A2.4 Determine a quadratic function from its roots and a point

Ontario Grade 11 Mathematics

Characteristics of Functions

MCR3U – Quadratic Functions | Expectation A2.4

When you know where a parabola crosses the x-axis, you already know a great deal about its equation. Those crossing points are called the roots of the quadratic function. However, knowing only the roots is not enough to write a unique equation — there are infinitely many parabolas that share the same two x-intercepts. Adding one extra point on the curve removes all ambiguity. In this lesson you will build a quadratic function step by step: start with the roots, set up the factored form, and then use the extra point to lock in the one missing value. Every step uses skills from Grade 10, so before diving in, a quick review of factored form will get everything in place.

What you will learn

Write a quadratic function in factored form using two given roots.
Use a known point on the parabola to find the value of the stretch/compression factor a.
Express the final quadratic function in both factored form and standard form.
Recognise that infinitely many parabolas share the same two roots, and that one additional point pins down exactly one function.

Prerequisite Bridge: Factored Form of a Quadratic

In Grade 10 you learned that a quadratic function can be written in factored form as

f(x) = a(x - r)(x - s)

, where

r

and

s

are the two x-intercepts (roots) of the function, and

a

is a real number that controls whether the parabola opens upward or downward and how wide or narrow it is.

To find the roots from factored form, you set each factor equal to zero:

x - r = 0

gives

x = r

, and

x - s = 0

gives

x = s

. This lesson reverses that process — you are given

r

and

s

first, and you must figure out

a

.

The value

a

is sometimes called the leading coefficient or the vertical stretch/compression factor. When

a > 0

the parabola opens upward; when

a < 0

it opens downward. Different values of

a

produce different parabolas, all passing through

x = r

and

x = s

. That is why a single extra point is needed to determine

a

uniquely.

f(x) = a(x - r)(x - s)

Factored form: $f(x) = a(x - r)(x - s)$ , where $r$ and $s$ are the roots.
Setting each factor to zero recovers the roots: $x = r$ and $x = s$ .
The factor $a$ cannot be determined from the roots alone.
One additional point that lies on the parabola provides an equation that can be solved for $a$ .

Why One Extra Point Is Enough

Picture two roots marked on a number line, say at

x = 1

and

x = 5

. You can draw a narrow parabola through those two points, or a wide one, or one that opens downward — all of them cross the x-axis at exactly

x = 1

and

x = 5

. Each of these different parabolas corresponds to a different value of

a

in

f(x) = a(x - 1)(x - 5)

.

Now suppose you are also told that the parabola passes through the point

(3, -8)

. There is only one value of

a

that makes this true. You substitute

x = 3

and

f(x) = -8

into the factored form and solve the resulting equation. Once

a

is found, the quadratic function is completely determined.

This idea is powerful: two roots give you the shape of the skeleton, and one more point gives you the exact size and direction. Together, three pieces of information uniquely identify one quadratic function.

Two roots alone correspond to infinitely many parabolas (one for each value of $a$ ).
A third point that is not an x-intercept creates a solvable equation for $a$ .
Substituting the known point into $f(x) = a(x - r)(x - s)$ and solving gives $a$ .
Once $a$ is known, the quadratic function is unique.

Step-by-Step Method

The process has four clean steps that you will use every time. Step 1 — Write the factored form skeleton. Place the two roots

r

and

s

into

f(x) = a(x - r)(x - s)

. Leave

a

as an unknown for now.

Step 2 — Substitute the known point. If the parabola passes through the point

(x_1, y_1)

, replace

x

with

x_1

and

f(x)

with

y_1

in your factored form. You now have a single equation in one unknown,

a

.

Step 3 — Solve for

a

. The right-hand side after substitution will be a number multiplied by

a

. Divide both sides by that number to isolate

a

.

Step 4 — Write the final function. Substitute the value of

a

back into

f(x) = a(x - r)(x - s)

. If the question asks for standard form, expand and collect like terms.

a = \frac{y_1}{(x_1 - r)(x_1 - s)}

Step 1: Write $f(x) = a(x - r)(x - s)$ using the given roots.
Step 2: Substitute the given point $(x_1, y_1)$ to get an equation in $a$ .
Step 3: Solve the equation to find the numerical value of $a$ .
Step 4: Write the complete factored form; expand to standard form if needed.

Connecting Factored Form to Standard Form

Once you have

f(x) = a(x - r)(x - s)

with a numerical value of

a

, you can expand it into standard form

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

. To expand, first multiply the two binomials

(x - r)(x - s)

using the distributive property (FOIL):

(x - r)(x - s) = x^2 - (r + s)x + rs

. Then multiply every term by

a

.

Standard form makes it easy to read off the y-intercept directly as the constant term

c

. Factored form makes it easy to read off the roots. Both forms describe the same function — they are simply two different ways of writing it. For this expectation, factored form is the natural starting point, but you should be comfortable converting to standard form when a question asks for it.

As a quick check after expanding, verify that the roots still work: substitute each root into your standard-form equation and confirm you get zero. Also substitute the given extra point to confirm the y-value matches.

(x - r)(x - s) = x^2 - (r + s)x + rs

$(x - r)(x - s) = x^2 - (r+s)x + rs$ after expanding.
Multiply through by $a$ to reach standard form $ax^2 + bx + c$ .
The constant term $c = a \cdot r \cdot s$ equals the y-intercept.
Always verify your answer by substituting the roots and the given point back into the final equation.

Checking Your Answer

A correct quadratic function built from roots

r

and

s

and a point

(x_1, y_1)

must satisfy three conditions simultaneously:

f(r) = 0

,

f(s) = 0

, and

f(x_1) = y_1

. Checking all three takes less than two minutes and immediately reveals any arithmetic slip.

If either root check fails, revisit whether you read the roots correctly and placed the correct signs in the factors. Remember that a root

x = 3

gives the factor

(x - 3)

, not

(x + 3)

. A sign error here is the single most common mistake in this topic.

If the point check fails but the root checks pass, the error is almost always in the arithmetic when solving for

a

. Re-substitute the point carefully, paying attention to negative numbers inside the factors.

Check $f(r) = 0$ and $f(s) = 0$ to confirm the roots are correct.
Check $f(x_1) = y_1$ to confirm $a$ was solved correctly.
A root $x = r$ produces the factor $(x - r)$ , not $(x + r)$ — watch the sign.
Three successful checks mean the function is fully verified.

How the Value of a Changes the Parabola (same roots x = 2 and x = 6)

Value of a	Direction parabola opens	Factored form	Passes through which point?
$a = 1$	Upward	$f(x) = (x-2)(x-6)$	$(4, -4)$
$a = -1$	Downward	$f(x) = -(x-2)(x-6)$	$(4, 4)$
$a = 2$	Upward, narrower	$f(x) = 2(x-2)(x-6)$	$(4, -8)$
$a = -3$	Downward, narrower	$f(x) = -3(x-2)(x-6)$	$(4, 12)$

Worked example

Example 1 – Roots Are Both Positive, Point Is Above the x-axis

A parabola has x-intercepts at

x = 2

and

x = 6

, and it passes through the point

(4, 12)

. Determine the quadratic function in both factored form and standard form.

Write the factored form skeleton
The roots are $r = 2$ and $s = 6$ . Substitute these into $f(x) = a(x - r)(x - s)$ to get the skeleton. The value $a$ is still unknown.
$f(x) = a(x - 2)(x - 6)$
Substitute the known point
The parabola passes through $(4, 12)$ , so $x = 4$ and $f(x) = 12$ . Replace $x$ with $4$ and $f(x)$ with $12$ in the skeleton equation.
$12 = a(4 - 2)(4 - 6)$
Simplify the right-hand side
Calculate each factor: $4 - 2 = 2$ and $4 - 6 = -2$ . Multiply: $2 \times (-2) = -4$ . The equation becomes $12 = a(-4)$ .
$12 = -4a$
Solve for a
Divide both sides by $-4$ to isolate $a$ .
$a = \frac{12}{-4} = -3$
Write the factored form
Substitute $a = -3$ back into the skeleton. This is the factored form of the function.
$f(x) = -3(x - 2)(x - 6)$
Expand to standard form
First expand the binomials: $(x-2)(x-6) = x^2 - 8x + 12$ . Then multiply every term by $-3$ .
$f(x) = -3x^2 + 24x - 36$

Answer: Factored form:

f(x) = -3(x - 2)(x - 6)

. Standard form:

f(x) = -3x^2 + 24x - 36

.

Check: Root check:

f(2) = -3(0)(-4) = 0

✓ and

f(6) = -3(4)(0) = 0

✓. Point check:

f(4) = -3(4-2)(4-6) = -3(2)(-2) = 12

✓. All three conditions are satisfied, so the answer is correct. Notice

a = -3 < 0

, meaning the parabola opens downward, which is consistent with the given point

(4, 12)

sitting above the x-axis between the two roots.

Worked example

Example 2 – One Negative Root, Point Below the x-axis

A quadratic function has zeros at

x = -1

and

x = 4

. The graph passes through the point

(2, -6)

. Find the quadratic function in factored form and in standard form.

Write the factored form skeleton
The roots are $r = -1$ and $s = 4$ . The factor for root $x = -1$ is $(x - (-1)) = (x + 1)$ , and the factor for root $x = 4$ is $(x - 4)$ . Write the skeleton with unknown $a$ .
$f(x) = a(x + 1)(x - 4)$
Substitute the known point
The curve passes through $(2, -6)$ , so substitute $x = 2$ and $f(x) = -6$ .
$-6 = a(2 + 1)(2 - 4)$
Simplify the right-hand side
Calculate each factor: $2 + 1 = 3$ and $2 - 4 = -2$ . Multiply: $3 \times (-2) = -6$ . The equation is $-6 = a(-6)$ .
$-6 = -6a$
Solve for a
Divide both sides by $-6$ to isolate $a$ .
$a = \frac{-6}{-6} = 1$
Write the factored form
Substitute $a = 1$ back into the skeleton. When $a = 1$ , the factor is usually written without the coefficient.
$f(x) = (x + 1)(x - 4)$
Expand to standard form
Expand the binomials using the distributive property: $(x+1)(x-4) = x^2 - 4x + x - 4 = x^2 - 3x - 4$ . Since $a = 1$ , no further multiplication is needed.
$f(x) = x^2 - 3x - 4$

Answer: Factored form:

f(x) = (x + 1)(x - 4)

. Standard form:

f(x) = x^2 - 3x - 4

.

Check: Root check:

f(-1) = (-1+1)(-1-4) = (0)(-5) = 0

✓ and

f(4) = (4+1)(4-4) = (5)(0) = 0

✓. Point check:

f(2) = (2+1)(2-4) = (3)(-2) = -6

✓. All three conditions pass. Since

a = 1 > 0

, the parabola opens upward, and the point

(2,-6)

sitting below the x-axis between the roots is consistent with an upward-opening curve that dips down between its intercepts.

Common mistakes and how to avoid them

Writing the factor as

(x + r)

instead of

(x - r)

when a root is positive. For example, using

(x + 2)

for the root

x = 2

.

Correction: A root

x = r

always produces the factor

(x - r)

. For

x = 2

, the correct factor is

(x - 2)

. Check by substituting the root:

(2 - 2) = 0

confirms zero output.

Forgetting to include

a

in the factored form and jumping straight to

f(x) = (x - r)(x - s)

, which forces

a = 1

without justification.

Correction: Always write

f(x) = a(x - r)(x - s)

with

a

as an unknown until you have solved for it using the given point.

Substituting the given point incorrectly — for example, substituting only the x-coordinate but not setting

f(x)

equal to the y-coordinate.

Correction: Both coordinates must be used. Replace

x

with

x_1

and simultaneously set

f(x) = y_1

before solving.

Arithmetic errors with negative numbers when evaluating the factors after substitution, leading to a wrong value of

a

.

Correction: Write out each factor separately, evaluate, then multiply. Do not rush the simplification. A quick check by substituting

a

and the point back into the function catches this error immediately.

Using one of the roots as the extra point, which makes the equation

0 = a \cdot 0 \cdot (\ )

, giving

0 = 0

— this is true for any value of

a

and tells you nothing.

Correction: The extra point must not be one of the x-intercepts. It must be a point where

f(x) \neq 0

so that the equation for

a

has a unique solution.

Lesson summary

A quadratic function can be written in factored form as $f(x) = a(x - r)(x - s)$ , where $r$ and $s$ are its roots.
Two roots alone do not uniquely determine a quadratic function — they fix the roots but leave the factor $a$ unknown.
Substituting one additional point $(x_1, y_1)$ into the factored form creates a solvable equation for $a$ .
Once $a$ is found, the factored form is complete and can be expanded into standard form $f(x) = ax^2 + bx + c$ .
Always verify the final function satisfies all three conditions: $f(r) = 0$ , $f(s) = 0$ , and $f(x_1) = y_1$ .
The sign of $a$ determines whether the parabola opens upward ( $a > 0$ ) or downward ( $a < 0$ ).

Check your understanding

Question 1

A parabola has roots at

x = -3

and

x = 5

. Which factored skeleton is correct before finding

a

?

$f(x) = a(x + 3)(x + 5)$
$f(x) = a(x - 3)(x + 5)$
$f(x) = a(x + 3)(x - 5)$
$f(x) = a(x - 3)(x - 5)$

Show answer and explanation

f(x) = a(x + 3)(x - 5)

A root

x = -3

gives the factor

(x - (-3)) = (x + 3)

, and a root

x = 5

gives the factor

(x - 5)

. So the correct skeleton is

f(x) = a(x + 3)(x - 5)

. Option A uses both wrong signs, option B has the signs reversed on both factors, and option D applies the wrong sign to the first root.

Question 2

A quadratic function has roots

x = 1

and

x = -2

and passes through the point

(0, 6)

. After substituting the point into

f(x) = a(x - 1)(x + 2)

, which equation do you get?

$6 = a(1)(2)$
$6 = a(-1)(2)$
$6 = a(1)(-2)$
$0 = a(-1)(2)$

Show answer and explanation

6 = a(-1)(2)

Substitute

x = 0

and

f(x) = 6

:

6 = a(0 - 1)(0 + 2) = a(-1)(2)

. This gives

6 = -2a

, so

a = -3

. Option A mistakes

0 - 1

as positive. Option C mistakes

0 + 2

as negative. Option D replaces the y-value with

0

, which would only be correct for a root, not this point.

Question 3

Using the result from the previous question (

a = -3

, roots

x = 1

and

x = -2

), what is the standard form of the quadratic function?

$f(x) = -3x^2 - 3x + 6$
$f(x) = -3x^2 + 3x + 6$
$f(x) = 3x^2 - 3x - 6$
$f(x) = -3x^2 + 3x - 6$

Show answer and explanation

f(x) = -3x^2 - 3x + 6

Start with

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

. Expand:

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

. Multiply by

-3

:

-3x^2 - 3x + 6

. Check:

f(0) = 6

✓,

f(1) = -3 - 3 + 6 = 0

✓,

f(-2) = -3(4) - 3(-2) + 6 = -12 + 6 + 6 = 0

✓.

Question 4

Why is it impossible to find a unique quadratic function if you are only given the two roots and no other information?

Because a quadratic function can have at most one root.
Because there are infinitely many values of $a$ that all produce parabolas through those two roots.
Because you need the vertex, not the roots, to write a quadratic function.
Because two roots always produce a linear function, not a quadratic.

Show answer and explanation

Because there are infinitely many values of

a

that all produce parabolas through those two roots.

The factored form

f(x) = a(x - r)(x - s)

contains the unknown

a

. Any non-zero value of

a

gives a different parabola that still passes through

x = r

and

x = s

. Without a third piece of information — a point on the curve that is not a root — there is no way to determine which value of

a

is correct.

Key terms

Root (of a quadratic function): An x-value where the function equals zero, i.e., where the graph crosses or touches the x-axis. Also called a zero or x-intercept.
Factored form: A way of writing a quadratic function as $f(x) = a(x - r)(x - s)$ , where $r$ and $s$ are the roots and $a$ is the leading coefficient.
Leading coefficient (a): The numerical factor in front of the highest-degree term of a polynomial. In factored form $f(x) = a(x-r)(x-s)$ , it is the value $a$ . It controls the direction and width of the parabola.
Standard form: A way of writing a quadratic function as $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are real numbers and $a \neq 0$ .
Vertical stretch/compression factor: Another name for the leading coefficient $a$ in a quadratic function. Values of $|a| > 1$ make the parabola narrower; values of $0 < |a| < 1$ make it wider.
Unique function: Exactly one function that satisfies all the given conditions. In this lesson, two roots plus one non-root point together determine a unique quadratic function.
Verify (check): Substitute values back into the final answer to confirm that every given condition is satisfied. For this topic, check both roots and the given point.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation A2.4. It is a study resource, not an official curriculum publication.

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A2.4 · Determine a quadratic function from its roots and a point

Ontario Grade 11 Mathematics

MCR3U – Quadratic Functions | Expectation A2.4

What you will learn

Prerequisite Bridge: Factored Form of a Quadratic

Why One Extra Point Is Enough

Step-by-Step Method

Connecting Factored Form to Standard Form

Checking Your Answer

How the Value of a Changes the Parabola (same roots x = 2 and x = 6)

Example 1 – Roots Are Both Positive, Point Is Above the x-axis

Example 2 – One Negative Root, Point Below the x-axis

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