Vertex Form Calculator
The Vertex Form Calculator instantly converts quadratic equations into vertex form, helping identify vertex, axis of symmetry, and parabola transformations.
a here. Leave expression blank below to use these.b for standard form ax^2 + bx + c.c. The y‑intercept equals c.x. This is parsed with and shown exactly (no LaTeX). If provided, it takes priority over the coefficient fields.Results
x^2 + 6x + 5(x - (-3))^2 - 4Helping notes
- Use either the coefficients
a, b, cor type the full expression; the expression takes priority. - For standard form
ax^2 + bx + c, vertex is ath = -b/(2a),k = f(h). - Decimal places affect displayed results only.
What is Vertex Form Calculator?
A Vertex Form Calculator is a focused tool that rewrites any quadratic function from standard form into the cleaner, insight-friendly vertex form. The standard representation is $$y = ax^2 + bx + c$$ while the vertex form is $$y = a(x - h)^2 + k,$$ where \((h, k)\) is the vertex of the parabola and \(a\) controls opening and vertical stretch. Converting reveals the parabola’s maximum or minimum point, the axis of symmetry, and how the graph shifts relative to \(y = ax^2\). Because vertex form exposes geometry directly, it helps with graphing, optimization, and interpretation in algebra, precalculus, physics, and data modeling.
About the Vertex Form Calculator
This calculator is designed to show the algebra and the result side by side, so learners and practitioners can verify each step. It accepts coefficients \(a\), \(b\), and \(c\) from the standard form $$y=ax^2+bx+c,$$ then returns the exact vertex form $$y=a(x-h)^2+k,$$ the vertex \((h,k)\), and the axis of symmetry \(x=h\). It also highlights the canonical vertex formulas $$h=-\frac{b}{2a},\qquad k=f(h)=a\,h^2+b\,h+c,$$ which provide a fast conversion without manual completing-the-square. When used with MathJax on your page, the formulas render crisply and wrap responsively to fit small screens.
How to Use this Vertex Form Calculator
- Enter \(a\), \(b\), and \(c\) from the quadratic \(y=ax^2+bx+c\).
- Compute \(h=-\dfrac{b}{2a}\), then evaluate \(k=f(h)=a\,h^2+b\,h+c\).
- Report the vertex form \(y=a(x-h)^2+k\) and the axis of symmetry \(x=h\).
- Interpret: the sign of \(a\) sets opening (up for \(a>0\), down for \(a<0\)); \(|a|\) sets vertical stretch.
If you prefer the algebraic route, complete the square: $$y=ax^2+bx+c=a\Big(x^2+\frac{b}{a}x\Big)+c =a\Big[\Big(x+\frac{b}{2a}\Big)^2-\Big(\frac{b}{2a}\Big)^2\Big]+c =a\Big(x-h\Big)^2+k,$$ confirming the same \((h,k)\) as above.
Examples (with the same formulas applied)
Example 1: \(y=2x^2-8x+5\). \(h=-\dfrac{-8}{2\cdot 2}=2,\; k=f(2)=2(4)-16+5=-3.\) Vertex form: $$y=2(x-2)^2-3,\quad \text{vertex }(2,-3),\ \text{axis }x=2.$$
Example 2: \(y=-x^2+6x-10\). \(h=-\dfrac{6}{2(-1)}=3,\; k=f(3)=-9+18-10=-1.\) Vertex form: $$y=-(x-3)^2-1,\quad \text{vertex }(3,-1),\ \text{axis }x=3.$$
Example 3: \(y=\tfrac12 x^2+x+4\). \(h=-\dfrac{1}{2\cdot \tfrac12}=-1,\; k=f(-1)=\tfrac12(1)-1+4=\tfrac{7}{2}.\) Vertex form: $$y=\tfrac12(x+1)^2+\tfrac{7}{2},\quad \text{vertex }(-1,\tfrac{7}{2}),\ \text{axis }x=-1.$$