General, standard or factored: which form of a quadratic function answers what

By Marius, private math tutor in Montreal · 2026-10-05

A quadratic function can be written in general form $a x^{2} + b x + c$ , standard form $a (x - h)^{2} + k$ (sometimes called vertex form) or factored form $a (x - x_{1}) (x - x_{2})$ , and each one hands you a different feature for free: the initial value, the vertex, or the zeroes. So choose the form from what the question asks for, and when you were given a different one, convert by expanding, completing the square or factoring.

One parabola, three rules

Here is a single function written three ways. Expand either of the last two and you get the first.

General form, $f (x) = - 2 x^{2} - 4 x + 16$ . The constant $16$ is the initial value $f (0)$ , where the graph crosses the $y$ -axis.
Standard form, $f (x) = - 2 (x + 1)^{2} + 18$ . The vertex is $(- 1, 18)$ , so the axis of symmetry is $x = - 1$ and the maximum is $18$ .
Factored form, $f (x) = - 2 (x + 4) (x - 2)$ . The zeroes are $- 4$ and $2$ , where the graph crosses the $x$ -axis.

The graph of

f

, with its axis of symmetry dashed. Each labelled point can be read straight off one of the three rules, without any calculation.

All three rules share the same $a = - 2$ . It sets the direction of opening (downward, since $a < 0$ ) and how narrow the parabola is, and it is the one parameter you never have to change form to find.

Four questions, and where each answer lives

Most questions about a quadratic function come down to four features. For each one, a single form shows it directly; the other two need one short calculation, which is often quicker than converting the whole rule.

The vertex and the axis of symmetry

Standard form: read it. $a (x - h)^{2} + k$ has vertex $(h, k)$ .
Factored form: the parabola is symmetric about its axis, so $h$ sits halfway between the zeroes, $h = \frac{x_{1} + x_{2}}{2}$ . Then $k = f (h)$ . For $f$ : $h = \frac{- 4 + 2}{2} = - 1$ and $k = - 2 (3) (- 3) = 18$ .
General form: $h = - \frac{b}{2 a}$ , then $k = f (h)$ . For $f$ : $h = - \frac{- 4}{2 (- 2)} = - 1$ . The formula comes from completing the square with letters.

The zeroes

Factored form: read them, with the signs reversed. $a (x - x_{1}) (x - x_{2})$ is zero at $x_{1}$ and $x_{2}$ , so the bracket $(x + 4)$ gives the zero $- 4$ .
Standard form: set the rule equal to zero and isolate the square, $(x - h)^{2} = - \frac{k}{a}$ , then take the square root with both signs. For $f$ : $(x + 1)^{2} = 9$ , so $x = - 1 \pm 3$ .
General form: factor, with the product-sum method, or use the quadratic formula. The discriminant $Δ = b^{2} - 4 a c$ tells you first whether there are two zeroes, one or none.

The initial value

General form: it is $c$ .
Standard form: $f (0) = a h^{2} + k$ . For $f$ : $- 2 (1)^{2} + 18 = 16$ .
Factored form: $f (0) = a (0 - x_{1}) (0 - x_{2}) = a x_{1} x_{2}$ . For $f$ : $- 2 (4) (- 2) = 16$ .

The maximum or minimum

Which one comes from the sign of $a$ , in any form. If $a > 0$ , the parabola opens upward and has a minimum; if $a < 0$ , it opens downward and has a maximum.
Its value is $k$ , the second coordinate of the vertex, so find the vertex by whichever route above is shortest. For $f$ , $a = - 2 < 0$ : a maximum of $18$ at $x = - 1$ , and the range is $] - \infty, 18]$ .

Moving from one form to another

Standard or factored to general: expand. In standard form, square the bracket before multiplying by $a$ : $- 2 (x + 1)^{2} + 18 = - 2 (x^{2} + 2 x + 1) + 18 = - 2 x^{2} - 4 x + 16$ .
General to standard: complete the square, or find $h = - \frac{b}{2 a}$ and $k = f (h)$ and write $a (x - h)^{2} + k$ with the same $a$ .
General to factored: find the zeroes by factoring or by the quadratic formula, then write $a (x - x_{1}) (x - x_{2})$ , again with the same $a$ .
Factored to standard: the vertex sits on the vertical line through the midpoint of the zeroes, as in the vertex section.
Standard to factored: solve $a (x - h)^{2} + k = 0$ for the zeroes, as in the zeroes section.

The two directions that end in factored form are not always available. The general and standard forms exist for every quadratic function, but a parabola that stays entirely above or entirely below the $x$ -axis has no zeroes, and so no factored form with real numbers.

The most common mistake. The same $a$ belongs in all three forms, so the zeroes alone do not determine a quadratic function. Once you have found the zeroes $- 4$ and $2$ of $- 2 x^{2} - 4 x + 16$ , its factored form is $- 2 (x + 4) (x - 2)$ . The trap is to write the two brackets and drop the $- 2$ . The brackets alone describe a different function, with the same zeroes but an upward opening, a different vertex and a different initial value. Expanding your answer back to general form catches it.

Check your understanding

No calculator. For each question, decide first which form you want.

Give the vertex of $g (x) = 4 (x + 3)^{2} - 1$ , and say whether $g$ has a maximum or a minimum.
Without expanding, find the initial value of $p (x) = 2 (x - 1) (x + 5)$ .
Find the vertex of $q (x) = - (x - 7) (x + 3)$ .
Write $r (x) = 2 (x + 2)^{2} - 32$ in factored form and in general form.
Find the maximum and the range of $s (x) = - x^{2} + 6 x + 1$ .
Maya finds that the zeroes of $2 x^{2} + 2 x - 12$ are $- 3$ and $2$ , and writes its factored form as $(x + 3) (x - 2)$ . What is wrong, and what is the correct factored form?
True or false: $u (x) = (x - 1) (x - 9)$ and $v (x) = 3 (x - 1) (x - 9)$ have the same vertex. Explain.
Explain why the vertex of $x^{2} - 6 x + c$ has $x$ -coordinate $3$ , whatever the value of $c$ .

Show the answers

$(- 3, - 1)$ , read from the standard form: $x + 3$ is $x - (- 3)$ , so $h = - 3$ , and $k = - 1$ . Since $a = 4 > 0$ , $g$ has a minimum, $- 1$ .
$- 10$ . The initial value is $p (0) = 2 (0 - 1) (0 + 5) = 2 (- 1) (5) = - 10$ .
$(2, 25)$ . The zeroes are $7$ and $- 3$ , so $h = \frac{7 + (- 3)}{2} = 2$ , and $q (2) = - (2 - 7) (2 + 3) = - (- 5) (5) = 25$ . Since $a = - 1 < 0$ , it is a maximum.
Factored form $r (x) = 2 (x - 2) (x + 6)$ : setting $2 (x + 2)^{2} - 32 = 0$ gives $(x + 2)^{2} = 16$ , so $x + 2 = \pm 4$ , that is $x = 2$ or $x = - 6$ . General form $r (x) = 2 x^{2} + 8 x - 24$ , from $2 (x^{2} + 4 x + 4) - 32$ . Both keep $a = 2$ .
A maximum of $10$ , at $x = 3$ , and the range is $] - \infty, 10]$ . From the general form, $h = - \frac{6}{2 (- 1)} = 3$ and $s (3) = - 9 + 18 + 1 = 10$ ; since $a = - 1 < 0$ , $10$ is the greatest value. In standard form, $s (x) = - (x - 3)^{2} + 10$ .
She dropped $a = 2$ . The product $(x + 3) (x - 2)$ expands to $x^{2} + x - 6$ : same zeroes, but a different function, with initial value $- 6$ instead of $- 12$ . The correct factored form is $2 (x + 3) (x - 2)$ .
False. Both have the zeroes $1$ and $9$ , so both vertices lie on the axis $x = 5$ . But $u (5) = (4) (- 4) = - 16$ and $v (5) = 3 (4) (- 4) = - 48$ , so the vertices are $(5, - 16)$ and $(5, - 48)$ . Multiplying by 3 stretches the parabola vertically, which moves every point except the zeroes.
From the general form, $h = - \frac{b}{2 a} = - \frac{- 6}{2 (1)} = 3$ , and $c$ does not appear in that calculation. Changing $c$ only slides the parabola up or down, which leaves its axis of symmetry where it is. The vertex is $(3, c - 9)$ .

Keep practising

The free Secondary 4 Quadratic Functions worksheet asks for the vertex, the zeroes, the $y$ -intercept and the maximum or minimum of rules given in different forms, and for conversions between the forms. For the technique behind each conversion, see how to complete the square (general to standard) and the product-sum method (general to factored).

Want one-on-one help with this? I offer Secondary 4 and 5 math tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.