Mathematics · Functions

Vertex Form of the Parabola

Q: How do you read the vertex from the vertex form?

For $y = a(x - d)^{2} + e$ the vertex is S(d|e): d sits inside the bracket, e behind it. The big trap is the sign of d, because the bracket reads x MINUS d. For $y = 2(x - 3)^{2} + 1$ the vertex is S(3|1), but for $y = 2(x + 3)^{2} + 1$ it is S(−3|1), since $x + 3 = x - (-3)$. Safe route: ask for which x the bracket becomes 0; the vertex sits there, and e is the corresponding function value. The e, in contrast, is read with its ordinary sign: ... − 4 means $e = -4$. Quick check: insert the vertex coordinates, $y = e$ must result.

Q: How does completing the square work?

The goal is to turn $x^{2} + bx$ into a perfect square. Recipe: square half of the x-coefficient, add it and subtract it again immediately. Example: $y = x^{2} + 6x + 5$. Halve 6 to 3, complete with $3^{2} = 9$: $y = x^{2} + 6x + 9 - 9 + 5 = (x + 3)^{2} - 4$, so S(−3|−4). If a factor sits before $x^{2}$, factor it out first: $y = 2x^{2} + 8x = 2(x^{2} + 4x) = 2((x + 2)^{2} - 4) = 2(x + 2)^{2} - 8$. The two classic mistakes: forgetting to subtract the completion term again and distributing a minus incorrectly when factoring out. Check: expanding must reproduce the original form.

Q: What does the parameter a do in the vertex form?

a controls the opening direction and shape of the parabola without moving the vertex. Sign: $a > 0$ opens upwards (vertex is a minimum), $a 1$ stretches the parabola in the y-direction, it looks narrower; $0 0)$.

Q: How do you find the roots from the vertex form?

Set $a(x - d)^{2} + e = 0$ and solve backwards: $(x - d)^{2} = -\frac{e}{a}$, so $x = d \pm{} \sqrt{(-\frac{e}{a})}$. Example: $2(x - 2)^{2} - 3 = 0$ leads to $(x - 2)^{2} = 1.5$ and $x = 2 \pm{} \sqrt{1.5} \approx{} 0.78$ and 3.22. From $-\frac{e}{a}$ you immediately see the number of roots: positive gives two, zero exactly one (the vertex lies on the x-axis), negative no real root. This is often faster than the detour via the standard form with the pq formula, and geometrically clear: if the vertex lies below the x-axis and the parabola opens upwards (e 0), it must cross the axis twice.

The vertex form shows the vertex S(d|e) of a parabola directly in the equation; the factor a controls opening and stretching.

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Formula

y = a (x - d)^{2} + e

Dimensionless (algebra)

Variables & units

Variables & units
Symbol	Meaning	Unit
$a$	Stretch factor (a > 0 opens upwards, $a < 0$ downwards)	dimensionless
$d$	x-coordinate of the vertex (mind the sign)	dimensionless
$e$	y-coordinate of the vertex	dimensionless
S(d\|e)	Vertex (extreme point of the parabola)	dimensionless

LaTeX

y = a(x - d)^{2} + e

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Dashed standard parabola at the origin and the highlighted shifted parabola with a = 1 and vertex S(d|e); guide lines mark d on the x-axis and e on the y-axis. — For a = 1 the vertex form shifts the standard parabola (dashed) by d to the right and by e upward; the vertex is S(d⁠|⁠e).

Derivation & background

From the basic parabola $y = ax^{2}$ the general parabola arises by shifting d to the right and e upwards. The transition from the standard form $y = ax^{2} + bx + c$ works by completing the square, the way back by expanding. The vertex is a minimum for $a > 0$ and a maximum for $a < 0$ ; in general $d = -\frac{b}{(2a)}$ . For $|a| > 1$ the parabola is stretched, for $|a| < 1$ compressed.

Exam blueprint

Validity range

Holds for all quadratic functions $(a \ne{} 0)$ ; every parabola can be written uniquely in vertex form. Mind the sign: $y = a(x + 3)^{2} + 1$ has its vertex at $d = -3$ .

Derivation steps

Completing the square compresses $x^{2} + bx$ into a perfect square.

1From $y = ax^{2} + bx + c$ factor out a, add and subtract $(\frac{b}{(2a)})^{2}$ .
2Combining yields $y = a(x +\frac{ b}{(2a)})^{2} + c -\frac{ b^{2}}{(4a)}$ , so $d = -\frac{b}{(2a)}$ and $e = c -\frac{ b^{2}}{(4a)}$ .

Rearrangements

Vertex from the standard form

d = -\frac{b}{2a}, \quad e = f(d)

Fast route without completing the square: compute d, then substitute.

Roots from the vertex form

x_{1,2} = d \pm \sqrt{-\frac{e}{a}}

Solvable only if $-\frac{e}{a} \ge{} 0$ ; otherwise no real roots.

Setting up from vertex and point

a = \frac{y_{P} - e}{(x_{P} - d)^{2}}

Insert the vertex, determine a with one further point.

Task variant

Bring $y = x^{2} + 6x + 5$ into vertex form and state S.

$y = (x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4$ , so S(−3|−4).

Check: $d = -\frac{6}{2} = -3$ , $f(-3) = 9 - 18 + 5 = -4$ ✓.

A parabola has vertex S(2|−3) and passes through P(4|5). Determine the equation.

$y = a(x - 2)^{2} - 3$ with P: $5 = a\cdot{}(4 - 2)^{2} - 3 = 4a - 3$ , so $a = 2$ .
Result: $y = 2(x - 2)^{2} - 3$ .

Common mistakes

Reading the sign of d wrongly: $(x - 3)^{2}$ means $d = 3$ , not −3.

The vertex lies where the bracket becomes 0.

Forgetting to subtract the completion term.

Whatever is added must be subtracted immediately, otherwise the function changes.

Not factoring out a before completing the square.

For $y = 2x^{2} + 8x$ first $2(x^{2} + 4x)$ , then complete inside the bracket.

Exam context

Extreme value tasks without derivative, modelling projectile and bridge parabolas, transformation tasks.

Formula cluster

Quadratic functions

Vertex form, standard form and factored form complement each other; pq formula and completing the square connect them.

x₁,₂ = −p/2 ± √((p/2)² − q)

pq Formula

x = (−b ± √D) / 2a

Quadratic Formula

(a ± b)² = a² ± 2ab + b²

Binomial Formulas

Worked example

$y = 2(x - 3)^{2} + 1$ has the vertex S(3|1) and opens upwards. Way back by completing the square: $x^{2} + 6x + 5 = (x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4$ , so S(−3|−4).

Applications

Extreme value and optimization tasks (throw height, profit maximum), modelling bridge and projectile parabolas, quick sketching of parabolas, function transformations

How do you calculate with Vertex Form of the Parabola?

Here is how to work through a typical Vertex Form of the Parabola (y = a(x − d)² + e) task step by step:

1
Task
Bring $y = x^{2} + 6x + 5$ into vertex form and state S.
Solution path
1. $y = (x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4$ , so S(−3|−4).
Check: $d = -\frac{6}{2} = -3$ , $f(-3) = 9 - 18 + 5 = -4$ ✓.
2
Task
A parabola has vertex S(2|−3) and passes through P(4|5). Determine the equation.
Solution path
1. $y = a(x - 2)^{2} - 3$ with P: $5 = a\cdot{}(4 - 2)^{2} - 3 = 4a - 3$ , so $a = 2$ .
2. Result: $y = 2(x - 2)^{2} - 3$ .

Frequently asked questions about Vertex Form of the Parabola

How do you read the vertex from the vertex form?+

For $y = a(x - d)^{2} + e$ the vertex is S(d|e): d sits inside the bracket, e behind it. The big trap is the sign of d, because the bracket reads x MINUS d. For $y = 2(x - 3)^{2} + 1$ the vertex is S(3|1), but for $y = 2(x + 3)^{2} + 1$ it is S(−3|1), since $x + 3 = x - (-3)$ . Safe route: ask for which x the bracket becomes 0; the vertex sits there, and e is the corresponding function value. The e, in contrast, is read with its ordinary sign: ... − 4 means $e = -4$ . Quick check: insert the vertex coordinates, $y = e$ must result.

How does completing the square work?+

The goal is to turn $x^{2} + bx$ into a perfect square. Recipe: square half of the x-coefficient, add it and subtract it again immediately. Example: $y = x^{2} + 6x + 5$ . Halve 6 to 3, complete with $3^{2} = 9$ : $y = x^{2} + 6x + 9 - 9 + 5 = (x + 3)^{2} - 4$ , so S(−3|−4). If a factor sits before $x^{2}$ , factor it out first: $y = 2x^{2} + 8x = 2(x^{2} + 4x) = 2((x + 2)^{2} - 4) = 2(x + 2)^{2} - 8$ . The two classic mistakes: forgetting to subtract the completion term again and distributing a minus incorrectly when factoring out. Check: expanding must reproduce the original form.

What does the parameter a do in the vertex form?+

a controls the opening direction and shape of the parabola without moving the vertex. Sign: $a > 0$ opens upwards (vertex is a minimum), $a < 0$ downwards (vertex is a maximum). Magnitude: $|a| > 1$ stretches the parabola in the y-direction, it looks narrower; $0 < |a| < 1$ compresses it, it looks wider; $a = 1$ is the basic parabola. Concretely: from S you go 1 to the right and a upwards (instead of 1 as with the basic parabola), then 2 to the right and 4a upwards. For extreme value tasks this means: the optimal value is e, and a decides whether it is a maximum $(a < 0)$ or a minimum $(a > 0)$ .

How do you find the roots from the vertex form?+

Set $a(x - d)^{2} + e = 0$ and solve backwards: $(x - d)^{2} = -\frac{e}{a}$ , so $x = d \pm{} \sqrt{(-\frac{e}{a})}$ . Example: $2(x - 2)^{2} - 3 = 0$ leads to $(x - 2)^{2} = 1.5$ and $x = 2 \pm{} \sqrt{1.5} \approx{} 0.78$ and 3.22. From $-\frac{e}{a}$ you immediately see the number of roots: positive gives two, zero exactly one (the vertex lies on the x-axis), negative no real root. This is often faster than the detour via the standard form with the pq formula, and geometrically clear: if the vertex lies below the x-axis and the parabola opens upwards (e < 0, a > 0), it must cross the axis twice.

How do you set up a parabola equation from the vertex and one point?+

The vertex provides d and e, the additional point determines a. Ansatz $y = a(x - d)^{2} + e$ , insert the point, solve for a: $a =\frac{ (y_{P} - e)}{((x_{P} - d)^{2})}$ . Example: vertex S(2|−3), point P(4|5): $5 = a\cdot{}(4 - 2)^{2} - 3 = 4a - 3$ , so $a = 2$ and $y = 2(x - 2)^{2} - 3$ . Exactly this scheme underlies modelling tasks: a ball reaches its highest point at S and passes a known point, the flight parabola is required. One vertex plus one further point determine the parabola uniquely; two arbitrary points alone do not suffice, for that you would need three.

Quanta exam set

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Which formula describes Vertex Form of the Parabola?

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How do you rearrange Vertex Form of the Parabola for Vertex from the standard form?

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Which common mistake happens with Vertex Form of the Parabola?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, College Algebra 2e, §5.1: Quadratic Functions

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Create a curated FSRS exam set for y = a(x − d)² + e: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Vertex Form of the Parabola

The vertex form shows the vertex S(d|e) of a parabola directly in the equation; the factor a controls opening and stretching.

BasicExam-relevantLower secondaryUpper secondary

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

a

Stretch factor (a > 0 opens upwards,

a < 0

downwards)

dimensionless

d

x-coordinate of the vertex (mind the sign)

dimensionless

e

y-coordinate of the vertex

dimensionless

S(d|e)

Vertex (extreme point of the parabola)

dimensionless

Derivation & background

Exam blueprint

Validity range

Holds for all quadratic functions $(a \ne{} 0)$ ; every parabola can be written uniquely in vertex form. Mind the sign: $y = a(x + 3)^{2} + 1$ has its vertex at $d = -3$ .

Derivation steps

Completing the square compresses $x^{2} + bx$ into a perfect square.

1From $y = ax^{2} + bx + c$ factor out a, add and subtract $(\frac{b}{(2a)})^{2}$ .
2Combining yields $y = a(x +\frac{ b}{(2a)})^{2} + c -\frac{ b^{2}}{(4a)}$ , so $d = -\frac{b}{(2a)}$ and $e = c -\frac{ b^{2}}{(4a)}$ .

How do you calculate with Vertex Form of the Parabola?

Here is how to work through a typical Vertex Form of the Parabola (y = a(x − d)² + e) task step by step:

Task

Bring $y = x^{2} + 6x + 5$ into vertex form and state S.

Solution path

$y = (x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4$ , so S(−3|−4).

Check: $d = -\frac{6}{2} = -3$ , $f(-3) = 9 - 18 + 5 = -4$ ✓.

Task

A parabola has vertex S(2|−3) and passes through P(4|5). Determine the equation.

Solution path

$y = a(x - 2)^{2} - 3$ with P: $5 = a\cdot{}(4 - 2)^{2} - 3 = 4a - 3$ , so $a = 2$ .
Result: $y = 2(x - 2)^{2} - 3$ .

Frequently asked questions about Vertex Form of the Parabola

How do you read the vertex from the vertex form?+

How does completing the square work?+

What does the parameter a do in the vertex form?+

How do you find the roots from the vertex form?+

How do you set up a parabola equation from the vertex and one point?+

Retain Vertex Form of the Parabola for exams

Create a curated FSRS exam set for y = a(x − d)² + e: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential