A2.1 Determine the number of zeros of a quadratic function

Ontario Grade 11 Mathematics

Characteristics of Functions

Using the Discriminant to Count x-Intercepts

Every quadratic function produces a parabola when graphed. One of the most useful questions you can ask about a parabola is: how many times does it cross or touch the x-axis? Those crossing or touching points are called the zeros of the function. In this lesson you will learn a fast, reliable way to answer that question without fully solving the equation — by calculating a single number called the discriminant. You will move from a picture of what zeros look like, to the algebra behind them, to a step-by-step method you can apply to any quadratic.

What you will learn

Explain what a zero of a quadratic function means graphically and algebraically.
Calculate the discriminant of any quadratic function written in standard form.
Use the value of the discriminant to determine whether a quadratic has two, one, or zero real zeros.
Interpret the number of zeros in a real-world or graphical context.

Prerequisite Bridge: Quadratics in Standard Form

A quadratic function is any function of the form

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

, where

a

,

b

, and

c

are real numbers and

a \neq 0

. The condition

a \neq 0

is essential: if

a

were zero, the

x^2

term would disappear and the function would no longer be quadratic.

You already know from Grade 10 that the graph of a quadratic is a U-shaped (or inverted U-shaped) curve called a parabola. When

a > 0

the parabola opens upward, and the vertex is the lowest point. When

a < 0

the parabola opens downward, and the vertex is the highest point. The vertex's position relative to the x-axis has a direct effect on how many zeros the function has.

You also learned the quadratic formula: if

ax^2 + bx + c = 0

, the solutions are

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

. This lesson focuses on the expression under the square root,

b^2 - 4ac

, because its value alone tells you how many real solutions — and therefore how many zeros — the quadratic has.

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

Standard form: $f(x) = ax^2 + bx + c$ , with $a \neq 0$ .
When $a > 0$ the parabola opens upward and the vertex is the minimum point.
When $a < 0$ the parabola opens downward and the vertex is the maximum point.
The expression $b^2 - 4ac$ inside the quadratic formula is the key to counting zeros.

What Is a Zero of a Function?

A zero of a function is any input value

x

for which the output

f(x)

equals zero. Geometrically, a zero is an x-intercept — the exact point where the parabola crosses or touches the x-axis. The word 'zero' and the phrase 'x-intercept' describe the same location from two different viewpoints: algebraic and graphical.

To find zeros algebraically you set

f(x) = 0

and solve for

x

. For a quadratic, that means solving

ax^2 + bx + c = 0

. Depending on the values of

a

,

b

, and

c

, this equation can have two different solutions, exactly one repeated solution, or no real solutions at all.

Think about it graphically. An upward-opening parabola whose vertex is below the x-axis must cross the axis at two separate points — two zeros. If that vertex sits exactly on the x-axis, the parabola just touches it once — one zero. If the vertex is above the x-axis and the parabola opens upward, the curve floats entirely above the axis and never reaches it — no zeros. For a downward-opening parabola the vertex is the highest point, so the pictures are flipped: a vertex above the x-axis gives two crossings, a vertex on the axis gives one, and a vertex below the axis means the whole parabola hangs below with no crossings.

f(x) = 0

A zero is a value of $x$ where $f(x) = 0$ .
Zeros appear as x-intercepts on the graph of the parabola.
A quadratic can have 0, 1, or 2 real zeros.
The vertex position relative to the x-axis, combined with the direction the parabola opens, determines the number of zeros.

The Discriminant: Your Counting Tool

The discriminant of the quadratic

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

is the expression

b^2 - 4ac

. It is usually written using the Greek capital letter delta:

\Delta = b^2 - 4ac

. You do not need to finish solving the quadratic formula to use it — you only need to evaluate

\Delta

and check its sign.

Here is why

\Delta

works as a counting tool. The quadratic formula contains

\sqrt{\Delta}

. If

\Delta > 0

, the square root is a positive real number, so the

\pm

in the formula produces two different values of

x

— two distinct real zeros. If

\Delta = 0

, the square root equals zero, the

\pm

adds or subtracts nothing, and both versions of the formula give the same value — one repeated real zero. If

\Delta < 0

, you would need the square root of a negative number, which has no real value — so there are no real zeros.

This rule works for every quadratic, regardless of what the numbers look like. You do not need to complete the square, factor, or finish the quadratic formula — just compute

b^2 - 4ac

and read the sign of the result.

\Delta = b^2 - 4ac

The discriminant is $\Delta = b^2 - 4ac$ .
$\Delta > 0$ : two distinct real zeros — the parabola crosses the x-axis twice.
$\Delta = 0$ : exactly one real zero — the parabola touches the x-axis at the vertex.
$\Delta < 0$ : no real zeros — the parabola does not reach the x-axis.

Reading the Graph Alongside the Discriminant

Connecting the discriminant to a sketch helps the rule stick. When

\Delta > 0

, picture an upward-opening parabola whose lowest point (the vertex) is below the x-axis; the curve must cross the axis at two separate points. When

\Delta = 0

, the vertex sits right on the x-axis and the parabola bounces off it at exactly one point. When

\Delta < 0

, the vertex of an upward-opening parabola is above the x-axis, so the whole curve floats above it.

For a downward-opening parabola (

a < 0

), the vertex is the highest point. When

\Delta > 0

, the vertex is above the x-axis, so the parabola still crosses it at two points. When

\Delta = 0

, the vertex is exactly on the x-axis — one touch point. When

\Delta < 0

, the vertex is below the x-axis, so the entire parabola lies below the axis with no crossings.

An important detail: when

\Delta = 0

, the single zero is sometimes called a repeated zero or a double root. The parabola does not cross the x-axis — it just grazes it at the vertex. Both descriptions (one zero, repeated zero) are acceptable in MCR3U.

Sketching the parabola reinforces what $\Delta$ predicts.
The sign of $a$ sets the direction of opening, but $\Delta$ counts the zeros the same way regardless.
For $a < 0$ : a vertex below the x-axis means $\Delta < 0$ and no zeros; a vertex above means $\Delta > 0$ and two zeros.
A repeated zero ( $\Delta = 0$ ) is located at the vertex of the parabola.

Applying the Method: A Step-by-Step Process

Follow these three steps every time you need to find the number of zeros. First, make sure the quadratic is in standard form

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

and identify

a

,

b

, and

c

by matching coefficients. Be careful with signs: in

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

, you have

a = 3

,

b = -5

, and

c = 2

. If the equation is not already in standard form — for example

3x^2 = 5x - 2

— rearrange it first to get

3x^2 - 5x + 2 = 0

, then read off the coefficients.

Second, substitute into

\Delta = b^2 - 4ac

and calculate carefully. It helps to write the substitution out in full before simplifying, especially when

b

is negative, since squaring a negative number always gives a positive result. For instance, if

b = -5

, then

b^2 = (-5)^2 = 25

, not

-25

.

Third, look at the sign of

\Delta

and state your conclusion: positive means two distinct real zeros, zero means one repeated real zero, and negative means no real zeros. That conclusion is the complete answer for expectation A2.1.

\Delta = b^2 - 4ac

Step 1: Write the quadratic in standard form and identify $a$ , $b$ , $c$ with their correct signs.
Step 2: Calculate $\Delta = b^2 - 4ac$ , writing the substitution out in full first.
Step 3: State the number of zeros based on the sign of $\Delta$ .
Watch the sign of $b$ carefully when squaring — a negative squared is always positive.

Discriminant Summary: Three Cases at a Glance

Value of $\Delta = b^2 - 4ac$	Number of Real Zeros	What the Graph Looks Like
$\Delta > 0$	2 distinct zeros	Parabola crosses the x-axis at two separate points
$\Delta = 0$	1 repeated zero	Parabola touches the x-axis at exactly one point (the vertex)
$\Delta < 0$	0 zeros	Parabola does not touch or cross the x-axis at any point

Worked example

Example 1: A Function with Two Zeros

Determine the number of zeros of

f(x) = 2x^2 - 6x + 1

.

Identify a, b, and c
Compare $f(x) = 2x^2 - 6x + 1$ with the standard form $f(x) = ax^2 + bx + c$ . Reading off the coefficients gives $a = 2$ , $b = -6$ , and $c = 1$ . Note that $b$ is negative — this matters when squaring in the next step.
$a = 2, \quad b = -6, \quad c = 1$
Substitute into the discriminant formula
Write out $\Delta = b^2 - 4ac$ and substitute $b = -6$ , $a = 2$ , $c = 1$ . Writing the substitution in full before simplifying makes it easier to spot sign errors.
$\Delta = (-6)^2 - 4(2)(1)$
Evaluate each part
Squaring a negative gives a positive, so $(-6)^2 = 36$ . Then $4 \times 2 \times 1 = 8$ . Subtract to get $\Delta = 36 - 8 = 28$ .
$\Delta = 36 - 8 = 28$
Interpret the result
Since $\Delta = 28 > 0$ , the quadratic has two distinct real zeros. Because $a = 2 > 0$ the parabola opens upward, which means its vertex is the lowest point. A positive discriminant tells us that lowest point must lie below the x-axis, so the parabola crosses the x-axis at two separate points.
$\Delta > 0 \Rightarrow \text{two distinct real zeros}$

Answer: The function

f(x) = 2x^2 - 6x + 1

has two distinct real zeros.

Check: Because

\Delta = 28 > 0

and

\sqrt{28}

is a positive real number, the quadratic formula would produce two different values of

x

(one using

+\sqrt{28}

and one using

-\sqrt{28}

). This is consistent with the conclusion of two distinct real zeros.

Worked example

Example 2: Determining Zeros of a Downward-Opening Quadratic in Context

A ball is launched and its height above a platform in metres is modelled by

h(t) = -4t^2 + 8t - 5

, where

t

is time in seconds. Determine the number of times the ball reaches a height of zero metres — that is, the number of zeros of

h

.

Confirm standard form and identify a, b, c
The function $h(t) = -4t^2 + 8t - 5$ is already written in standard form $at^2 + bt + c$ . Reading the coefficients: $a = -4$ , $b = 8$ , $c = -5$ . Because $a = -4 < 0$ , the parabola opens downward, which means the vertex is the highest point the ball reaches.
$a = -4, \quad b = 8, \quad c = -5$
Write and substitute into the discriminant formula
Write $\Delta = b^2 - 4ac$ and substitute $b = 8$ , $a = -4$ , $c = -5$ . Keep the negatives attached to the coefficients inside the brackets so no sign is lost.
$\Delta = (8)^2 - 4(-4)(-5)$
Evaluate each part
First, $8^2 = 64$ . Then evaluate $4 \times (-4) \times (-5)$ : multiplying the two negative numbers together gives a positive, so $(-4) \times (-5) = 20$ , and $4 \times 20 = 80$ . Therefore $\Delta = 64 - 80 = -16$ .
$\Delta = 64 - 80 = -16$
Interpret the result
Since $\Delta = -16 < 0$ , the quadratic has no real zeros. In context, this means the ball's height never equals zero — it never reaches the platform level. Because $a < 0$ the parabola opens downward, so the vertex is the maximum height. With $\Delta < 0$ , that maximum height is negative (below zero), meaning the entire parabola lies below the x-axis with no crossings.
$\Delta < 0 \Rightarrow \text{no real zeros}$

Answer: The function

h(t) = -4t^2 + 8t - 5

has no real zeros. The ball never reaches a height of zero metres.

Check: Because

a = -4 < 0

, the parabola opens downward and the vertex is the maximum point. With

\Delta = -16 < 0

, that maximum output is negative. A downward-opening parabola whose peak is below the x-axis lies entirely below the axis — confirming there are no zeros.

Common mistakes and how to avoid them

Forgetting to square

b

: writing

\Delta = b - 4ac

instead of

\Delta = b^2 - 4ac

.

Correction: Always square the entire value of

b

first. Write

b^2

as a separate calculation, then subtract

4ac

.

Treating a negative

b

as negative after squaring, for example writing

(-6)^2 = -36

.

Correction: Squaring any real number always gives a non-negative result:

(-6)^2 = 36

, not

-36

. Write the bracket and the square explicitly to remind yourself.

Misreading the sign of

c

when the quadratic contains a subtraction, such as identifying

c = 7

in

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

instead of

c = -7

.

Correction: Write the function in full standard form and attach the sign in front of each term to its coefficient before identifying

a

,

b

, and

c

.

Confusing

\Delta = 0

(one repeated zero) with

\Delta < 0

(no zeros).

Correction: When

\Delta = 0

the function has exactly one zero located at the vertex. No zeros only occurs when

\Delta < 0

.

Forgetting to rearrange the equation into standard form before identifying

a

,

b

, and

c

, for example reading

a

,

b

,

c

directly from

3x^2 = 5x - 2

.

Correction: Rearrange to

3x^2 - 5x + 2 = 0

first, then read off

a = 3

,

b = -5

,

c = 2

.

Lesson summary

A zero of a quadratic function $f(x)$ is a value of $x$ where $f(x) = 0$ ; it appears as an x-intercept on the graph.
The discriminant $\Delta = b^2 - 4ac$ tells you the number of real zeros without fully solving the equation.
If $\Delta > 0$ , the quadratic has two distinct real zeros and the parabola crosses the x-axis at two points.
If $\Delta = 0$ , the quadratic has exactly one repeated real zero and the parabola touches the x-axis at its vertex.
If $\Delta < 0$ , the quadratic has no real zeros and the parabola does not intersect the x-axis at any point.
Always write the quadratic in standard form and identify $a$ , $b$ , $c$ carefully — especially the signs — before computing $\Delta$ .

Check your understanding

Question 1

What is the value of the discriminant for

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

?

$\Delta = 32$
$\Delta = 0$
$\Delta = -16$
$\Delta = 8$

Show answer and explanation

\Delta = 0

Here

a = 1

,

b = -4

,

c = 4

. So

\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0

. The discriminant is zero, meaning the quadratic has exactly one repeated zero.

Question 2

A quadratic has discriminant

\Delta = -9

. How many real zeros does it have?

Two distinct real zeros
Exactly one real zero
No real zeros
Three real zeros

Show answer and explanation

No real zeros

Because

\Delta = -9 < 0

, the square root inside the quadratic formula would involve the square root of a negative number, which has no real value. The quadratic therefore has no real zeros.

Question 3

Which set of values gives a quadratic with two distinct real zeros?

$a = 1,\ b = 2,\ c = 5$
$a = 1,\ b = -2,\ c = 1$
$a = 2,\ b = -7,\ c = 3$
$a = 3,\ b = 0,\ c = 4$

Show answer and explanation

a = 2,\ b = -7,\ c = 3

Check each discriminant. Option A:

\Delta = 4 - 20 = -16 < 0

(no zeros). Option B:

\Delta = 4 - 4 = 0

(one zero). Option C:

\Delta = 49 - 24 = 25 > 0

(two zeros — correct). Option D:

\Delta = 0 - 48 = -48 < 0

(no zeros).

Question 4

The function

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

is being analyzed. What does the discriminant tell you about its graph?

The parabola crosses the x-axis at two separate points.
The parabola touches the x-axis at exactly one point.
The parabola lies entirely below the x-axis and never touches it.
The parabola lies entirely above the x-axis and never touches it.

Show answer and explanation

The parabola touches the x-axis at exactly one point.

With

a = -3

,

b = 6

,

c = -3

:

\Delta = 6^2 - 4(-3)(-3) = 36 - 36 = 0

. Since

\Delta = 0

, there is exactly one repeated zero and the parabola touches the x-axis at exactly one point — its vertex. Because

a = -3 < 0

the parabola opens downward, so the vertex is the highest point and the rest of the curve sits below the x-axis.

Key terms

Zero of a function: A value of $x$ for which $f(x) = 0$ ; it appears as an x-intercept on the graph of the function.
Discriminant: The expression $\Delta = b^2 - 4ac$ calculated from the coefficients of a quadratic in standard form; its sign determines the number of real zeros.
Standard form: The way of writing a quadratic function as $f(x) = ax^2 + bx + c$ , where $a \neq 0$ .
Parabola: The U-shaped (or inverted U-shaped) curve that is the graph of any quadratic function.
Repeated zero: A zero that occurs when $\Delta = 0$ ; the parabola touches but does not cross the x-axis at this single point, which is the vertex.
x-intercept: The point where a graph crosses or touches the x-axis; for a quadratic, x-intercepts are the same locations as the zeros of the function.
Coefficient: The numerical factor in front of a variable term; in $-6x$ , the coefficient is $-6$ .
Quadratic formula: The formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ used to find the zeros of any quadratic equation $ax^2 + bx + c = 0$ .

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation A2.1. It is a study resource, not an official curriculum publication.

Official curriculum reference

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A2.1 · Determine the number of zeros of a quadratic function

Ontario Grade 11 Mathematics

Using the Discriminant to Count x-Intercepts

What you will learn

Prerequisite Bridge: Quadratics in Standard Form

What Is a Zero of a Function?

The Discriminant: Your Counting Tool

Reading the Graph Alongside the Discriminant

Applying the Method: A Step-by-Step Process

Discriminant Summary: Three Cases at a Glance

Example 1: A Function with Two Zeros

Example 2: Determining Zeros of a Downward-Opening Quadratic in Context

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