Quadratic Functions and Graphing Techniques

A quadratic function is a polynomial of degree 2. Its graph is a parabola, a U-shaped curve that opens upward or downward depending on the leading coefficient. Rewriting the function in standard form reveals the vertex directly and makes graphing straightforward.

Quick Reference

Concept	Formula
General form	$P (x) = a x^{2} + b x + c$ , $a \neq 0$
Standard form	$P (x) = a (x - h)^{2} + k$
Vertex	$(h, k)$ where $h = - \frac{b}{2 a}$ , $k = P (h)$
Opens upward	$a > 0$ (vertex is minimum)
Opens downward	$a < 0$ (vertex is maximum)
$y$ -intercept	$(0, c)$
$x$ -intercepts	Solve $a x^{2} + b x + c = 0$

Definition

A polynomial function of degree 2 is called a quadratic function. It has the general form

P (x) = a x^{2} + b x + c where a \neq 0.

Quadratic functions produce parabolic graphs: U-shaped curves that open upward when $a > 0$ and downward when $a < 0$ .

Standard Form and Completing the Square

The standard form of a quadratic function is

P (x) = a (x - h)^{2} + k .

Here $(h, k)$ is the vertex of the parabola: the lowest point when $a > 0$ , and the highest point when $a < 0$ .

Expressing a quadratic in standard form has two main advantages: it makes the graph easier to sketch, and it reveals the vertex immediately.

To convert from general form to standard form, we use a process called completing the square:

How to complete the square:

Start with $P (x) = a x^{2} + b x + c$ .
Factor out $a$ from the quadratic and linear terms: $P (x) = a (x^{2} + \frac{b}{a} x) + c$ .
Inside the parentheses, add and subtract ${(\frac{b}{2 a})}^{2}$ , the square of half the coefficient of $x$ .
Regroup to write the perfect square trinomial as a binomial square, and adjust the constant outside.

Rewrite $P (x) = 2 x^{2} - 8 x + 5$ in standard form.

Solution

Factor 2 from the first two terms:

P (x) = 2 (x^{2} - 4 x) + 5

Complete the square inside the parentheses by adding and subtracting 4:

= 2 (x^{2} - 4 x + 4 - 4) + 5 = 2 ((x - 2)^{2} - 4) + 5

Distribute the 2:

= 2 (x - 2)^{2} - 8 + 5 = 2 (x - 2)^{2} - 3

The standard form is

P (x) = 2 (x - 2)^{2} - 3

, and the vertex is

(2, - 3)

Graphing from Standard Form

The graph of $P (x) = a (x - h)^{2} + k$ is obtained from the basic parabola $f (x) = x^{2}$ by four transformations:

Horizontal shift: Replace $x$ with $x - h$ to shift the graph right by $h$ units (or left if $h < 0$ ).
Vertical stretch or compression: Multiplying by $| a |$ makes the parabola narrower when $| a | > 1$ and wider when $0 < | a | < 1$ .
Reflection: If $a < 0$ , the parabola flips over the $x$ -axis and opens downward.
Vertical shift: Adding $k$ moves the graph up by $k$ units (or down if $k < 0$ ).

Parabola opening downward illustrating P(x) = a(x−h)² + k with a less than 0. — $P (x) = a (x - h)^{2} + k$ with $a < 0$ : the parabola opens downward and $(h, k)$ is the maximum point.

Parabola opening upward illustrating P(x) = a(x−h)² + k with a greater than 0. — $P (x) = a (x - h)^{2} + k$ with $a > 0$ : the parabola opens upward and $(h, k)$ is the minimum point.

Graph $P (x) = 3 (x - 1)^{2} + 4$ starting from $f (x) = x^{2}$ .

Solution

Shift the graph of $f$ one unit to the right (because $h = 1$ ).
Stretch vertically by a factor of 3, making the parabola narrower.
Shift the result 4 units upward (because $k = 4$ ).

The vertex is $(1, 4)$, the parabola opens upward, and the minimum value is 4.

Why $(h, k)$ Is the Vertex

The squared term $(x - h)^{2}$ is always non-negative. Its smallest value is 0, which occurs at $x = h$ . At that point, $P (h) = a \cdot 0 + k = k$ .

If $a > 0$ : moving $x$ away from $h$ increases $P (x)$ , so $(h, k)$ is a minimum.
If $a < 0$ : moving $x$ away from $h$ decreases $P (x)$ , so $(h, k)$ is a maximum.

Rewrite $f (x) = 3 x^{2} - 18 x + 27$ in standard form and identify the vertex.

Solution

The standard form is

f (x) = 3 (x - 3)^{2}

and the vertex is $(3, 0)$.

Maximum and Minimum Values

For $P (x) = a x^{2} + b x + c$ , the vertex coordinates are given by:

h = - \frac{b}{2 a}, k = P (- \frac{b}{2 a}) .

The minimum (if $a > 0$ ) or maximum (if $a < 0$ ) value of $P$ is $k$ , attained at $x = h$ .

Find the minimum value of $f (x) = 2 x^{2} - 4 x + 1$ .

Solution

Here

a = 2

and

b = - 4

. The minimum occurs at:

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

The minimum value is:

f (1) = 2 (1)^{2} - 4 (1) + 1 = 2 - 4 + 1 = - 1.

The minimum value is

- 1

, attained at

x = 1

Finding Intercepts

To find the $x$ -intercepts, set $P (x) = 0$ and solve. To find the $y$ -intercept, evaluate $P (0) = c$ .

Find the $x$ - and $y$ -intercepts of $P (x) = 2 x^{2} - 4 x - 6$ .

Solution

$y$ -intercept:

P (0) = 2 (0)^{2} - 4 (0) - 6 = - 6

. The

y

-intercept is

(0, - 6)

. $x$ -intercepts: Solve

2 x^{2} - 4 x - 6 = 0

. By the quadratic formula with

a = 2

b = - 4

c = - 6

x = \frac{4 + 8}{4} = 3

and

x = \frac{4 - 8}{4} = - 1

. The

x

-intercepts are $(3, 0)$ and

(- 1, 0)

Frequently Asked Questions

What is the difference between general form and standard form?

General form

P (x) = a x^{2} + b x + c

shows the coefficients directly but hides the vertex. Standard form

P (x) = a (x - h)^{2} + k

reveals the vertex

(h, k)

at a glance and makes transformations easy to read off. Completing the square converts between them.

How do I find the vertex without completing the square?

Use the formula

h = - b / (2 a)

directly from the general form, then compute

k = P (h)

. This is faster when you only need the vertex, but standard form is more useful for graphing.

Can a parabola have no x-intercepts?

Yes. If the discriminant

b^{2} - 4 a c < 0

, the quadratic has no real roots and the parabola does not cross the

x

-axis. It lies entirely above the

x

-axis (if

a > 0

) or below it (if

a < 0

What does the value of a control beyond opening direction?

The absolute value

| a |

controls the "width" of the parabola. Large

| a |

makes the parabola narrow (steeply curved); small

| a |

near zero makes it wide (gently curved). The sign of

a

determines whether it opens up or down.

Is completing the square always the best method for finding the vertex?

Not always. For simple coefficients, the formula

h = - b / (2 a)

is fastest. Completing the square is most useful when you want the full standard form, for example when graphing or solving optimization problems.

Quadratic Functions and Graphing Techniques

Quick Reference

Definition

Standard Form and Completing the Square

Graphing from Standard Form

Why $(h, k)$ Is the Vertex

Maximum and Minimum Values

Finding Intercepts

Frequently Asked Questions

Quick Links

Social Media

Quadratic Functions and Graphing Techniques

Quick Reference

Definition

Standard Form and Completing the Square

Graphing from Standard Form

Why ( h , k ) Is the Vertex

Maximum and Minimum Values

Finding Intercepts

Frequently Asked Questions

Quick Links

Social Media

Why $(h, k)$ Is the Vertex