$a < 0$ : parabola opens downward (inverted U, sad face)

The larger $|a|$ , the narrower the parabola. $|a| < 1$ makes it wider than $y = x^2$ .

The axis of symmetry is the vertical line:

x = -\frac{b}{2a}

The vertex is the point on this axis:

\text{Vertex} = \left(-\frac{b}{2a}, \; -\frac{D}{4a}\right)

where $D = b^2 - 4ac$ (the discriminant).

Alternatively, substitute $x = -b/(2a)$ back into the equation to find the $y$ -coordinate.

The vertex is the minimum point when $a > 0$ and the maximum point when $a < 0$ .

Set $y = 0$ : solve $ax^2 + bx + c = 0$ .

x = \frac{-b \pm \sqrt{D}}{2a}

$D > 0$ : two distinct real roots (parabola crosses x-axis at 2 points)
$D = 0$ : one repeated root (parabola touches x-axis at vertex)
$D < 0$ : no real roots (parabola does not touch x-axis)

The y-intercept is simply $c$ (put $x = 0$ ).

Example: Sketch $y = x^2 - 4x + 3$ .

Here $a = 1$ (upward), $b = -4$ , $c = 3$ .

Axis of symmetry: $x = -(-4)/(2 \times 1) = 2$

Vertex: $y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$ . So vertex = $(2, -1)$ .

$D = 16 - 12 = 4 > 0$ : two roots.

Roots: $x = \frac{4 \pm 2}{2} = 1$ and $3$ .

y-intercept: $c = 3$ , so point $(0, 3)$ .

Plot these points — vertex at $(2, -1)$ , crosses x-axis at $(1, 0)$ and $(3, 0)$ , y-intercept at $(0, 3)$ — and draw a smooth U-shaped curve.

flowchart TD
    A["y = ax2 + bx + c"] --> B["Check sign of a: up or down?"]
    B --> C["Find axis: x = minus b over 2a"]
    C --> D["Find vertex y-coordinate: substitute x back"]
    D --> E["Find discriminant D = b2 minus 4ac"]
    E --> F{"D value?"}
    F -->|"D greater than 0"| G["Two x-intercepts: use quadratic formula"]
    F -->|"D = 0"| H["One x-intercept at vertex"]
    F -->|"D less than 0"| I["No x-intercepts"]
    G --> J["Plot vertex, roots, y-intercept, draw curve"]
    H --> J
    I --> J

Why This Works

Every quadratic function is a parabola because the $x^2$ term creates a symmetric curve. Completing the square converts $y = ax^2 + bx + c$ into $y = a(x - h)^2 + k$ form, where $(h, k)$ is the vertex. The $-b/(2a)$ formula for the axis of symmetry comes directly from this completion.

The discriminant determines how the parabola interacts with the x-axis because it tells us whether the quadratic equation has real solutions.

Alternative Method

Instead of memorising the vertex formula, complete the square directly. For $y = x^2 - 4x + 3$ : take half of $b$ (which is $-2$ ), square it ( $4$ ), add and subtract: $y = (x^2 - 4x + 4) - 4 + 3 = (x-2)^2 - 1$ . The vertex is immediately visible as $(2, -1)$ . This method works for any quadratic and builds deeper understanding.

Common Mistake

Students forget to include the sign of $a$ in the vertex formula. They calculate $x = -b/2a$ correctly but then find the y-coordinate by computing $-D/4a$ and forget that $a$ could be negative. A safer approach: just substitute $x = -b/(2a)$ back into the original equation to get the y-coordinate. This avoids the sign confusion entirely.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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