Q7 Sketch a quadratic graph from vertex form | Ontario Grade 10 Mathematics

Ontario Grade 10 Mathematics

Quadratic Relations

Reading the shape, position, and direction of a parabola directly from y = a(x − h)² + k

In Grade 9 you graphed y = x² by plotting points and noticed the U-shaped curve called a parabola. You also learned that changing numbers in an equation shifts or stretches a graph. In this lesson you will go further: you will read a special form of the quadratic equation — called vertex form — and use it to sketch the full parabola without making a large table of values. Every feature you need is already hidden inside the equation, waiting to be decoded.

What you will learn

Identify the vertex, axis of symmetry, direction of opening, and vertical stretch or compression from vertex form.
Explain how each parameter a, h, and k changes the position or shape of a parabola.
Plot a parabola accurately by finding the vertex, two symmetric points, and the direction of opening.
Distinguish between a vertical stretch and a vertical compression based on the value of a.

Grade 9 Bridge — The Basic Parabola y = x²

Before working with vertex form, recall the simplest quadratic:

y = x^2

. Its graph is a parabola that opens upward, with its lowest point at the origin

(0, 0)

. That lowest point is called the vertex. The graph is perfectly symmetric: the left side is a mirror image of the right side. The imaginary vertical line running through the vertex is called the axis of symmetry, and for

y = x^2

that line is

x = 0

.

Two key pairs of points help you sketch

y = x^2

quickly. When

x = 1

,

y = 1

; when

x = -1

,

y = 1

. When

x = 2

,

y = 4

; when

x = -2

,

y = 4

. Each pair sits at the same height on opposite sides of the axis of symmetry. This symmetry property is something you will use constantly when sketching any parabola.

The graph of every quadratic equation is a parabola.
The vertex is the highest or lowest point of the parabola.
The axis of symmetry is a vertical line through the vertex.
Symmetric points are always the same distance left and right of the axis of symmetry.

Introducing Vertex Form

Vertex form is a way of writing a quadratic equation so that the vertex is immediately visible. The general vertex form is

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

, where

a

,

h

, and

k

are constants — meaning they are fixed numbers in a specific equation.

Here is what each constant does. The value of

h

moves the parabola left or right: the vertex sits at

x = h

. Notice the subtraction sign in

(x - h)^2

— this is important. If the equation shows

(x - 3)^2

, then

h = 3

and the vertex is at

x = 3

. If the equation shows

(x + 5)^2

, rewrite it as

(x - (-5))^2

, so

h = -5

and the vertex is at

x = -5

.

The value of

k

moves the parabola up or down: the vertex sits at

y = k

. Together,

(h, k)

is the vertex. The value of

a

controls two things at once: its sign tells you the direction of opening, and its size tells you how wide or narrow the parabola is. If

a > 0

, the parabola opens upward (vertex is the minimum point). If

a < 0

, it opens downward (vertex is the maximum point). If

|a| > 1

, the parabola is narrower than

y = x^2

— this is called a vertical stretch. If

0 < |a| < 1

, the parabola is wider than

y = x^2

— this is called a vertical compression.

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

Vertex form is $y = a(x - h)^2 + k$ ; the vertex is at $(h, k)$ .
Watch the sign carefully: $(x + 5)^2$ means $h = -5$ , not $h = 5$ .
$a > 0$ → opens up; $a < 0$ → opens down.
$|a| > 1$ → vertical stretch (narrower); $0 < |a| < 1$ → vertical compression (wider).
The axis of symmetry is always the vertical line $x = h$ .

A Reliable Sketching Strategy

Rather than making a full table of values, use a four-step strategy to sketch any parabola from vertex form. This strategy uses what you already know about symmetry to keep the work short and accurate.

Step 1 — Find and plot the vertex. Read

h

and

k

directly from the equation and plot the point

(h, k)

. Step 2 — Draw the axis of symmetry. Draw a dashed vertical line through

x = h

. This line is a guide, not part of the final graph. Step 3 — Find two more points. Choose one

x

-value that is a small whole number away from

h

, substitute it into the equation, and find

y

. Then use symmetry to immediately get the matching point on the other side. You now have three plotted points. Step 4 — Draw the parabola. Connect the three points with a smooth U-shaped (or ∩-shaped) curve, making the arms continue outward.

Choosing the right

x

-value in Step 3 matters. Pick a value that is 1 or 2 units away from

h

so the arithmetic stays simple. If

a

is a fraction, sometimes choosing 2 units away gives a nicer whole-number

y

-value — experiment with one unit first and switch if the result is messy.

Plot the vertex $(h, k)$ first — it anchors everything else.
The axis of symmetry $x = h$ gives you a free second point by reflection.
Choosing $x = h + 1$ or $x = h + 2$ keeps arithmetic manageable.
You need only three points to produce a reasonable sketch.
Check the sign of $a$ to confirm whether the arms point up or down before drawing.

How a Changes the Width — A Closer Look

When you move 1 unit away from the vertex horizontally, the

y

-value rises (or falls) by exactly |a| units, because

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

and the change in

y

equals

a \times 1 = a

. When you move 2 units away, the change is

a \times 4

. This is why a large |a| makes the parabola climb steeply — it is stretched vertically — and a small |a| makes it rise gently, appearing wider.

For example, compare

y = 3(x - 1)^2 + 2

and

y = \frac{1}{3}(x - 1)^2 + 2

. Both have vertex

(1, 2)

and open upward. At

x = 2

(one unit right of the vertex): for the first equation,

y = 3(1)^2 + 2 = 5

, so the point is

(2, 5)

; for the second,

y = \frac{1}{3}(1)^2 + 2 = 2\frac{1}{3}

, so the point is much closer to the vertex. The first parabola is clearly narrower.

Moving 1 unit from the vertex changes $y$ by $a \times 1 = a$ .
Moving 2 units from the vertex changes $y$ by $a \times 4$ .
Large |a| $→ steep sides (narrow); small$ |a| → gentle sides (wide).
$a = 1$ reproduces the standard $y = x^2$ shape.

Connecting the Sketch to Key Features

Once you have sketched the parabola, you can read off several key features directly. The vertex

(h, k)

is either the minimum value (when

a > 0

) or the maximum value (when

a < 0

) of the relation. The minimum or maximum value itself is the

y

-coordinate

k

. The axis of symmetry is

x = h

.

The domain of any parabola sketched here is all real numbers — the arms extend left and right without end. The range depends on the direction. When

a > 0

, the range is

y \geq k

(the parabola goes upward from the vertex). When

a < 0

, the range is

y \leq k

(the parabola goes downward from the vertex). Being able to state these features from a sketch — not just from a calculation — is an important Grade 10 skill.

Vertex $(h, k)$ gives the minimum or maximum value $k$ .
Axis of symmetry: $x = h$ .
Domain is always all real numbers for these parabolas.
Range is $y \geq k$ when $a > 0$ , and $y \leq k$ when $a < 0$ .

What Each Parameter in y = a(x − h)² + k Controls

Parameter	What it controls	Effect when value increases	Quick reading rule
$a$ (sign)	Direction of opening	$a > 0$ : opens up; $a < 0$ : opens down	Positive $a$ → U-shape; negative $a$ → ∩-shape
$a$ (size)	Width of parabola	$\|a\| > 1$ : narrower (stretch); $0 < \|a\| < 1$ : wider (compression)	Bigger \|a\| means steeper sides
$h$	Horizontal position of vertex	Vertex shifts right as $h$ increases	Read directly: vertex is at $x = h$
$k$	Vertical position of vertex	Vertex shifts up as $k$ increases	Read directly: vertex is at $y = k$

Worked example

Example 1 — Upward-Opening Parabola with a Stretch

Sketch the parabola

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

and state the vertex, axis of symmetry, direction of opening, and range.

Read the parameters
Compare the equation with $y = a(x - h)^2 + k$ . The value $a = 2$ , $h = 3$ , and $k = -5$ . Because $a = 2 > 0$ , the parabola opens upward. Because $|a| = 2 > 1$ , there is a vertical stretch — the parabola will be narrower than $y = x^2$ .
$a = 2,\ h = 3,\ k = -5$
Identify and plot the vertex
The vertex is at $(h, k) = (3, -5)$ . Plot this point. It is the minimum point because the parabola opens upward.
$(h,\ k) = (3,\ -5)$
Draw the axis of symmetry
Draw a dashed vertical line at $x = 3$ . Every point on the left of this line has a matching point the same distance to the right at the same height.
$x = 3$
Find a point one unit right of the vertex
Substitute $x = 4$ (one unit to the right of $h = 3$ ) into the equation. Compute $(4 - 3)^2 = 1$ , then multiply by 2 to get 2, then add $-5$ to get $-3$ . So the point $(4, -3)$ is on the parabola.
$y = 2(4 - 3)^2 - 5 = 2(1) - 5 = -3$
Use symmetry to get the matching point
The point $(4, -3)$ is 1 unit to the right of the axis $x = 3$ . Its mirror image is 1 unit to the left, at $x = 2$ , with the same $y$ -value. Plot $(2, -3)$ .
$(2,\ -3)$
Find a point two units right of the vertex
Substitute $x = 5$ to get a second pair. Compute $(5 - 3)^2 = 4$ , multiply by 2 to get 8, then add $-5$ to get $3$ . The point $(5, 3)$ is on the parabola, and by symmetry so is $(1, 3)$ .
$y = 2(5 - 3)^2 - 5 = 2(4) - 5 = 3$
Sketch and state the key features
Plot all five points — $(1,3)$ , $(2,-3)$ , $(3,-5)$ , $(4,-3)$ , $(5,3)$ — and connect them with a smooth upward-opening curve. The vertex is $(3, -5)$ , the axis of symmetry is $x = 3$ , the parabola opens upward, and the range is $y \geq -5$ because the vertex is the lowest point.

Answer: Vertex:

(3, -5)

; axis of symmetry:

x = 3

; opens upward (vertical stretch by factor 2); range:

y \geq -5

.

Check: Substitute

(5, 3)

back into

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

:

2(5-3)^2 - 5 = 2(4) - 5 = 8 - 5 = 3

. ✓ Substitute

(1, 3)

:

2(1-3)^2 - 5 = 2(4) - 5 = 3

. ✓

Worked example

Example 2 — Downward-Opening Parabola with a Compression

Sketch the parabola

y = -\frac{1}{2}(x + 2)^2 + 6

and state the vertex, axis of symmetry, direction of opening, and range.

Read the parameters carefully
Rewrite the equation as $y = -\frac{1}{2}(x - (-2))^2 + 6$ to match the form $y = a(x - h)^2 + k$ . So $a = -\frac{1}{2}$ , $h = -2$ , and $k = 6$ . Because $a < 0$ , the parabola opens downward. Because $|a| = \frac{1}{2} < 1$ , there is a vertical compression — the parabola will be wider than $y = x^2$ .
$a = -\frac{1}{2},\ h = -2,\ k = 6$
Identify and plot the vertex
The vertex is at $(h, k) = (-2, 6)$ . Plot this point. It is the maximum point because the parabola opens downward.
$(-2,\ 6)$
Draw the axis of symmetry
Draw a dashed vertical line at $x = -2$ .
$x = -2$
Find a point two units right of the vertex
Choose $x = 0$ (two units to the right of $h = -2$ ) because it gives easy arithmetic. Compute $(0 - (-2))^2 = 4$ , multiply by $-\frac{1}{2}$ to get $-2$ , then add $6$ to get $4$ . So $(0, 4)$ is on the parabola. By symmetry, $(-4, 4)$ is also on the parabola.
$y = -\frac{1}{2}(0 + 2)^2 + 6 = -\frac{1}{2}(4) + 6 = -2 + 6 = 4$
Find a point four units right of the vertex
Choose $x = 2$ (four units right of $h = -2$ ). Compute $(2 + 2)^2 = 16$ , multiply by $-\frac{1}{2}$ to get $-8$ , then add $6$ to get $-2$ . So $(2, -2)$ is on the parabola, and by symmetry $(-6, -2)$ is also on the parabola.
$y = -\frac{1}{2}(2 + 2)^2 + 6 = -\frac{1}{2}(16) + 6 = -8 + 6 = -2$
Sketch and state the key features
Plot all five points — $(-6,-2)$ , $(-4,4)$ , $(-2,6)$ , $(0,4)$ , $(2,-2)$ — and connect them with a smooth downward-opening curve. Notice how wide the curve is compared with a standard parabola — this is the effect of the vertical compression. The vertex is $(-2, 6)$ , the axis of symmetry is $x = -2$ , the parabola opens downward, and the range is $y \leq 6$ because the vertex is the highest point.

Answer: Vertex:

(-2, 6)

; axis of symmetry:

x = -2

; opens downward (vertical compression by factor

\frac{1}{2}

); range:

y \leq 6

.

Check: Substitute

(0, 4)

into

y = -\frac{1}{2}(x+2)^2 + 6

:

-\frac{1}{2}(0+2)^2 + 6 = -\frac{1}{2}(4) + 6 = -2 + 6 = 4

. ✓ Substitute

(-4, 4)

:

-\frac{1}{2}(-4+2)^2 + 6 = -\frac{1}{2}(4) + 6 = 4

. ✓

Common mistakes and how to avoid them

Reading the sign of h incorrectly. For example, seeing

(x + 4)^2

and writing

h = 4

instead of

h = -4

.

Correction: Rewrite the bracket in the form

(x - h)^2

. Since

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

, you can clearly see

h = -4

.

Forgetting that the range depends on the sign of

a

. Writing

y \geq k

even when the parabola opens downward.

Correction: Check the sign of

a

first. If

a > 0

, the vertex is a minimum so the range is

y \geq k

. If

a < 0

, the vertex is a maximum so the range is

y \leq k

.

Treating |a| as a horizontal stretch or shift instead of a vertical one, and placing points too far left or right.

Correction:

a

only scales the

y

-values. To find extra points, substitute an

x

-value into the equation; do not move the vertex left or right by

a

.

Plotting the symmetric point at the wrong position, for example reflecting across the

x

-axis instead of the axis of symmetry.

Correction: The axis of symmetry is the vertical line

x = h

. Count how many units the original point is from this line, then go the same number of units to the other side at the same height.

Connecting plotted points with straight lines instead of a smooth curve.

Correction: A parabola is always a smooth, rounded curve. After plotting the points, draw the curve freehand so it gradually changes direction through the vertex.

Lesson summary

Vertex form $y = a(x - h)^2 + k$ makes the vertex $(h, k)$ immediately visible — no calculation needed.
The sign of $a$ tells you the direction: positive means the parabola opens up, negative means it opens down.
The size of |a| tells you the shape: greater than 1 gives a vertical stretch (narrower curve), between 0 and 1 gives a vertical compression (wider curve).
To sketch, plot the vertex, draw the axis of symmetry $x = h$ , find one extra point by substitution, then reflect it across the axis for a fourth point.
The range is $y \geq k$ when the parabola opens up, and $y \leq k$ when it opens down; the domain is always all real numbers.
Always rewrite $(x + c)^2$ as $(x - (-c))^2$ to avoid sign errors when reading $h$ .

Check your understanding

Question 1

What is the vertex of the parabola

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

?

$(1, 7)$
$(-1, 7)$
$(-1, -7)$
$(3, 7)$

Show answer and explanation

(-1, 7)

Rewrite as

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

, so

h = -1

and

k = 7

. The vertex is

(-1, 7)

. The common error is reading

(x + 1)^2

as

h = 1

, but the subtraction in the definition means

h = -1

.

Question 2

The equation

y = \frac{1}{4}(x - 5)^2 - 3

produces a parabola that, compared with

y = x^2

, is _.

narrower and opens downward
wider and opens downward
narrower and opens upward
wider and opens upward

Show answer and explanation

wider and opens upward

a = \frac{1}{4} > 0

so the parabola opens upward. Because

|a| = \frac{1}{4} < 1

, this is a vertical compression, making the parabola wider than

y = x^2

.

Question 3

For the parabola

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

, what is the

y

-value when

x = 6

?

$5$
$9$
$17$
$3$

Show answer and explanation

9

Substitute

x = 6

:

y = 2(6 - 4)^2 + 1 = 2(2)^2 + 1 = 2(4) + 1 = 8 + 1 = 9

.

Question 4

A parabola has equation

y = -5(x - 2)^2 + 8

. What is its range?

All real numbers
$y \geq 8$
$y \leq 2$
$y \leq 8$

Show answer and explanation

y \leq 8

Because

a = -5 < 0

, the parabola opens downward and the vertex

(2, 8)

is the maximum point. The

y

-values can never exceed 8, so the range is

y \leq 8

.

Key terms

Parabola: The U-shaped (or ∩-shaped) curve that is the graph of every quadratic relation.
Vertex: The highest or lowest point of a parabola; in vertex form it is the point $(h, k)$ .
Vertex form: The equation $y = a(x - h)^2 + k$ , where the vertex is directly visible as $(h, k)$ .
Axis of symmetry: The vertical line $x = h$ that divides the parabola into two mirror-image halves.
Vertical stretch: A transformation where $|a| > 1$ makes the parabola narrower than $y = x^2$ by multiplying all $y$ -values by |a|.
Vertical compression: A transformation where $0 < |a| < 1$ makes the parabola wider than $y = x^2$ by multiplying all $y$ -values by |a|.
Minimum value: The lowest $y$ -value of the parabola, equal to $k$ when $a > 0$ .
Maximum value: The highest $y$ -value of the parabola, equal to $k$ when $a < 0$ .

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic Q7. It is a study resource, not an official curriculum publication.

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Q7 · Sketch a quadratic graph from vertex form

Ontario Grade 10 Mathematics

Reading the shape, position, and direction of a parabola directly from y = a(x − h)² + k

What you will learn

Grade 9 Bridge — The Basic Parabola y = x²

Introducing Vertex Form

A Reliable Sketching Strategy

How a Changes the Width — A Closer Look

Connecting the Sketch to Key Features

What Each Parameter in y = a(x − h)² + k Controls

Example 1 — Upward-Opening Parabola with a Stretch

Example 2 — Downward-Opening Parabola with a Compression

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MPM2D

About this lesson