Completing the Square, Axis and Vertex

In $y = a(x - h)^2 + k$ the squared bracket can never be negative. If $a > 0$ the term $a(x-h)^2$ is at its smallest, namely zero, exactly when $x = h$, and then $y = k$.

So the vertex is $(h, k)$ and the parabola is symmetric about the vertical line $x = h$. If $a < 0$ the same reasoning gives a maximum instead: $y$ is largest when the bracket is zero.

The signs behave as they did for circles: $y = (x - 4)^2 + 1$ has vertex $(4, 1)$, while $y = (x + 4)^2 + 1$ has vertex $(-4, 1)$.

In $y = a(x-h)^2 + k$ the vertex is $(h,k)$ and the axis of symmetry is $x = h$, because a squared bracket is never negative and is zero exactly at $x = h$. With $a > 0$ the vertex is a minimum and the range is $y \geq k$; with $a < 0$ it is a maximum.

Pause, copy completed-square form with the vertex and axis identified, the never-negative reasoning, and the range statement, into your book.

We just saw how to read a vertex once a quadratic is in completed-square form. That raises a question: how do you get it there when the coefficient of $x^2$ is not 1? This card answers it → factor that coefficient out of the $x$ terms first, then complete the square inside the bracket.

Start from $y = 2x^2 + 8x + 3$. Factor 2 out of the $x$ terms only: $y = 2(x^2 + 4x) + 3$. Leave the constant outside.

Complete the square inside: half of 4 is 2, and $2^2 = 4$, so $x^2 + 4x = (x+2)^2 - 4$. That gives $y = 2[(x+2)^2 - 4] + 3$.

Expand the outer factor: $y = 2(x+2)^2 - 8 + 3 = 2(x+2)^2 - 5$. Vertex $(-2, -5)$, axis $x = -2$, and since $a = 2 > 0$ it is a minimum.

Factor $a$ out of the $x$ terms only, leaving the constant outside. Complete the square inside the bracket by halving and squaring. Then distribute $a$ back through, remembering it multiplies the subtracted term as well.

Pause, copy the full worked $y = 2x^2 + 8x + 3$ becoming $2(x+2)^2 - 5$, and the warning that the subtracted term is multiplied by $a$, into your book.

We just saw the method on a specific quadratic. That raises a question: what happens if you run it on the general quadratic $ax^2 + bx + c$? This card answers it → you get the axis-of-symmetry formula, which is what the syllabus asks you to show.

Take $y = ax^2 + bx + c$ and factor $a$ from the $x$ terms: $y = a\left(x^2 + \dfrac{b}{a}x\right) + c$.

Complete the square inside. Half of $\dfrac{b}{a}$ is $\dfrac{b}{2a}$, so $x^2 + \dfrac{b}{a}x = \left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a^2}$.

That gives $y = a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a} + c$. Comparing with $a(x-h)^2 + k$ shows $h = -\dfrac{b}{2a}$, so the axis of symmetry is $x = -\dfrac{b}{2a}$.

Completing the square on $ax^2+bx+c$ gives $a\left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$, so the axis of symmetry is $x = -\frac{b}{2a}$ and that is also the $x$-coordinate of the vertex. It holds for every $a$, and the derivation itself is examinable.

Pause, copy the three-line derivation in full, and the note that $-\frac{b}{2a}$ is both the axis and the vertex $x$-coordinate, into your book.

1. State the vertex of $y = (x - 7)^2 + 3$.
2. Write $y = x^2 + 4x + 1$ in completed-square form.
3. Find the axis of symmetry of $y = 2x^2 - 8x + 1$.
4. State the minimum value of $y = (x+1)^2 - 6$.