What Are the Three Forms of a Quadratic Function (Standard, Factored, Vertex)?

The three forms are standard form $f(x)=ax^2+bx+c$, factored form $f(x)=a(x-p)(x-q)$, and vertex form $f(x)=a(x-h)^2+k$, where $a\neq 0$. Each form represents the same type of parabola but makes different features immediately visible.

This is part of IB Mathematics: Analysis and Approaches, SL 2.6: Quadratic function, and applies to both Paper 1 and Paper 2.

In every form, $a$ controls the parabola's direction and width. If $a>0$, it opens upwards; if $a<0$, it opens downwards.

The forms are equivalent, so conversion does not change the graph. To move from standard to factored form, solve $f(x)=0$ or factor algebraically. To move from standard to vertex form, complete the square. Expanding either factored or vertex form returns standard form.

For example, to convert $x^2-6x+5$ into vertex form, group the first two terms and add and subtract $9$: $x^2-6x+9-9+5=(x-3)^2-4$. This preserves equality while creating a perfect square. In standard form, the discriminant determines whether there are two distinct, one repeated, or no real roots.

The axis can also be found from standard form using $x=-\frac{b}{2a}$. Substituting this value into $f$ gives the $y$-coordinate of the vertex. For real roots and , the axis lies halfway between them: .

Consider $f(x)=x^2-6x+5$. Factorizing gives $f(x)=(x-1)(x-5)$, so its roots are and . Completing the square gives , so its vertex is and its axis is . The midpoint of the roots is also , confirming the result.

A common misconception is reading the signs directly: $(x-3)^2$ gives an $x$-coordinate of $3$, not $-3$.

For IB exams, use factored form for roots, vertex form for extrema, and standard form for the $y$-intercept or discriminant. Show all conversion algebra because an unsupported answer may not receive full marks.