Why Completing the Square Matters for Vertex Form and the Turning Point

A quadratic function and its turning point. Link to graph: https://www.desmos.com/calculator/fktyfs12st

A quadratic function is any function of the form f(x) = ax² + bx + c with a ≠ 0. Its graph is a parabola, and every parabola has exactly one turning point (also called the vertex). Completing the square is fundamental because it rewrites the quadratic as a shifted square, which makes the turning point immediately visible.

Vertex form: the turning point is built in

The vertex form of a quadratic is:

f(x) = a(x − h)² + k

This form is powerful because it exposes two facts at once:

  • (x − h)² ≥ 0 for all real x (a square is never negative).
  • (x − h)² = 0 happens exactly when x = h.

So:

  • If a > 0, then a(x − h)² ≥ 0, so the smallest possible value of f(x) is k, achieved at x = h (a minimum).
  • If a < 0, then a(x − h)² ≤ 0, so the largest possible value of f(x) is k, achieved at x = h (a maximum).

Therefore, in vertex form, the turning point is simply:

Turning point = (h, k)

Why standard form hides the turning point

In standard form ax² + bx + c, the turning point is not obvious because the expression does not visibly contain a single square like (x − h)². But geometrically, every parabola is just a stretched and shifted version of the basic square function y = x². The “shifted square” structure is real — it is just not exposed yet.

Completing the square is the algebraic process that exposes it.

The key pattern behind a perfect square

Expand a shifted square:

(x − h)² = x² − 2hx + h²

Notice the constraint: once the x-term is chosen, the constant term is forced. A square is not “x² plus some x-term”; it is “x² plus an x-term and the matching constant that completes the square”. That is exactly why completing the square works: it adds and subtracts the constant needed to make a perfect square, without changing the function overall.

Completing the square in the simplest case

Start with:

x² + px

Half the coefficient of x is p/2. Add and subtract its square:

x² + px = x² + px + (p/2)² − (p/2)²

Now the first three terms form a perfect square:

= (x + p/2)² − (p/2)²

This is the essence of the method: manufacture a square, then correct with a constant so nothing has actually changed.

Completing the square for f(x) = ax² + bx + c

Start with:

f(x) = ax² + bx + c

Factor out a from the x-terms:

f(x) = a(x² + (b/a)x) + c

Inside the brackets, this is x² + px with p = b/a. Half of that is b/(2a). Add and subtract its square inside the brackets:

f(x) = a[(x + b/(2a))² − (b/(2a))²] + c

Distribute a and simplify the constant term:

f(x) = a(x + b/(2a))² + (c − b²/(4a))

This is vertex form:

f(x) = a(x − h)² + k

where:

  • h = −b/(2a)
  • k = c − b²/(4a)

Why this instantly gives the turning point

Once you have:

f(x) = a(x − h)² + k

the turning point is immediate because (x − h)² is minimised at 0 (or maximised after multiplying by a negative a) exactly when x = h. At that x-value, the function equals k.

So the turning point is:

(h, k)

Summary: why completing the square is fundamental

  • Vertex form is a quadratic written as “a shifted square plus a constant”.
  • The turning point is where the square term is as small as it can be (or as large after scaling by a negative).
  • Completing the square is the algebra that converts ax² + bx + c into that shifted-square structure.

In short: vertex form makes the turning point obvious, and completing the square is the method that reveals vertex form.

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