A quadratic in vertex form y=a(x−h)²+k, with a≠0, has vertex (h,k) and symmetry axis x=h.
A quadratic function in vertex form (with a≠0) y=a(x−h)²+k, with a≠0, has vertex (h,k) and symmetry axis x=h. If a is positive it opens upward; if negative, downward. The domain of a quadratic polynomial is all real x unless the situation imposes additional restrictions. Vertex form displays location and orientation directly.
For y=2(x−3)²−1, the vertex is (3,−1), the axis is x=3, and the graph opens upward. At x=4, y=2(1)²−1=1. At x=2 the value is also 1, confirming symmetry around x=3. Substitution checks the calculation and illustrates how the squared distance from h controls paired points.
The coefficient a cannot be zero here: that would remove the quadratic term and no longer produce a parabola. A larger absolute value of a makes the graph narrower than the corresponding unit parabola; a negative sign reflects it downward. In an application, a time or length variable may have a restricted meaningful domain even though the polynomial is defined more widely.
Expanding gives ax²−2ahx+ah²+k, the standard polynomial form. Completing the square can recover the vertex when the quadratic coefficient is nonzero. The vertex does not by itself tell how many x-intercepts exist: one must solve y=0. A parabola may meet the x-axis twice, once, or not at all, depending on its position and shape. Una tabella di valori simmetrici rispetto a h rende riconoscibile l’asse del grafico.
- In y=a(x−h)²+k, il vertice è (h,k) e l’asse è x=h.
- Se a>0 la parabola apre verso l’alto; se a<0 verso il basso.
- La simmetria assiale implica valori uguali per x equidistanti da h.
Metti alla prova quello che hai letto
Vertice di y=2(x−3)²−1?
- (−3,−1)
- (3,1)
- (3,−1)
- (2,−3)
Soluzione e spiegazione
(3,−1). La forma y=a(x−h)²+k dà vertice (h,k).
Fonti
- 5.1 Quadratic Functions — College Algebra 2e — OpenStax
- Vertex Form of Quadratic Equation — MathBitsNotebook
Nota editoriale
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