A2.2 Find a quadratic maximum or minimum algebraically

Ontario Grade 11 Mathematics

Characteristics of Functions

MCR3U – A2.2 | Quadratic Functions

Every time you throw a ball in the air, track profit over time, or design an arch, the situation follows a curved path called a parabola. The highest or lowest point on that curve — called the vertex — holds the maximum or minimum value of the function. In this lesson you will learn how to find that exact point using only algebra, without graphing. The key tool is a technique called completing the square, which rewrites a quadratic function in a form that makes the vertex easy to read off directly.

What you will learn

Complete the square to convert a quadratic function from standard form to vertex form.
Identify the vertex of a quadratic function from its vertex form.
State whether the vertex represents a maximum or a minimum value, and explain why.
Apply the vertex to solve a real-world problem involving an optimal value.

Prerequisite Bridge: Standard Form and Vertex Form

A quadratic function written as

f(x) = ax^2 + bx + c

is in standard form. You worked with this in Grade 10. The values

a

,

b

, and

c

are constants, and

a \neq 0

.

The same function can also be written as

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

, which is called vertex form. In vertex form the vertex of the parabola is the point

(h, k)

. The vertex is either the highest point (maximum) or the lowest point (minimum) of the parabola.

The sign of

a

tells you which one it is. When

a > 0

the parabola opens upward, so the vertex is the lowest point and

k

is the minimum value. When

a < 0

the parabola opens downward, so the vertex is the highest point and

k

is the maximum value. This lesson focuses on converting from standard form to vertex form so you can find

h

and

k

algebraically.

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

Standard form: $f(x) = ax^2 + bx + c$
Vertex form: $f(x) = a(x - h)^2 + k$ , vertex at $(h, k)$
If $a > 0$ : parabola opens up, vertex is a minimum.
If $a < 0$ : parabola opens down, vertex is a maximum.

The Core Method: Completing the Square

Completing the square is a step-by-step algebraic process that rewrites

ax^2 + bx + c

in vertex form. Each step is reversible, so the rewritten function is exactly equal to the original — nothing changes except how it looks.

Here is the general process. First, if

a \neq 1

, factor

a

out of the first two terms only, leaving

c

outside the brackets. Second, look at the coefficient of

x

inside the brackets. Take half of it, then square that result. Add and subtract that square inside the brackets so the overall expression stays equal. Third, the first three terms inside the brackets now form a perfect-square trinomial, which you can write as a squared binomial. Finally, multiply the subtracted value by

a

and combine it with

c

to get

k

.

A perfect-square trinomial has the pattern

x^2 + 2mx + m^2 = (x + m)^2

. Recognising this pattern is the heart of the method. You are engineering that pattern by choosing exactly the right number to add and subtract.

x^2 + 2mx + m^2 = (x + m)^2

Factor $a$ from the first two terms before completing the square.
Half the $x$ -coefficient, then square it — that is the number you add and subtract.
Adding and subtracting the same value keeps the expression equivalent.
Group the perfect-square trinomial and write it as a squared binomial.
Collect remaining constants to find $k$ .

Reading the Vertex and Stating the Optimal Value

Once you have vertex form

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

, reading the vertex is straightforward. The

x

-coordinate of the vertex is

h

, which is the value that makes the bracket equal to zero. The

y

-coordinate is

k

, which is the value of

f(x)

at that point.

Be careful with the sign of

h

. If vertex form shows

(x - 3)^2

, then

h = 3

. If it shows

(x + 3)^2

, rewrite it as

(x - (-3))^2

, so

h = -3

.

The optimal value is simply

k

. You state the maximum or minimum value of the function as

k

, and it occurs when

x = h

. In a word problem you may also need to interpret what

h

and

k

represent in context — for example, the time at which height is greatest, and what that greatest height is.

Vertex is $(h, k)$ ; optimal value is $k$ .
Watch the sign: $(x + p)^2$ means $h = -p$ .
State whether $k$ is a maximum or minimum based on the sign of $a$ .
In context problems, include units and interpret both coordinates.

Connecting Vertex Form Back to Standard Form

It is good practice to verify your algebra by expanding vertex form and checking that you recover the original standard form. Expand

a(x - h)^2 + k

and collect like terms. If the result matches

ax^2 + bx + c

, your completing-the-square work is correct.

You can also use the vertex coordinates as a partial check: substitute

x = h

into the original standard-form function and confirm that the output equals

k

. This is a fast, reliable arithmetic check that catches sign errors.

For example, if completing the square gives vertex

(2, -5)

, substitute

x = 2

into the original function. The output should be exactly

-5

. If it is not, re-examine every step for a sign or arithmetic error.

Expand vertex form to verify it matches the original standard form.
Substitute $x = h$ into the original function; the result must equal $k$ .
This two-step check catches most completing-the-square errors.

Summary: Reading the Vertex from Vertex Form

Feature	When a > 0 (opens up)	When a < 0 (opens down)
Direction of parabola	Upward	Downward
Vertex is a …	Minimum	Maximum
Optimal value	$k$ is the minimum value	$k$ is the maximum value
Where it occurs	At $x = h$	At $x = h$
Value of $(x - h)^2$ at vertex	$0$ (smallest possible)	$0$ (makes $k$ as large as possible)

Worked example

Example 1 — Finding a Minimum Value (a > 0)

Find the minimum value of

f(x) = x^2 - 6x + 11

algebraically. State the vertex and explain what it tells you.

Check the leading coefficient
The coefficient of $x^2$ is $1$ , so $a = 1 > 0$ . The parabola opens upward, meaning the vertex will be a minimum. Because $a = 1$ , there is no need to factor anything out before completing the square.
$a = 1 > 0$
Identify the x-term coefficient and halve it
The coefficient of $x$ is $-6$ . Half of $-6$ is $-3$ . Squaring $-3$ gives $9$ . This value, $9$ , is what you will add and subtract inside the expression.
$(\frac{-6}{2})^2 = (-3)^2 = 9$
Add and subtract 9 to keep the expression equivalent
Write the function with $+9$ and $-9$ inserted after the $x$ -term. Adding and subtracting the same number changes nothing about the value of the expression — it is the same as adding zero.
$f(x) = x^2 - 6x + 9 - 9 + 11$
Factor the perfect-square trinomial
The first three terms $x^2 - 6x + 9$ form a perfect-square trinomial. They factor as $(x - 3)^2$ . Collect the remaining constants: $-9 + 11 = 2$ .
$f(x) = (x - 3)^2 + 2$
Read the vertex and state the minimum
Vertex form is $f(x) = (x - 3)^2 + 2$ , so $h = 3$ and $k = 2$ . The vertex is $(3, 2)$ . Because $a > 0$ , this vertex is a minimum. The minimum value of the function is $2$ , and it occurs when $x = 3$ .
$(h, k) = (3, 2)$

Answer: The minimum value of

f(x)

is

2

, occurring at

x = 3

. The vertex is

(3, 2)

.

Check: Substitute

x = 3

into the original function:

f(3) = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2

. This matches

k = 2

, confirming the answer.

Worked example

Example 2 — Finding a Maximum in Context (a < 0)

A model rocket is launched from the ground. Its height in metres after

t

seconds is given by

h(t) = -5t^2 + 40t

. Find the maximum height the rocket reaches and when it reaches that height.

Check the leading coefficient
The coefficient of $t^2$ is $-5$ , so $a = -5 < 0$ . The parabola opens downward, so the vertex will be a maximum — exactly what is needed to find the greatest height.
$a = -5 < 0$
Factor out -5 from the first two terms
There is no constant term $c$ here, but the process is the same. Factor $-5$ from $-5t^2 + 40t$ . Dividing each term inside by $-5$ : $-5t^2 \div (-5) = t^2$ and $40t \div (-5) = -8t$ .
$h(t) = -5(t^2 - 8t)$
Find the value to add and subtract inside the brackets
The coefficient of $t$ inside the brackets is $-8$ . Half of $-8$ is $-4$ . Squaring $-4$ gives $16$ . Add and subtract $16$ inside the brackets.
$(\frac{-8}{2})^2 = (-4)^2 = 16$
Insert +16 and -16 inside the brackets
Write the expression with $+16$ and $-16$ placed inside the brackets. The overall value has not changed because $+16 - 16 = 0$ .
$h(t) = -5(t^2 - 8t + 16 - 16)$
Separate the subtracted term and factor the trinomial
Bring $-16$ outside the brackets by multiplying it by $-5$ : $-5 \times (-16) = +80$ . The remaining trinomial $t^2 - 8t + 16$ factors as $(t - 4)^2$ .
$h(t) = -5(t - 4)^2 + 80$
Read the vertex and answer in context
Vertex form is $h(t) = -5(t - 4)^2 + 80$ , giving vertex $(4, 80)$ . So $h = 4$ and $k = 80$ . Because $a < 0$ , this is a maximum. The rocket reaches a maximum height of $80$ metres at $t = 4$ seconds.
$(h, k) = (4, 80)$

Answer: The rocket reaches a maximum height of

80

metres at

t = 4

seconds.

Check: Substitute

t = 4

into the original function:

h(4) = -5(4)^2 + 40(4) = -5(16) + 160 = -80 + 160 = 80

. This matches

k = 80

, confirming the answer.

Common mistakes and how to avoid them

Forgetting to multiply the added/subtracted constant by

a

when

a \neq 1

. For example, adding

16

inside the brackets of

-5(t^2 - 8t + 16 - 16)

but writing

+16

instead of

+80

when moving the term outside.

Correction: Whatever you add inside the brackets must be multiplied by the factor outside when you bring it out. If you add

16

inside

-5(\ldots)

, the equivalent outside the bracket is

-5 \times 16 = -80

, so you add

+80

to compensate for the

-80

carried outside.

Getting the sign of

h

wrong. Students often read

(x - h)^2

and write

h

as negative when the bracket shows subtraction.

Correction: Set the bracket equal to zero to find

h

. For

(x - 3)^2

:

x - 3 = 0

gives

x = 3

, so

h = 3

. For

(x + 3)^2 = (x - (-3))^2

:

x + 3 = 0

gives

x = -3

, so

h = -3

.

Adding the completing-the-square value without also subtracting it, changing the value of the expression.

Correction: Always add and subtract the same number in the same step. Adding

9

and subtracting

9

is the same as adding zero, so the function stays equivalent.

Concluding that the vertex is a maximum just because the value of

k

is large, or a minimum because

k

is small.

Correction: The type of optimal value (maximum or minimum) depends only on the sign of

a

, not on the size of

k

. Always check whether

a > 0

or

a < 0

first.

Skipping the check step and not verifying by substituting

x = h

into the original function.

Correction: Always substitute

x = h

back into the original standard-form function. If the output is not equal to

k

, there is an error in the completing-the-square steps.

Lesson summary

Completing the square converts $f(x) = ax^2 + bx + c$ into vertex form $f(x) = a(x - h)^2 + k$ , revealing the vertex $(h, k)$ algebraically.
Factor $a$ from the first two terms, then add and subtract the square of half the inner $x$ -coefficient.
When you move the subtracted constant outside the brackets, multiply it by $a$ .
If $a > 0$ the vertex is a minimum; if $a < 0$ the vertex is a maximum. The optimal value is always $k$ .
Verify every answer by substituting $x = h$ into the original function and confirming the output equals $k$ .
In context problems, interpret both $h$ (the input at which the optimal value occurs) and $k$ (the optimal value itself) using the units of the problem.

Check your understanding

Question 1

What is the vertex of

f(x) = 2(x - 5)^2 - 3

, and does it represent a maximum or a minimum?

Vertex $(5, -3)$ ; minimum because $a = 2 > 0$
Vertex $(-5, -3)$ ; minimum because $a = 2 > 0$
Vertex $(5, -3)$ ; maximum because $a = 2 > 0$
Vertex $(5, 3)$ ; minimum because $a = 2 > 0$

Show answer and explanation

Vertex

(5, -3)

; minimum because

a = 2 > 0

In

f(x) = 2(x - 5)^2 - 3

, we have

h = 5

and

k = -3

, giving vertex

(5, -3)

. Since

a = 2 > 0

the parabola opens upward, so the vertex is a minimum.

Question 2

Which step comes immediately after factoring out

a

from the first two terms when completing the square?

Factor the resulting trinomial as a perfect square.
Add and subtract the square of half the inner $x$ -coefficient.
Expand the expression to check it.
Move the constant $c$ inside the brackets.

Show answer and explanation

Add and subtract the square of half the inner

x

-coefficient.

After factoring out

a

, you look at the coefficient of

x

inside the brackets, take half of it, square the result, and then add and subtract that square inside the brackets. This is what creates the perfect-square trinomial.

Question 3

Complete the square to find the maximum or minimum value of

g(x) = x^2 + 4x - 1

.

Minimum value of $-5$ at $x = -2$
Minimum value of $-1$ at $x = 4$
Maximum value of $-5$ at $x = -2$
Minimum value of $3$ at $x = 2$

Show answer and explanation

Minimum value of

-5

at

x = -2

Half of

4

is

2

; squaring gives

4

. So

g(x) = x^2 + 4x + 4 - 4 - 1 = (x + 2)^2 - 5

. Vertex is

(-2, -5)

. Since

a = 1 > 0

, this is a minimum. Check:

g(-2) = (-2)^2 + 4(-2) - 1 = 4 - 8 - 1 = -5

. ✓

Question 4

A function is written as

f(x) = -3(x + 1)^2 + 7

after completing the square. What is the maximum value of

f(x)

and at what

x

does it occur?

Maximum value of $7$ at $x = 1$
Maximum value of $-3$ at $x = -1$
Maximum value of $7$ at $x = -1$
Minimum value of $7$ at $x = -1$

Show answer and explanation

Maximum value of

7

at

x = -1

Written as

f(x) = -3(x - (-1))^2 + 7

, the vertex is

(-1, 7)

. Since

a = -3 < 0

the parabola opens downward, making the vertex a maximum. The maximum value is

7

and it occurs at

x = -1

.

Key terms

Standard form: A quadratic function written as $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants and $a \neq 0$ .
Vertex form: A quadratic function written as $f(x) = a(x - h)^2 + k$ , where the vertex of the parabola is the point $(h, k)$ .
Vertex: The point on a parabola where it changes direction — the highest point if the parabola opens down, or the lowest point if it opens up.
Completing the square: An algebraic technique that rewrites a quadratic expression in standard form as a perfect-square binomial plus a constant, revealing the vertex directly.
Perfect-square trinomial: A three-term polynomial that can be written as the square of a binomial, following the pattern $x^2 + 2mx + m^2 = (x + m)^2$ .
Maximum value: The greatest output value of a function. For a downward-opening parabola, it equals $k$ in vertex form.
Minimum value: The smallest output value of a function. For an upward-opening parabola, it equals $k$ in vertex form.
Optimal value: A general term for either the maximum or the minimum value of a function, depending on context.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation A2.2. It is a study resource, not an official curriculum publication.

Official curriculum reference

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A2.2 · Find a quadratic maximum or minimum algebraically

Ontario Grade 11 Mathematics

MCR3U – A2.2 | Quadratic Functions

What you will learn

Prerequisite Bridge: Standard Form and Vertex Form

The Core Method: Completing the Square

Reading the Vertex and Stating the Optimal Value

Connecting Vertex Form Back to Standard Form

Summary: Reading the Vertex from Vertex Form

Example 1 — Finding a Minimum Value (a > 0)

Example 2 — Finding a Maximum in Context (a < 0)

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

About this lesson