Graphing Quadratic Functions: Vertex, Axis and Intercepts

The shape is always a parabola

$$y = ax^2 + bx + c \qquad (a \neq 0)$$

Every such function graphs as a parabola: a symmetric curve with a single turning point called the vertex.

$a$	Opens	Vertex is
$a > 0$	upward	the minimum
$a < 0$	downward	the maximum

$$y = x^2 - 4x + 3$$

• y = x^2 - 4x + 3
• x = 2

The dashed line is the axis of symmetry. Fold the page along it and the two halves of the curve land on each other.

Why the axis sits at −b/(2a)

The two $x$-intercepts are the same distance from the axis, so the axis is halfway between them.

The quadratic formula gives those roots:

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

Write $D$ for $b^2 - 4ac$. The two roots differ only in the sign in front of $\sqrt{D}$, so averaging them cancels that part entirely:

$$\frac{1}{2}\left(\frac{-b + \sqrt{D}}{2a} + \frac{-b - \sqrt{D}}{2a}\right) = \frac{1}{2} \cdot \frac{-2b}{2a} = -\frac{b}{2a}$$

The formula for the axis is the average of the roots with the radical gone. That is why it works even when there are no real roots — the average of the two complex solutions still lands there.

For $y = x^2 - 4x + 3$:

$$x = -\frac{-4}{2(1)} = 2$$

Then substitute to get the vertex height:

$$y = 4 - 8 + 3 = -1 \quad\Rightarrow\quad \text{vertex } (2, -1)$$

Three forms, three things revealed

The same parabola can be written three ways, and each hands you a different fact without any work.

Form	Looks like	Reveals
standard	$ax^2 + bx + c$	the $y$-intercept, $c$
factored	$a(x - p)(x - q)$	the zeros, $p$ and $q$
vertex	$a(x - h)^2 + k$	the vertex, $(h, k)$

For our parabola all three exist:

$$x^2 - 4x + 3 = (x - 1)(x - 3) = (x - 2)^2 - 1$$

Read them off:

• Standard: the curve crosses the $y$-axis at $3$.
• Factored: it crosses the $x$-axis at $1$ and $3$.
• Vertex: its lowest point is $(2, -1)$.

Choose the form that answers the question you were asked. Hunting for the vertex in factored form is work; converting to vertex form first is less.

To get	Do this
standard → factored	factor
standard → vertex	complete the square
factored or vertex → standard	expand

Note the shortcut hiding in factored form: the zeros are symmetric about the axis, so the axis is their average. For $(x-1)(x-3)$ that is $\tfrac{1+3}{2} = 2$, agreeing with $-\tfrac{b}{2a}$ and needing no formula.

How many x-intercepts

An upward parabola with its vertex below the axis has to cross twice on its way back up. One sitting on the axis touches once. One above never crosses.

Vertex position (upward parabola)	$x$-intercepts
below the $x$-axis	two
on the $x$-axis	one
above the $x$-axis	none

The discriminant $b^2 - 4ac$ says the same thing in symbols: positive gives two roots, zero gives one, negative gives none.

Sketching one

Four facts are enough for a usable sketch.

1. Which way it opens — the sign of $a$.
2. The $y$-intercept — read $c$ straight off.
3. The vertex — $x = -\tfrac{b}{2a}$, then substitute.
4. The $x$-intercepts, if any — factor or use the formula.

Sketch $y = -x^2 + 6x - 5$.

Opens downward, since $a = -1$. Crosses the $y$-axis at $-5$.

$$x = -\frac{6}{2(-1)} = 3 \qquad y = -9 + 18 - 5 = 4$$

Vertex $(3, 4)$, a maximum. Factoring gives $-(x-1)(x-5)$, so it crosses at $1$ and $5$ — symmetric about $3$, as they must be.

Reading a parabola in context

A ball is thrown, and its height in feet after $t$ seconds is $h = -16t^2 + 64t + 5$.

Every feature of the graph answers a question about the throw:

Feature	Question it answers
$y$-intercept, $5$	the height it left the hand
vertex $t$-value	when it reached its highest point
vertex height	how high it got
positive $x$-intercept	when it hit the ground
$a < 0$	that it comes back down

$$t = -\frac{64}{2(-16)} = 2 \text{ s} \qquad h = -64 + 128 + 5 = 69 \text{ ft}$$

The negative $t$-intercept is arithmetic, not physics. The model only describes the throw from $t = 0$ onward, and reading a value outside that window is where sensible algebra produces nonsense.

Worked examples

Common mistakes

Practice problems