Vertex Form and How to Graph a Parabola

Every quadratic graphs as a U-shaped curve called a parabola. Its turning point, the vertex, is the most useful point on the graph, and vertex form puts that point right in the equation.

A ideia em resumo

In vertex form y = a(x − h)² + k, the vertex is (h, k). From standard form y = ax² + bx + c, the vertex sits at x = −b/(2a).

Passo a passo

1. Find the x-coordinate of the vertex with x = −b/(2a), or read h from vertex form.
2. Substitute that x into the equation to get the y-coordinate.
3. Check the sign of a: positive opens up, negative opens down.
4. Plot the vertex, the y-intercept, and any x-intercepts, then draw a smooth symmetric curve.

Exemplo resolvido

Graph y = x² − 4x + 3 and write it in vertex form.

Solução completa

Explicação passo a passo

Passo 1

Here a = 1 and b = −4, so the vertex is at x = 4/2 = 2.

Passo 2

Substitute x = 2: y = 4 − 8 + 3 = −1. The vertex is (2, −1).

Passo 3

So y = (x − 2)² − 1. It opens up, crosses the y-axis at 3, and crosses the x-axis at x = 1 and x = 3.

Erro comum

Watch the sign of h. In y = (x + 3)² − 2 the vertex is at x = −3, not 3, because x + 3 is the same as x − (−3).

Exercícios para praticar

1. Find the vertex of y = (x + 1)² + 4.
2. Find the vertex of y = x² + 6x + 5.
3. Does y = −2(x − 1)² + 3 open up or down, and what is its maximum value?

Gabarito

1. (−1, 4)
2. (−3, −4)
3. It opens down; the maximum value is 3, at x = 1.

Peça ao PennPaper para explicar ao vivo

Se os passos fazem sentido mas você ainda travou, comece uma aula no PennPaper e peça ao tutor para desenhar o problema na lousa. Ver os símbolos se movendo passo a passo costuma ser o que faz a ficha cair.

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