Completing the square and the graph

Complete the square for monic and non-monic quadratics, read the vertex from vertex form, and sketch any parabola, with a live vertex-form graph.

Why complete the square?

Factoring only works when the zeroes are nice numbers. Completing the square works for every quadratic. It rewrites $y = ax^2 + bx + c$ in , which shows the vertex directly.

Vertex form

$$y = a(x - h)^2 + k \qquad \text{vertex } (h, k), \quad \text{axis } x = h$$

Changing $a$ only changes how wide the parabola is and which way it opens. So all the parabolas $y = a(x + 3)^2 + 2$ form a family that shares the vertex $(-3, 2)$.

Vertex form playground

Try this

Change $h$. Which way does the graph move when $h$ increases? Notice the sign inside the bracket is the opposite.
Change $k$. Find the value of $k$ where the zeroes disappear.
Set $a = -1$, $h = -2$, $k = 4$, and write down the expanded form.

Completing the square: $a = 1$

The trick is that $(x + p)^2 = x^2 + 2px + p^2$. So to make $x^2 + bx$ into a perfect square, add the square of half the coefficient of $x$, then take it away again so nothing changes.

Example 1●

Write $y = x^2 - 6x + 2$ in vertex form, and state the vertex.

Half of $-6$ is $-3$; squared is $9$.

Half of −6 is −3, and (−3)² = 9.

$y = x^2 - 6x + 9 - 9 + 2$

Adding 9 and taking 9 away leaves the expression unchanged.

$y = (x - 3)^2 - 7$

x² − 6x + 9 is the perfect square (x − 3)².

Vertex $(3, -7)$, axis $x = 3$.

Completing the square: $a \ne 1$

When the coefficient of $x^2$ is not $1$, take it out of the $x^2$ and $x$ terms first, then complete the square inside the bracket.

Example 2●●

Write $y = 2x^2 + 12x + 7$ in vertex form, then sketch it.

$y = 2(x^2 + 6x) + 7$

Factor 2 out of the first two terms only. The constant stays outside.

$y = 2(x^2 + 6x + 9 - 9) + 7$

Half of 6 is 3, so add and subtract 9 inside the bracket.

$y = 2(x + 3)^2 - 18 + 7$

The −9 is inside a bracket multiplied by 2, so it comes out as −18.

$y = 2(x + 3)^2 - 11$. Vertex $(-3, -11)$, concave up.

Zeroes: $(x + 3)^2 = \tfrac{11}{2}$, so $x = -3 \pm \sqrt{\tfrac{11}{2}} \approx -0.65$ or $-5.35$.

Set y = 0 and solve using square roots.

$y$-intercept: $7$. Sketch with the vertex, both zeroes and the $y$-intercept labelled.

Example 2 sketch

Flashcards

Vertex form

$y = a(x - h)^2 + k$, vertex $(h, k)$.

Number to add for x² + bx

$\left(\dfrac{b}{2}\right)^2$

First step when a ≠ 1

Factor $a$ out of the $x^2$ and $x$ terms only.

Vertex of $y = (x + 2)^2 - 3$?

$(-2, -3)$

Range from vertex form

$y \ge k$ if $a > 0$; $y \le k$ if $a < 0$.

Practice

The vertex of $y = (x + 4)^2 + 1$ is:
To complete the square on $x^2 + 10x$, you add and subtract:
$x^2 + 8x + 3 = (x + 4)^2 + k$. Find $k$.
The range of $y = -3(x - 1)^2 + 5$ is:
$x^2 - 10x + 7 = (x - 5)^2 + k$. Find $k$.
The vertex of $y = 2(x - 3)^2 - 8$ is:
The zeroes of $y = (x - 1)^2 - 9$ are:
Write $y = 3x^2 - 12x + 5$ as $a(x - h)^2 + k$. Find $k$.