Axis of Symmetry

Q: Why does $x = -\dfrac{b}{2a}$ give the vertex?

The two $x$-intercepts of a parabola (when they exist) sit at equal distances from the axis of symmetry, and $-\dfrac{b}{2a}$ is exactly the midpoint of the quadratic formula's two solutions. The vertex lies on that mirror line, so its $x$-coordinate is $-\dfrac{b}{2a}$ — then you substitute back in for the $y$-coordinate.

A quadratic function has the form $y = ax^2 + bx + c$ , and its graph is a parabola — a symmetric U-shaped curve. Every parabola has a turning point called the vertex, and a vertical mirror line through that vertex called the axis of symmetry. Once you locate the vertex, the rest of the graph is just a few plotted points and their mirror images.

The whole process runs on one formula: the axis of symmetry is $x = -\dfrac{b}{2a}$ . Find that $x$ -value, substitute it back in to get the vertex, plot two or three more points, and mirror them across the axis. Four steps, same every time.

What a, b, and c tell you

Before plotting anything, read the equation. The sign of $a$ sets the direction: if $a$ is positive the parabola opens up (the vertex is the lowest point); if $a$ is negative it opens down (the vertex is the highest point). The size of $a$ sets the width — a larger absolute value of $a$ makes a narrower parabola, and a value between $0$ and $1$ makes a wider one.

The constant $c$ is a free point: it is the $y$ -intercept, because substituting $x = 0$ into $ax^2 + bx + c$ leaves just $c$ . The middle coefficient $b$ has no shortcut meaning by itself — it works together with $a$ to position the vertex, through the formula in the next section.

The vertex and the axis of symmetry

The axis of symmetry of $y = ax^2 + bx + c$ is the vertical line $x = -\dfrac{b}{2a}$ . The vertex sits on that line, so its $x$ -coordinate is $-\dfrac{b}{2a}$ ; substitute that value into the function to get the vertex's $y$ -coordinate.

Take $y = x^2 - 2x - 3$ , graphed below. Here $a = 1$ and $b = -2$ , so the axis of symmetry is $x = -\dfrac{-2}{2(1)} = 1$ . Substituting $x = 1$ gives $y = 1 - 2 - 3 = -4$ , so the vertex is $(1, -4)$ . Since $a = 1$ is positive, the parabola opens up from that lowest point — and the graph confirms it, crossing the $x$ -axis at $-1$ and $3$ and the $y$ -axis at $-3$ .

Plot the vertex, then mirror

Symmetry does half your work. Plot the vertex first. Then pick one or two $x$ -values on one side of the axis, compute their $y$ -values, and plot them. Every point you plot has a free twin the same distance from the axis on the other side, at the same height.

The $y$ -intercept is usually the easiest extra point. For $y = x^2 - 2x - 3$ , the $y$ -intercept $(0, -3)$ sits $1$ unit left of the axis $x = 1$ , so its mirror $(2, -3)$ sits $1$ unit right. That's five points — vertex, two points, two mirrors — which is plenty for a clean parabola.

Worked examples

Example 1: graph from standard form

Find the key features of $y = x^2 - 2x - 3$ and describe its graph.

Identify the coefficients

a = 1, \quad b = -2, \quad c = -3

Axis of symmetry

x = -\dfrac{b}{2a} = -\dfrac{-2}{2(1)} = 1

Vertex: substitute

x = 1

y = (1)^2 - 2(1) - 3 = -4

The

y

-intercept is

c

, at

(0, -3)

; its mirror across

x = 1

(2, -3)

Since

a > 0

, the parabola opens up from the vertex

(1, -4)

Answer: Vertex $(1, -4)$ , axis of symmetry $x = 1$ , opens up, $y$ -intercept $(0, -3)$

Example 2: a parabola that opens down

Find the vertex of $y = -x^2 + 4x - 1$ and tell which way it opens.

Identify the coefficients

a = -1, \quad b = 4, \quad c = -1

Axis of symmetry

x = -\dfrac{b}{2a} = -\dfrac{4}{2(-1)} = 2

Substitute

x = 2

y = -(2)^2 + 4(2) - 1

Simplify

y = -4 + 8 - 1 = 3

Since

a < 0

, the parabola opens down — the vertex is its highest point

Answer: Vertex $(2, 3)$ , opening down

Example 3: graph from vertex form

Find the key features of $y = (x - 3)^2 + 1$ .

Vertex form is

y = a(x - h)^2 + k

with vertex

(h, k)

h = 3, \quad k = 1

Read the vertex directly

(3, 1)

Axis of symmetry is the vertical line through the vertex

x = 3

Extra points:

x = 2

and

x = 4

both give

y = 1 + 1 = 2

— mirror twins

Since

a = 1

is positive, the parabola opens up

Answer: Vertex $(3, 1)$ , axis of symmetry $x = 3$ , opens up

Example 4: a coefficient bigger than 1

Find the vertex of $y = 2x^2 + 8x + 5$ .

Identify the coefficients

a = 2, \quad b = 8, \quad c = 5

Axis of symmetry

x = -\dfrac{b}{2a} = -\dfrac{8}{2(2)} = -2

Substitute

x = -2

y = 2(-2)^2 + 8(-2) + 5

Simplify

y = 8 - 16 + 5 = -3

Answer: Vertex $(-2, -3)$

Try one yourself

Common questions

Why does

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

give the vertex?

The two $x$ -intercepts of a parabola (when they exist) sit at equal distances from the axis of symmetry, and $-\dfrac{b}{2a}$ is exactly the midpoint of the quadratic formula's two solutions. The vertex lies on that mirror line, so its $x$ -coordinate is $-\dfrac{b}{2a}$ — then you substitute back in for the $y$ -coordinate.

How many points do I need to plot?

Five is the standard: the vertex, two points on one side of the axis, and their two mirror images. The $y$ -intercept $(0, c)$ is usually the easiest of those points, and its mirror comes free.

What if the parabola never crosses the

x

-axis?

That's fine — it just means the function has no real zeros, like $y = x^2 + 2$ . You graph it exactly the same way: vertex, axis of symmetry, a few mirrored points. $x$ -intercepts are a bonus feature, not a requirement for graphing.

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Axis of Symmetry

What a, b, and c tell you

The vertex and the axis of symmetry

Plot the vertex, then mirror

Worked examples

Example 1: graph from standard form

Example 2: a parabola that opens down

Example 3: graph from vertex form

Example 4: a coefficient bigger than 1

Try one yourself

Common questions

Keep learning

Quadratic Formula

Completing the Square

Factoring Trinomials (a = 1)