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 x² + bx into a perfect square. Recipe: square half of the x-coefficient, add it and subtract it again immediately. Example: y = x² + 6x + 5. Halve 6 to 3, complete with 3² = 9: y = x² + 6x + 9 − 9 + 5 = (x + 3)² − 4, so S(−3|−4). If a factor sits before x², factor it out first: y = 2x² + 8x = 2(x² + 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).

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

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

Dimensionless (algebra)

Variables & units – Vertex Form of the Parabola

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 – Vertex Form of the Parabola

From the basic parabola y = ax² the general parabola arises by shifting d to the right and e upwards. The transition from the standard form y = ax² + 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 = −b/(2a). For |a| > 1 the parabola is stretched, for |a| < 1 compressed.

Exam blueprint

Validity range

Holds for all quadratic functions (a ≠ 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 x² + bx into a perfect square.

1From y = ax² + bx + c factor out a, add and subtract (b/(2a))².
2Combining yields y = a(x + b/(2a))² + c − b²/(4a), so d = −b/(2a) and e = c − b²/(4a).

Rearrangements

Vertex from the standard form

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

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

Roots from the vertex form

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

Solvable only if −e/a ≥ 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 = x² + 6x + 5 into vertex form and state S.

y = (x + 3)² − 9 + 5 = (x + 3)² − 4, so S(−3|−4). Check: d = −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·(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 = 2x² + 8x first 2(x² + 4x), then complete inside the bracket.

Exam context

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

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Formula cluster

Quadratic functions

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

x_{1, 2} = - \frac{p}{2} \pm (\frac{p}{2})^{2} - q

pq Formula

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

Quadratic Formula

(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}, (a + b) (a - b) = a^{2} - b^{2}

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: x² + 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

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Which formula describes Vertex Form of the Parabola?

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How do you rearrange y = a(x − d)² + e for Vertex from the standard form?

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Which common mistake happens with Vertex Form of the Parabola?

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Scientific sources

[1]Bronstein, I.N. et al. (2020). Taschenbuch der Mathematik. Springer.

Common notations & search queries

y=a(x-d)^2+eScheitelpunktformScheitelform ParabelScheitelpunkt ablesenquadratische Ergänzung Scheitelpunktvertex form parabolaScheitelpunkt berechnen FormelNormalform in Scheitelpunktform

Related formulas

x_{1, 2} = - \frac{p}{2} \pm (\frac{p}{2})^{2} - q

pq Formula

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

Quadratic Formula

(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}, (a + b) (a - b) = a^{2} - b^{2}

Binomial Formulas

More Mathematics formulas

a^{2} + b^{2} = c^{2}

Pythagorean Theorem

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

Quadratic Formula

e^{i φ} = cos φ + i \cdot sin φ

Euler's Formula

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

Product Rule of Differentiation

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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 x² + bx into a perfect square. Recipe: square half of the x-coefficient, add it and subtract it again immediately. Example: y = x² + 6x + 5. Halve 6 to 3, complete with 3² = 9: y = x² + 6x + 9 − 9 + 5 = (x + 3)² − 4, so S(−3|−4). If a factor sits before x², factor it out first: y = 2x² + 8x = 2(x² + 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 = −e/a, so x = d ± √(−e/a). Example: 2(x − 2)² − 3 = 0 leads to (x − 2)² = 1.5 and x = 2 ± √1.5 ≈ 0.78 and 3.22. From −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 = (y_P − e)/((x_P − d)²). Example: vertex S(2|−3), point P(4|5): 5 = a·(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.

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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 = x² + 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 = −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
y = a(x − 2)² − 3 with P: 5 = a·(4 − 2)² − 3 = 4a − 3, so a = 2. Result: y = 2(x − 2)² − 3.