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Vertex Form of a Parabola

Vertex form y = a(x−h)² + k gives vertex (h,k), axis x = h, and direction from a. a > 0: opens up; a < 0: opens down. Converts to standard form by expanding.

Concept Fundamentals
y = a(x−h)² + k
Form
(h, k)
Vertex
x = h
Axis
a > 0 up, a < 0 down
Direction

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Vertex (h,k) is max if a<0, min if a>0. Axis of symmetry: x = h. Completing square converts standard → vertex.

Key quantities
y = a(x−h)² + k
Form
Key relation
(h, k)
Vertex
Key relation
x = h
Axis
Key relation
a > 0 up, a < 0 down
Direction
Key relation

Ready to run the numbers?

Why: Vertex form directly reveals the vertex and axis of symmetry. Used in optimization (vertex = max/min), projectile motion, and graphing parabolas.

How: Vertex (h,k) from the equation. Expand a(x−h)² + k to get standard form ax²+bx+c. Axis of symmetry x = h. Y-intercept: set x=0.

Vertex (h,k) is max if a<0, min if a>0.Axis of symmetry: x = h.
Sources:Khan Academy — Vertex FormWolfram MathWorld — Parabola

Run the calculator when you are ready.

Analyze Vertex FormEnter a, h, k

Sample Examples

Input (y = a(x-h)² + k)

Results

Vertex

(0, 0)

Axis of symmetry: x = 0

Opens: up

Standard form: y = 1x² + -0x + 0

Y-intercept: (0, 0)

Zeros: 0

Visualization

For educational and informational purposes only. Verify with a qualified professional.

🧮 Fascinating Math Facts

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y = a(x−h)² + k; vertex (h,k).

— Vertex Form

V

Axis of symmetry: x = h.

— Property

Key Takeaways

  • Vertex form: y = a(x-h)² + k; vertex at (h,k)
  • Axis of symmetry: x = h
  • a > 0: opens up; a < 0: opens down
  • Standard form: y = ax² + bx + c; b = -2ah, c = ah² + k
  • Zeros from 0 = a(x-h)² + k → x = h ± √(-k/a)

Did You Know?

  • Vertex form reveals transformations of y = x²
  • Projectile motion: h(t) = -½gt² + v₀t + h₀
  • Completing the square converts standard → vertex
  • Optimal values at vertex (max/min)
  • Used in optimization, physics, economics
  • Parabola focus/directrix from vertex form

Understanding

Vertex form makes the vertex (h,k) and axis x=h explicit.

y=a(x−h)2+k⇒Vertex (h,k)y = a(x-h)^2 + k \quad \Rightarrow \quad \text{Vertex } (h, k)
y=ax2+bx+cwhere b=−2ah,  c=ah2+ky = ax^2 + bx + c \quad \text{where } b = -2ah, \; c = ah^2 + k

Expert Tips

  • |a| > 1: narrower; 0 < |a| < 1: wider
  • Y-intercept: set x=0 → y = ah² + k
  • Zeros exist when k/a ≤ 0
  • Completing square: h = -b/(2a), k = c - b²/(4a)

FAQ

Q: Why vertex form?
A: Directly shows vertex, axis, direction.
Q: How to find zeros?
A: Solve 0 = a(x-h)² + k; x = h ± √(-k/a).
Q: When no real zeros?
A: When -k/a < 0 (a and k same sign).
Q: Applications?
A: Projectiles, profit max, optics.
Q: From standard form?
A: Complete the square or use h=-b/(2a), k=c-b²/(4a).
Q: a=0?
A: Not quadratic; linear.
Q: Direction?
A: a>0 up, a<0 down.

How to Use

  1. Enter a, h, k for y = a(x-h)² + k
  2. Get vertex, axis, standard form, intercepts, zeros
  3. Visualize the parabola

Disclaimer

a ≠ 0 for quadratic. Zeros are real when -k/a ≥ 0.

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