Graphing Quadratic Functions: Vertex & Axis

Graphing quadratic functions means drawing the U-shaped curve, called a parabola, for $y = ax^2 + bx + c$ . Find the vertex with $x = -\frac{b}{2a}$ , draw the axis of symmetry through it, and plot the y-intercept and its mirror point. If $a \gt 0$ the parabola opens up; if $a \lt 0$ it opens down.

Before this lesson: You should know function notation, key features of functions and multiplying binomials.

What Does a Quadratic Function Look Like?

A quadratic function has an $x^2$ term and no higher power. Its standard form is $y = ax^2 + bx + c$ , where $a \ne 0$ .

Its graph is a parabola, a U shape. Think of the path of a ball tossed in the air, or the curve of a satellite dish.

Every parabola has a turning point called the vertex. A vertical line through the vertex, the axis of symmetry, splits the parabola into two mirror halves.

The parabola y = x² − 4x + 3 on a grid from −2 to 6. It opens up. The vertex (2, −1) is the lowest point. A dashed green vertical line x = 2 is the axis of symmetry. The y-intercept (0, 3) and its mirror point (4, 3) are marked, and the x-intercepts are (1, 0) and (3, 0). — The vertex (2, −1) sits on the axis of symmetry, x = 2. Points on each side are mirror images.

Key Parts of a Parabola

Feature	How to find it in y = ax² + bx + c
Direction	Opens up if $a \gt 0$ . Opens down if $a \lt 0$ .
Axis of symmetry	$x = -\frac{b}{2a}$
Vertex	Use $x = -\frac{b}{2a}$ , then plug that x into the function to get y.
y-intercept	$(0, c)$ , because $y = c$ when $x = 0$ .
Width	If $\|a\| \gt 1$ , it is narrower than $y = x^2$ . If $\|a\| \lt 1$ , it is wider.

Steps for graphing a quadratic function

Decide if it opens up or down from the sign of $a$ .
Find the axis of symmetry: $x = -\frac{b}{2a}$ .
Plug that x-value in to find the vertex.
Plot the y-intercept $(0, c)$ and its mirror point across the axis.
Plot one or two more points if needed, then draw a smooth U through them.

Worked Examples

Example 1: The parent function y = x²

Graph $y = x^2$ with a table of values.

Step 1. Pick x-values on both sides of 0: $-2, -1, 0, 1, 2$ .

Step 2. Square each x-value to get y (see evaluating algebraic expressions). For example, $(-2)^2 = 4$ and $(-1)^2 = 1$ .

$x$	−2	−1	0	1	2
$y$	4	1	0	1	4

Step 3. Plot the five points and join them with a smooth U-shaped curve.

Step 4. The lowest point is $(0, 0)$ , so that is the vertex. The y-values match on each side: 1 and 1, 4 and 4. So the axis of symmetry is $x = 0$ , the y-axis.

Answer: The graph is a U-shape with vertex $(0, 0)$ and axis of symmetry $x = 0$ .

Example 2: A parabola that opens down

Describe the graph of $y = -2x^2 + 3$ .

Step 1. Find $a$ , $b$ and $c$ : $a = -2$ , $b = 0$ and $c = 3$ .

Step 2. $a$ is negative, so the parabola opens down.

Step 3. Find the axis of symmetry with the formula.

x = -\frac{b}{2a} = -\frac{0}{2(-2)} = 0

Step 4. Substitute $x = 0$ to find the y-value of the vertex.

y = -2(0)^2 + 3 = 0 + 3 = 3

Step 5. The vertex is $(0, 3)$ . It opens down, so it is a maximum.

Step 6. Compare the width. $|-2| = 2$ , and 2 is greater than 1, so the graph is narrower than $y = x^2$ (see absolute value).

Answer: It opens down, has a maximum at $(0, 3)$ and axis $x = 0$ , and is narrower than $y = x^2$ .

Example 3: Find the vertex and graph

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

Step 1. Find $a$ and $b$ : $a = 1$ and $b = -4$ . Use the formula for the axis of symmetry.

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

Step 2. Substitute $x = 2$ to find the y-value of the vertex.

y = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1

Step 3. The vertex is $(2, -1)$ . Since $a = 1$ is positive, it opens up and the vertex is a minimum.

Step 4. Find the y-intercept by substituting $x = 0$ .

y = 0^2 - 4(0) + 3 = 3

Step 5. The y-intercept $(0, 3)$ is 2 units left of the axis $x = 2$ . Its mirror point is 2 units right, at $x = 4$ .

y = 4^2 - 4(4) + 3 = 16 - 16 + 3 = 3

Step 6. Plot $(0, 3)$ , $(2, -1)$ and $(4, 3)$ , and draw the U-shape through them. The graph is shown above.

Answer: Vertex $(2, -1)$ , axis $x = 2$ , y-intercept $(0, 3)$ and mirror point $(4, 3)$ .

Example 4: A negative a with a b-term

Find the vertex, axis and y-intercept of $y = -x^2 - 2x + 8$ .

Step 1. Here $a = -1$ and $b = -2$ . Find the axis of symmetry.

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

Step 2. Substitute $x = -1$ . Square first, then multiply, then add (see order of operations).

y = -(-1)^2 - 2(-1) + 8 = -1 + 2 + 8 = 9

Step 3. The vertex is $(-1, 9)$ . Since $a$ is negative, it opens down and the vertex is a maximum.

Step 4. Find the y-intercept by substituting $x = 0$ .

y = -0^2 - 2(0) + 8 = 8

Step 5. $(0, 8)$ is 1 unit right of the axis $x = -1$ , so its mirror point is 1 unit left, at $x = -2$ .

y = -(-2)^2 - 2(-2) + 8 = -4 + 4 + 8 = 8

Step 6. The graph crosses the x-axis at $x = -4$ and $x = 2$ . Check each one: the y-value is 0.

\begin{aligned} -(-4)^2 - 2(-4) + 8 &= -16 + 8 + 8 = 0 \\ -(2)^2 - 2(2) + 8 &= -4 - 4 + 8 = 0 \end{aligned}

Answer: Vertex $(-1, 9)$ (a maximum), axis $x = -1$ , y-intercept $(0, 8)$ , mirror point $(-2, 8)$ , and x-intercepts $x = -4$ and $x = 2$ .

Example 5: Word problem

Water from a fountain follows the path $h = -2x^2 + 8x$ , where $x$ is the distance from the spout and $h$ is the height, both in feet. How high does the water go? Where does it land?

Step 1. Here $a = -2$ and $b = 8$ . The highest point is the vertex. Find its x-value.

x = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2

Step 2. Substitute $x = 2$ to find the height.

h = -2(2)^2 + 8(2) = -2(4) + 16 = -8 + 16 = 8

Step 3. The vertex is $(2, 8)$ . The water starts at $(0, 0)$ , 2 units left of the axis. By symmetry it lands 2 units right of the axis, at $x = 4$ .

Step 4. Check that the height is 0 at $x = 4$ .

-2(4)^2 + 8(4) = -32 + 32 = 0

Answer: The water reaches 8 feet, 2 feet from the spout. It lands 4 feet from the spout.

Common Mistakes

Mistake	Example of the error	Fix
Dropping the negative in the formula	For $y = x^2 - 4x + 3$ : $x = \frac{-4}{2} = -2$	Use $x = -\frac{b}{2a} = -\frac{-4}{2} = 2$ .
Using the x-value as the vertex	Vertex of $y = x^2 - 4x + 3$ is 2	The vertex is a point. Plug in to find y: (2, −1).
Squaring a negative wrong	$-(-1)^2 = 1$	Square first, then apply the minus: $-(-1)^2 = -1$ .
Wrong direction	$y = -x^2 + 4$ opens up	$a \lt 0$ , so it opens down.
Connecting points with straight lines	A V shape instead of a U	Draw a smooth curve that is rounded at the vertex.

How a Changes the Graph

$y = ax^2$	Direction	Shape compared with y = x²
$y = 3x^2$	Up	Narrower
$y = \tfrac{1}{2}x^2$	Up	Wider
$y = -x^2$	Down	Same width, flipped
$y = -4x^2$	Down	Narrower, flipped

The y-intercept and vertex give you the key features of functions right away: the maximum or minimum, where the graph increases and decreases, and the range. Vertex form, the next lesson, shows the vertex without any formula.

Practice Problems

Find the vertex and axis of symmetry of $y = x^2 + 2x - 3$ .
Find the vertex of $y = x^2 - 6x + 5$ . Does it open up or down?
Find the vertex of $y = -x^2 + 4x$ . Is it a maximum or a minimum?
Find the vertex of $y = 2x^2 - 8x + 1$ .
Find the vertex and y-intercept of $y = -3x^2 + 6x + 1$ .
Describe the graph of $y = x^2 - 9$ : direction, vertex and x-intercepts.
For $y = x^2 + 2x - 3$ , the y-intercept is $(0, -3)$ . Find its mirror point.
Which is narrower, $y = 5x^2$ or $y = \tfrac{1}{5}x^2$ ?
A parabola has vertex $(3, -2)$ and passes through $(1, 6)$ . Name another point on it.
A ball’s height is $h = -16t^2 + 32t + 4$ feet after $t$ seconds. Find its greatest height and when it happens.

Show answers

Find the axis with $a = 1$ and $b = 2$ : $x = -\frac{2}{2(1)} = -1$
Substitute $x = -1$ : $y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$
Vertex $(-1, -4)$ , axis $x = -1$ .
Find the axis with $a = 1$ and $b = -6$ : $x = -\frac{-6}{2(1)} = 3$
Substitute $x = 3$ : $y = 3^2 - 6(3) + 5 = 9 - 18 + 5 = -4$
Vertex $(3, -4)$ . It opens up, since $a = 1 \gt 0$ .
Find the axis with $a = -1$ and $b = 4$ : $x = -\frac{4}{2(-1)} = 2$
Substitute $x = 2$ : $y = -(2)^2 + 4(2) = -4 + 8 = 4$
Vertex $(2, 4)$ . It is a maximum, since $a \lt 0$ .
Find the axis with $a = 2$ and $b = -8$ : $x = -\frac{-8}{2(2)} = 2$
Substitute $x = 2$ : $y = 2(2)^2 - 8(2) + 1 = 8 - 16 + 1 = -7$
Vertex $(2, -7)$ .
Find the axis with $a = -3$ and $b = 6$ : $x = -\frac{6}{2(-3)} = 1$
Substitute $x = 1$ : $y = -3(1)^2 + 6(1) + 1 = -3 + 6 + 1 = 4$
Substitute $x = 0$ for the y-intercept: $y = -3(0)^2 + 6(0) + 1 = 1$
Vertex $(1, 4)$ , y-intercept $(0, 1)$ .
$a = 1 \gt 0$ , so it opens up. With $b = 0$ , the axis is $x = 0$
Substitute $x = 0$ : $y = 0 - 9 = -9$ . The vertex is $(0, -9)$
For x-intercepts, set $y = 0$ : $x^2 - 9 = 0$
Add 9 to both sides: $x^2 = 9$
Both 3 and $-3$ square to 9 (see square roots): $x = 3 \text{ or } x = -3$
Opens up, vertex $(0, -9)$ , x-intercepts $(-3, 0)$ and $(3, 0)$ .
From problem 1, the axis is $x = -1$
$(0, -3)$ is 1 unit right of the axis, so the mirror point is 1 unit left, at $x = -2$
Check it: $(-2)^2 + 2(-2) - 3 = 4 - 4 - 3 = -3$
The mirror point is $(-2, -3)$ .
Compare $|a|$ : $|5| = 5$ and $|\tfrac{1}{5}| = \tfrac{1}{5}$
The larger $|a|$ gives the narrower graph
$y = 5x^2$ is narrower.
The axis goes through the vertex, so it is $x = 3$
$(1, 6)$ is $3 - 1 = 2$ units left of the axis
Its mirror point is 2 units right, at $x = 3 + 2 = 5$ , with the same y-value
$(5, 6)$ is also on the parabola.
The greatest height is at the vertex. Find its t-value with $a = -16$ and $b = 32$ : $t = -\frac{32}{2(-16)} = -\frac{32}{-32} = 1$
Substitute $t = 1$ : $h = -16(1)^2 + 32(1) + 4 = -16 + 32 + 4 = 20$
The greatest height is 20 feet, at 1 second.

FAQ

How do you find the vertex of a quadratic function?

Use $x = -\frac{b}{2a}$ to find the x-coordinate. Then substitute it into the function to find the y-coordinate.

What is the axis of symmetry?

It is the vertical line through the vertex, $x = -\frac{b}{2a}$ . The parabola is a mirror image on each side of it.

How do you know if a parabola opens up or down?

Look at $a$ , the number in front of $x^2$ . If it is positive, the parabola opens up. If it is negative, it opens down.

What is the y-intercept of a quadratic?

It is $(0, c)$ , the constant term. For $y = x^2 - 4x + 3$ , the y-intercept is $(0, 3)$ .

Summary

A quadratic $y = ax^2 + bx + c$ graphs as a parabola.
$a \gt 0$ : opens up, with a minimum. $a \lt 0$ : opens down, with a maximum.
Axis of symmetry: $x = -\frac{b}{2a}$ . Plug in to get the vertex.
The y-intercept is $(0, c)$ . Reflect it across the axis for a second point.
Draw a smooth U, not a V.

Previous lesson: Rationalizing the Denominator

Next lesson: Vertex Form

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