Q8 Determine a vertex-form equation from a parabola graph

Ontario Grade 10 Mathematics

Quadratic Relations

MPM2D · Quadratic Relations · Study Topic Q8

You have already graphed parabolas by plotting points and by using transformations (MTH1W). In this lesson you go in the opposite direction: you start with the graph and write the equation. This is a key skill in MPM2D because a single, neat equation in vertex form captures everything about a parabola — its turning point, its width, and which way it opens. By the end of this lesson you will be able to look at any labelled parabola graph and write its equation with confidence.

What you will learn

Identify the vertex, direction of opening, and a second point from a parabola graph.
Write the vertex-form equation $y = a(x - h)^2 + k$ by substituting the vertex and a known point.
Determine the value of $a$ by solving a one-variable equation.
Verify a vertex-form equation by checking that its vertex and a second point match the graph.

Bridge from Grade 9 — What You Already Know About Parabolas

In Grade 9 you learned that a parabola is the U-shaped curve produced by a quadratic relation. You plotted parabolas from equations such as

y = x^2

,

y = 2x^2

, and

y = -x^2 + 4

. You also learned the idea of a vertex — the single point where the parabola changes direction — and you noticed that the curve is symmetric about a vertical line through that vertex.

You also saw how the number in front of

x^2

controls the shape. A large positive value makes the parabola narrow; a small positive value makes it wide. A negative value flips it upside down. This lesson builds directly on those observations.

The vertex is the highest or lowest point on the parabola.
The parabola is symmetric about a vertical line called the axis of symmetry.
The coefficient of $x^2$ controls both width and direction of opening.

Vertex Form — The Equation We Are Building

Vertex form is a way of writing a quadratic equation so the vertex stands out immediately. The general vertex-form equation is

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

, where

(h, k)

is the vertex of the parabola and

a

is a non-zero number called the stretch or compression factor.

Reading the vertex form: the vertex is the point

(h, k)

. Notice the subtle sign — the form uses

(x - h)

, so if the vertex is at

x = 3

the term inside the brackets is

(x - 3)

. If the vertex is at

x = -2

the term becomes

(x - (-2)) = (x + 2)

. Students often flip this sign, so pause and double-check every time.

The value of

a

tells you two things at once. Its sign tells you direction:

a > 0

means the parabola opens upward (a valley shape), and

a < 0

means it opens downward (a hill shape). Its size tells you width:

|a| > 1

gives a narrower curve than

y = x^2

, while

0 < |a| < 1

gives a wider curve. When

a = 1

you get the standard parabola

y = x^2

shifted to the vertex.

Your job in this topic is to find all three values —

h

,

k

, and

a

— from information shown on the graph.

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

Vertex form: $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex.
Sign rule: vertex at $x = h$ means the bracket is $(x - h)$ , not $(x + h)$ .
$a > 0$ : opens up; $a < 0$ : opens down.
$|a| > 1$ : narrower than $y = x^2$ ; $|a| < 1$ : wider.

Step-by-Step Strategy for Reading the Graph

Every time you are given a parabola graph, follow these four steps in order. Rushing past Step 1 is the most common source of errors.

Step 1 — Locate the vertex. Look for the turning point of the parabola. Read its coordinates carefully from the grid. Call them

(h, k)

. Write

h

and

k

down before doing anything else.

Step 2 — Write the partial equation. Substitute

h

and

k

into

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

. You now have an equation with only one unknown:

a

.

Step 3 — Find a second point. Pick any other clearly labelled point on the graph. This gives you a known

x

value and a known

y

value. Avoid the vertex itself — substituting the vertex always gives

0 = 0

, which tells you nothing about

a

. A point on the axis of symmetry is also useless; choose a point clearly off-centre.

Step 4 — Solve for

a

. Substitute the second point's coordinates into your partial equation. Simplify the bracket, then divide both sides to isolate

a

. Write the final equation with the numerical value of

a

filled in.

Always read the vertex first and write it down before forming any equation.
Substitute the vertex to get a partial equation with $a$ as the only unknown.
Choose a second point that is clearly readable from the graph grid.
Substitute the second point and solve for $a$ algebraically.

Choosing and Checking Your Second Point

A good second point sits on an exact grid intersection — a place where you can read both coordinates as integers without guessing. If the graph shows the

y

-intercept at a clean integer, that is often the easiest choice because you simply set

x = 0

.

Once you have found

a

, it is worth a quick check: substitute both the vertex and the second point into your completed equation and confirm you get the correct

y

value each time. If the check fails, re-read the vertex sign or re-read the second point — those are the two most common slip points.

A graphical sense-check also helps. If your

a

is positive but the parabola on the graph clearly opens downward, you made a sign error somewhere. Stop and retrace your steps rather than continuing with a wrong

a

.

Choose a second point at a clear grid intersection with integer coordinates.
The $y$ -intercept ( $x = 0$ ) is often the most convenient second point.
Always verify by substituting both the vertex and the second point into the final equation.
Use the direction of opening as a quick sign-check for $a$ .

What the Value of a Reveals About the Graph

Once you have solved for

a

, take a moment to interpret it. This builds the habit of connecting algebra back to the visual graph, which is exactly what examiners test.

If

a = 2

, the parabola is twice as steep as

y = x^2

near the vertex. Moving one unit left or right of the vertex raises (or lowers) the curve by

2

units instead of the usual

1

unit. If

a = \frac{1}{3}

, the curve is wider — one unit from the vertex produces only

\frac{1}{3}

of a unit of rise. If

a = -4

, the parabola opens downward and is four times steeper than the basic downward parabola

y = -x^2

.

This interpretation also gives you a self-check opportunity. Look at the graph: is the parabola clearly narrower or wider than the standard

y = x^2

shape? Does your calculated |a| agree with that visual impression? If the curve looks much steeper than

y = x^2

but you calculated

a = 0.1

, something went wrong.

$|a| > 1$ : parabola is steeper (narrower) than $y = x^2$ .
$0 < |a| < 1$ : parabola is shallower (wider) than $y = x^2$ .
Negative $a$ flips the parabola to open downward.
Connect the numerical value of $a$ back to the visual shape as a self-check.

Quick Reference: Reading Vertex Form $y = a(x - h)^2 + k$ from a Graph

What you read from the graph	What it gives you	Where it goes in the equation
Turning point (vertex) coordinates	$h$ and $k$	Inside the bracket as $(x - h)$ , and added at the end as $+ k$
Direction of opening (up or down)	Sign of $a$	$a > 0$ for up, $a < 0$ for down
A second labelled point $(x_1, y_1)$	Numerical value of $a$	Substitute into partial equation, then solve

Worked example

Example 1 — Upward-Opening Parabola with a Positive Vertex

A parabola graph shows a vertex at

(2, -3)

and passes through the point

(4, 5)

. The parabola opens upward. Write the equation in vertex form.

Identify the vertex
The vertex is given as $(2, -3)$ , so $h = 2$ and $k = -3$ .
$h = 2, k = -3$
Write the partial equation
Substitute $h = 2$ and $k = -3$ into the vertex form $y = a(x - h)^2 + k$ . The value $a$ is still unknown.
$y = a(x - 2)^2 - 3$
Substitute the second point
The graph also passes through $(4, 5)$ , so substitute $x = 4$ and $y = 5$ into the partial equation.
$5 = a(4 - 2)^2 - 3$
Simplify inside the bracket
Calculate $4 - 2 = 2$ , then square it to get $4$ .
$5 = a(2)^2 - 3 = 4a - 3$
Solve for a
Add $3$ to both sides to isolate the $4a$ term, then divide both sides by $4$ .
$8 = 4a \implies a = 2$
Write the final equation
Replace $a$ with $2$ in the partial equation to get the complete vertex-form equation.
$y = 2(x - 2)^2 - 3$

Answer:

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

Check: Vertex check: substitute

x = 2

:

y = 2(2-2)^2 - 3 = 2(0) - 3 = -3

. ✓ The vertex

(2, -3)

is correct. Second-point check: substitute

x = 4

:

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

. ✓ The point

(4, 5)

lies on the curve. Since

a = 2 > 0

, the parabola opens upward, which matches the graph description.

Worked example

Example 2 — Downward-Opening Parabola with a Negative Vertex x-coordinate

A parabola graph shows a vertex at

(-3, 4)

and passes through the point

(0, -5)

. The parabola opens downward. Write the equation in vertex form.

Identify the vertex
The vertex is $(-3, 4)$ , so $h = -3$ and $k = 4$ . Be careful: $h$ is negative here.
$h = -3, k = 4$
Write the partial equation
Substitute $h = -3$ and $k = 4$ into $y = a(x - h)^2 + k$ . Because $h = -3$ , the bracket becomes $x - (-3) = x + 3$ .
$y = a(x + 3)^2 + 4$
Substitute the second point
The point $(0, -5)$ is the $y$ -intercept — a clean grid point. Substitute $x = 0$ and $y = -5$ .
$-5 = a(0 + 3)^2 + 4$
Simplify inside the bracket
Calculate $0 + 3 = 3$ , then square it to get $9$ .
$-5 = 9a + 4$
Solve for a
Subtract $4$ from both sides, then divide both sides by $9$ .
$-9 = 9a \implies a = -1$
Write the final equation
Replace $a$ with $-1$ in the partial equation. The negative value confirms the parabola opens downward, which matches the graph.
$y = -(x + 3)^2 + 4$

Answer:

y = -(x + 3)^2 + 4

Check: Vertex check: substitute

x = -3

:

y = -(-3+3)^2 + 4 = -(0) + 4 = 4

. ✓ The vertex

(-3, 4)

is correct. Second-point check: substitute

x = 0

:

y = -(0+3)^2 + 4 = -9 + 4 = -5

. ✓ The

y

-intercept

(0, -5)

lies on the curve. Since

a = -1 < 0

, the parabola opens downward. ✓

Common mistakes and how to avoid them

Writing

(x + h)

instead of

(x - h)

when the vertex has a positive

x

-coordinate. For example, writing

(x + 2)

for a vertex at

x = 2

.

Correction: Always use

(x - h)

. If the vertex is at

x = 2

, the bracket is

(x - 2)

. Substitute

h

directly and keep the minus sign.

Choosing the vertex itself as the second point when solving for

a

, which always gives

0 = 0

and tells you nothing.

Correction: Pick any other clearly labelled grid point that is not the vertex. The

y

-intercept is usually the easiest choice.

Forgetting to square the bracket before solving for

a

. For example, treating

(x - h)^2

as

(x - h)

and getting the wrong value of

a

.

Correction: After substituting the second point, evaluate the bracket first, then square the result before dividing to find

a

.

Getting

a

with the wrong sign, resulting in an equation that opens in the opposite direction to the graph.

Correction: Before writing the final equation, check whether the graph opens up or down and confirm your

a

has the matching sign.

Misreading the vertex coordinates from the graph, especially when the vertex is in a negative quadrant.

Correction: Read each axis carefully. Count grid squares from the origin and confirm the sign of each coordinate before substituting.

Lesson summary

Vertex form $y = a(x - h)^2 + k$ expresses a quadratic relation so the vertex $(h, k)$ is immediately visible.
To find the equation from a graph: read the vertex $(h, k)$ , substitute it to make a partial equation, then use a second graph point to solve for $a$ .
The sign of $a$ tells you the direction of opening; its absolute value tells you whether the parabola is narrower or wider than $y = x^2$ .
A vertex with a negative $x$ -coordinate produces a positive value inside the bracket, for example vertex $(-3, 4)$ gives $(x + 3)^2$ .
Always verify your final equation by substituting both the vertex and the second point and confirming the correct $y$ values result.
A quick visual check — does your $a$ match the direction and approximate width shown on the graph — catches most arithmetic errors before they cost marks.

Check your understanding

Question 1

A parabola has its vertex at

(1, -4)

and passes through

(3, 4)

. Which equation correctly describes this parabola?

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

Show answer and explanation

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

The vertex

(1, -4)

gives

h = 1

and

k = -4

, so the partial equation is

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

. Substituting

(3, 4)

:

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

, so

8 = 4a

and

a = 2

. The equation is

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

. Option B has the wrong bracket sign; option C uses the wrong

a

; option D has the wrong

k

.

Question 2

A parabola opens downward with vertex at

(0, 6)

and passes through

(2, -2)

. What is the value of

a

?

$a = 2$
$a = -2$
$a = -4$
$a = 4$

Show answer and explanation

a = -2

The vertex

(0, 6)

gives

h = 0

and

k = 6

, so the partial equation is

y = a(x - 0)^2 + 6 = ax^2 + 6

. Substituting

(2, -2)

:

-2 = a(4) + 6

, so

-8 = 4a

and

a = -2

. The negative sign confirms the downward opening.

Question 3

The vertex of a parabola is at

(-5, 3)

. How should the bracket be written in vertex form?

$(x - 5)^2$
$(x + 5)^2$
$(x - 3)^2$
$(x + 3)^2$

Show answer and explanation

(x + 5)^2

Vertex form uses

(x - h)

. Here

h = -5

, so the bracket is

x - (-5) = x + 5

, giving

(x + 5)^2

. A common error is writing

(x - 5)^2

, which would place the vertex at

x = 5

, not

x = -5

.

Question 4

After finding

a = 3

for a parabola equation, you notice the graph clearly shows a wide, nearly flat curve. What should you do?

Accept $a = 3$ because the algebra is correct.
Re-examine your work — a wide curve suggests $|a| < 1$ , not $|a| = 3$ .
Change $a$ to $-3$ because the curve must be reflected.
Rewrite the equation in standard form to double-check.

Show answer and explanation

Re-examine your work — a wide curve suggests

|a| < 1

, not

|a| = 3

.

A wide, flat parabola has

|a| < 1

. Getting

a = 3

contradicts the visual —

|a| = 3

would produce a narrow curve. This mismatch is a signal to re-read the vertex or the second point from the graph, because an arithmetic or reading error has likely occurred.

Key terms

Vertex: The turning point of a parabola — the highest point if it opens downward, or the lowest point if it opens upward.
Vertex form: The equation $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex and $a$ controls direction and width.
Axis of symmetry: The vertical line $x = h$ that divides the parabola into two mirror-image halves.
Stretch/compression factor ( $a$ ): The coefficient $a$ in vertex form. Its sign gives direction of opening; its absolute value gives the width of the parabola relative to $y = x^2$ .
Direction of opening: Whether the parabola curves upward ( $a > 0$ , a valley shape) or downward ( $a < 0$ , a hill shape).
y-intercept: The point where the parabola crosses the $y$ -axis, found by setting $x = 0$ in the equation. Often used as a convenient second point when solving for $a$ .
Partial equation: The vertex-form equation after $h$ and $k$ have been substituted but before $a$ has been found — it contains only the unknown $a$ .

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic Q8. It is a study resource, not an official curriculum publication.

Official curriculum reference

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Q8 · Determine a vertex-form equation from a parabola graph

Ontario Grade 10 Mathematics

MPM2D · Quadratic Relations · Study Topic Q8

What you will learn

Bridge from Grade 9 — What You Already Know About Parabolas

Vertex Form — The Equation We Are Building

Step-by-Step Strategy for Reading the Graph

Choosing and Checking Your Second Point

What the Value of a Reveals About the Graph

Quick Reference: Reading Vertex Form $y = a(x - h)^2 + k$ from a Graph

Example 1 — Upward-Opening Parabola with a Positive Vertex

Example 2 — Downward-Opening Parabola with a Negative Vertex x-coordinate

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