Quadratic Functions - General, Vertex and Factored Form

A quadratic function is the first function in this course whose graph genuinely curves — a parabola, not a straight line. It is also the first function for which the shape of the formula itself carries useful information: depending how it is written, the same quadratic can instantly reveal its intercepts, its turning point, or neither. This chapter develops the three standard forms of a quadratic, derives the vertex formula by completing the square, and builds fluency converting between forms — the essential groundwork for Chapter 5's inequalities and nonlinear systems.

You need confident expansion and factorisation of quadratic expressions, and the linear-function toolkit from Chapter 3 (gradient as rate of change will reappear as the tangent gradient at the vertex, in spirit, once calculus arrives in Block D).

1. The Three Standard Forms

A quadratic function has the form $f(x) = ax^2+bx+c$ for constants $a\neq0, b,c$. This is the general form. Two other forms express exactly the same function, chosen to expose different features at a glance.

Definition — the three forms.

• General form: $f(x)=ax^2+bx+c$ — the constant term $c=f(0)$ is the $y$-intercept, read off immediately.
• Factored form: $f(x)=a(x-p)(x-q)$ — the roots (zeros) $p,q$ are read off immediately, since $f(p)=f(q)=0$.
• Vertex form: $f(x)=a(x-h)^2+k$ — the vertex (turning point) $(h,k)$ is read off immediately.

In every form, the leading coefficient $a$ is the same number, and it controls the parabola's width and direction: $a>0$ opens upward (a minimum at the vertex), $a<0$ opens downward (a maximum).

2. From General Form to Vertex Form: Completing the Square

Vertex form is not a separate fact to memorise — it is derived from general form by completing the square, a technique worth doing from scratch rather than recalling a formula.

Worked example 1. Convert $f(x) = 2x^2-12x+7$ to vertex form.

Factor the leading coefficient out of the first two terms only:

Complete the square inside the brackets: take half the coefficient of $x$ (which is $-6$, so half is $-3$), square it ($9$), add and subtract it inside the bracket:

Distribute the $2$ back through, being careful it multiplies both terms inside:

So the vertex is $(3,-11)$, and since $a=2>0$, this is a minimum.

Warning. The constant subtracted inside the bracket must be multiplied by $a$ when it is moved outside — forgetting this (moving $-9$ straight to $+7$ without multiplying by $2$ first) is the single most common error in this technique.

3. The Vertex Formula

Completing the square on the general form $f(x)=ax^2+bx+c$ once, symbolically, produces a reusable formula:

so the vertex is at $h=-\dfrac{b}{2a}$, and $k=f(h)$ is most reliably found by substituting $h$ back into the original function rather than memorising the expression for $k$ above.

Worked example 2. Find the vertex of $f(x)=-3x^2+6x+1$ using the formula.

Vertex: $(1,4)$, a maximum since $a=-3<0$.

4. From General Form to Factored Form

Factored form requires finding the roots of $f(x)=0$ — by factoring directly when possible, or by the quadratic formula otherwise:

Worked example 3. Convert $f(x)=x^2-x-12$ to factored form.

Look for two numbers multiplying to $-12$ and summing to $-1$: these are $-4$ and $3$. So $f(x)=(x-4)(x+3)$, with roots $x=4$ and $x=-3$.

Worked example 4 (the edge case — no real factorisation). Attempt to write $f(x)=x^2+2x+5$ in factored form over the reals.

The discriminant $\Delta = b^2-4ac = 4-20=-16 <0$. Since $\Delta<0$ gives no real square root, there is no real factorisation — the parabola never crosses the $x$-axis. This is not a failure of the method; it is the method correctly reporting that factored form over $\mathbb{R}$ does not exist for this quadratic. (The complex roots $x=-1\pm2i$ exist, but are outside this chapter's scope.)

Key distinction. Every quadratic has a vertex form and a general form, but not every quadratic has a real factored form — factored form exists over $\mathbb{R}$ exactly when $\Delta=b^2-4ac\geq0$.

5. From Factored or Vertex Form Back to General Form

Going the other direction is pure expansion.

Worked example 5. Expand $f(x)=3(x-2)(x+5)$ to general form.

Worked example 6. Expand $f(x)=-2(x+1)^2+6$ to general form.

6. The Axis of Symmetry and Reading Off Features

The vertical line $x=h$ through the vertex is the axis of symmetry: the parabola is a mirror image of itself across this line, since $f(h-t)=f(h+t)$ for every $t$. This gives a fast way to find a second root once one root and the vertex's $x$-coordinate are known: the two roots are symmetric about $x=h$.

Worked example 7. A parabola has a root at $x=-1$ and axis of symmetry $x=3$. Find the other root.

The other root is the reflection of $-1$ across $x=3$: distance from $-1$ to $3$ is $4$, so the other root is $3+4=7$.

Common Exam Pitfalls

• Forgetting to multiply the completed-square constant by $a$ before moving it outside the bracket. The constant correction is $-a\times(\text{half the inner coefficient})^2$, not just that squared number alone.
• Reading the vertex straight off general form. The vertex is not $(b,c)$ or any direct reading of the general-form coefficients — it requires completing the square or the formula $h=-\frac{b}{2a}$.
• Assuming every quadratic factors over the reals. Check the discriminant first; $\Delta<0$ means no real roots and no real factored form.
• Sign errors expanding $(x-p)(x-q)$, especially when $p$ or $q$ is itself negative — rewrite $(x-(-3))$ as $(x+3)$ explicitly before expanding.
• Confusing the sign of $a$'s effect: $a>0$ gives a minimum (opens upward), $a<0$ gives a maximum (opens downward) — stating the wrong one after correctly finding the vertex coordinates.
• Using the vertex formula for $k$ from memory incorrectly. It is far safer to compute $h=-\frac{b}{2a}$ and then substitute into the original $f$ to get $k=f(h)$, rather than recalling the symbolic expression for $k$.

Summary

• Three equivalent forms: general $ax^2+bx+c$ (shows $y$-intercept $c$), factored $a(x-p)(x-q)$ (shows roots $p,q$), vertex $a(x-h)^2+k$ (shows vertex $(h,k)$).
• Vertex form is derived from general form by completing the square; the vertex is also given directly by $h=-\dfrac{b}{2a}$, $k=f(h)$.
• The sign of $a$ determines the parabola's direction: $a>0$ minimum, $a<0$ maximum; $|a|$ controls its width.
• Factored form exists over $\mathbb{R}$ exactly when the discriminant $\Delta=b^2-4ac\geq0$; the roots are $x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$.
• The axis of symmetry $x=h$ reflects the parabola onto itself; given one root and the axis, the other root is the mirror image across $x=h$.
• Converting to general form from either other form is pure expansion; converting from general form requires completing the square (vertex) or factoring/the quadratic formula (factored).