How to find the rule of a quadratic function

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

Start from the form of the rule that already holds what you were given: with the vertex $(h, k)$ , use standard form $f (x) = a (x - h)^{2} + k$ ; with the two zeroes $x_{1}$ and $x_{2}$ , use factored form $f (x) = a (x - x_{1}) (x - x_{2})$ . Only $a$ is left unknown, so substitute the one other point, solve for $a$ , and finish by checking the rule against every point you were given.

Why one more point is enough

Written in general form, $f (x) = a x^{2} + b x + c$ , a quadratic rule has three unknowns. Each piece of information pins down one of them, so the trick is to pick a form in which most of the work is already done for you.

Standard form (vertex form) $f (x) = a (x - h)^{2} + k$ has the unknowns $a$ , $h$ and $k$ . A vertex hands you $h$ and $k$ at once.
Factored form $f (x) = a (x - x_{1}) (x - x_{2})$ has the unknowns $a$ , $x_{1}$ and $x_{2}$ . Two zeroes hand you $x_{1}$ and $x_{2}$ at once.

Either way, one unknown is left: $a$ . One more point gives one equation, which is exactly enough. It has to be a new point, though. Substituting the vertex into a rule you built from that vertex, or a zero into a rule you built from that zero, only repeats what you already wrote down.

Information sometimes arrives in disguise, and it fits the same two cases:

a $y$ -intercept of $c$ is the point $(0, c)$ ;
a zero at $x_{1}$ is the point $(x_{1}, 0)$ , so a vertex plus one zero is the first case;
a maximum or minimum of $k$ reached at $x = h$ is the vertex $(h, k)$ .

Case 1: the vertex and one point

Write standard form with $h$ and $k$ filled in. The form subtracts $h$ , so a vertex at $x = - 4$ gives the bracket $(x + 4)$ .
Replace $x$ and $f (x)$ with the coordinates of the other point.
Solve the equation for $a$ .
Write the rule, then check it at every given point.

Worked example 1

Find the rule of the quadratic function whose vertex is $(- 4, 6)$ and whose graph passes through $(2, - 12)$ .

Start from standard form with the vertex in place:

f (x) = a (x + 4)^{2} + 6

Substitute the point $(2, - 12)$ :

- 12 = a (2 + 4)^{2} + 6

- 12 = 36 a + 6

- 18 = 36 a

a = - \frac{1}{2}

The rule is

f (x) = - \frac{1}{2} (x + 4)^{2} + 6

If the question wants general form, expand the square first and multiply by $a$ second:

f (x) = - \frac{1}{2} (x^{2} + 8 x + 16) + 6 = - \frac{1}{2} x^{2} - 4 x - 2

Check:

At the point: $f (2) = - \frac{1}{2} (6)^{2} + 6 = - 18 + 6 = - 12$ . ✓
At the vertex: $f (- 4) = - \frac{1}{2} (0)^{2} + 6 = 6$ . ✓
The general form gives the same value at $x = 2$ : $- 2 - 8 - 2 = - 12$ , so the expansion is right too. ✓

There is a free sanity check in the sign of $a$ . The point $(2, - 12)$ is below the vertex, so the parabola must open downward and $a$ must be negative. If your $a$ comes out positive when the point is below the vertex, or negative when it is above, look for the slip before going on.

Case 2: two zeroes and one point

Write factored form with the zeroes filled in. Each zero $x_{1}$ gives a bracket $(x - x_{1})$ , so a zero at $- 1$ gives $(x + 1)$ .
Replace $x$ and $f (x)$ with the coordinates of the other point.
Solve for $a$ .
Write the rule, then check it at both zeroes and at the point.

Worked example 2

A quadratic function has zeroes $- 1$ and $7$ , and its graph passes through $(5, 18)$ . Find its rule, then write it in standard form.

f (x) = a (x + 1) (x - 7)

Substitute $(5, 18)$ :

18 = a (5 + 1) (5 - 7)

18 = - 12 a

a = - \frac{3}{2}

The rule in factored form is

f (x) = - \frac{3}{2} (x + 1) (x - 7)

Check: $f (5) = - \frac{3}{2} (6) (- 2) = 18$ , $f (- 1) = - \frac{3}{2} (0) (- 8) = 0$ and $f (7) = - \frac{3}{2} (8) (0) = 0$ . ✓

Standard form is now one short step away, because a parabola is symmetric and its vertex sits halfway between the zeroes:

h = \frac{- 1 + 7}{2} = 3 k = f (3) = - \frac{3}{2} (4) (- 4) = 24

f (x) = - \frac{3}{2} (x - 3)^{2} + 24

So the function has a maximum of $24$ , reached at $x = 3$ .

Check every given point, not only the one you used

The point you used to find $a$ passes the check almost by itself: $a$ was chosen to make it work. The check that catches mistakes is the one at the other given points, the vertex or the two zeroes.

Here is what that looks like on Worked example 1. Suppose the bracket had been written with $x - 4$ in it instead of $x + 4$ . Solving for $a$ with the point $(2, - 12)$ then gives $a = - \frac{9}{2}$ , and that rule really does give $- 12$ at $x = 2$ , so checking the point shows nothing wrong. At $x = - 4$ , though, it gives $- 282$ instead of $6$ . Only the vertex reveals that the parabola was built around $(4, 6)$ instead of $(- 4, 6)$ .

The most common mistake: the sign in the bracket. Standard form subtracts $h$ , so a vertex with $h = - 4$ gives $(x + 4)$ and a vertex with $h = 4$ gives $(x - 4)$ . Factored form works the same way: a zero at $- 1$ gives $(x + 1)$ . The trap is copying the coordinate into the bracket sign and all, and it survives any check made only at the point used to find $a$ . Substitute the vertex, or both zeroes, into your finished rule every time.

Check your understanding

No calculator needed. Questions 5 to 7 are about reasoning rather than computing.

Find the rule, in standard form, of the quadratic function with vertex $(1, 3)$ whose graph passes through $(4, - 15)$ .
Find the rule, in factored form, of the quadratic function with zeroes $- 2$ and $4$ whose graph passes through $(2, - 16)$ .
A parabola has vertex $(- 3, - 2)$ and $y$ -intercept $7$ . Find its rule in standard form and in general form.
A quadratic function has zeroes $1$ and $9$ and $y$ -intercept $- 18$ . Find its rule and its maximum value.
A parabola has zeroes $- 1$ and $4$ , and you are also told that it passes through $(4, 0)$ . Why is that not enough to find $a$ ?
True or false: some quadratic function has vertex $(2, 3)$ and passes through $(2, - 5)$ . Explain.
Find the error. A parabola has vertex $(5, - 2)$ and passes through $(3, 14)$ . A student writes $f (x) = a (x + 5)^{2} - 2$ , substitutes the point to get $14 = 64 a - 2$ , and concludes that $a = \frac{1}{4}$ .
A footbridge has a parabolic arch. Its two feet stand on the ground $24$ m apart, and $3$ m in from one foot the arch is $4.2$ m high. With the origin at the left foot, find a rule for the height of the arch, then its greatest height.

Show the answers

$f (x) = - 2 (x - 1)^{2} + 3$ . Substituting the point: $- 15 = a (3)^{2} + 3$ , so $9 a = - 18$ and $a = - 2$ . Check: $f (4) = - 2 (9) + 3 = - 15$ .
$f (x) = 2 (x + 2) (x - 4)$ . Substituting: $- 16 = a (4) (- 2) = - 8 a$ , so $a = 2$ . Check: $f (2) = 2 (4) (- 2) = - 16$ .
$f (x) = (x + 3)^{2} - 2 = x^{2} + 6 x + 7$ . The $y$ -intercept is the point $(0, 7)$ : $7 = a (3)^{2} - 2$ gives $a = 1$ . The constant term of the general form, $7$ , is the $y$ -intercept, as it should be.
$f (x) = - 2 (x - 1) (x - 9)$ , with a maximum of $32$ . The point $(0, - 18)$ gives $- 18 = a (- 1) (- 9) = 9 a$ , so $a = - 2$ . The vertex is halfway between the zeroes, at $x = 5$ , and $f (5) = - 2 (4) (- 4) = 32$ ; it is a maximum because $a < 0$ .
$(4, 0)$ is one of the zeroes, so it is not new information. Substituting it gives $0 = a (5) (0)$ , which is true for every value of $a$ : every parabola with those zeroes passes through that point.
False. At $x = 2$ , standard form gives $f (2) = a (0)^{2} + 3 = 3$ whatever $a$ is. A function has only one value at $x = 2$ , so it cannot also be $- 5$ .
The bracket should be $(x - 5)$ , since $h = 5$ . Then $14 = a (- 2)^{2} - 2$ , so $a = 4$ and $f (x) = 4 (x - 5)^{2} - 2$ . The student's rule does pass through $(3, 14)$ , but at the vertex it gives $f (5) = 23$ instead of $- 2$ : checking the vertex catches it.
$h (x) = - \frac{1}{15} x (x - 24)$ , and the greatest height is $9.6$ m. The feet are the zeroes $0$ and $24$ , and the point $(3, 4.2)$ gives $4.2 = a (3) (- 21) = - 63 a$ , so $a = - \frac{1}{15}$ . The top is halfway between the feet, at $x = 12$ : $h (12) = - \frac{1}{15} (12) (- 12) = 9.6$ .

If questions 1 to 4 went smoothly, the method is in place. If question 7 caught you, make the vertex check a habit before anything else.

Practise it

The free Secondary 4 Quadratic Functions worksheet has a sheet on finding the rule, alongside the three forms, the zeroes, the properties of a parabola and its inverse. Do the rule question first, then rewrite your answer in all three forms.

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.