Find the equation from given features

In 7 of 20 recent exams
1% of the marks
Paper 1, no calculator
1 to 5 marks

How it is set in the exam

Mostly Paper 1 (no calculator), 1–5 marks per part. The information often comes from a graph printed with the question. A common shape has one unknown constant, fixed by a fact about the vertex or the roots, and then used in a later part.

Spot it in the exam

Exam phrases to spot (4)

“the graph of $f$ passes through…”
“has x-intercepts…”
“has vertex…”
“find the values of $a$ , $b$ and $c$ ”

In disguise (5)

a sketch with labelled points and no equation
"the axis of symmetry is $x = 3$ ", which silently gives you a second root when one root is known
two points with the same y-coordinate, or equal values in a table: the axis is halfway between them
a context: "a ball reaches its maximum height of $20$ m at $t = 2$ and lands at $t = 4$ " gives the vertex $(2, 20)$ and a root $t = 4$
one unknown constant and a fact about the vertex, such as "its vertex is on the line $y = 3x - 6$ ". If the unknown is only in $c$ , $-\frac{b}{2a}$ gives the vertex's x-coordinate straight away; then substitute that point, and only that point

Worked example

Roots and y-intercept

Paper 1 (no calculator)

Question

$f$ is a quadratic function. The graph of $y = f(x)$ crosses the x-axis at $x = -1$ and $x = 5$ and passes through $(0, -10)$ . Find $f(x)$ in the form $ax^2 + bx + c$ . [4]

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Which method?

both x-intercepts $p, q$
Start from
$y = a(x-p)(x-q)$
Then
one more point finds $a$
the vertex $(h, k)$
Start from
$y = a(x-h)^2 + k$
Then
one more point finds $a$
the axis of symmetry + one root
Start from
reflect the root in the axis to get the other root
Then
then row 1
two points with equal y-values
Start from
the axis is at the midpoint of their x-values, so $-\frac{b}{2a} =$ that value
Then
one more fact finds the rest
the y-intercept and two other points
Start from
$y = ax^2 + bx + c$ , with $c$ = the y-intercept
Then
substitute the other two points: 2 simultaneous equations in $a$ and $b$ . GDC on : solve simultaneous linear equations
$a$ and $b$ , one unknown constant, and a fact about the vertex
Start from
the vertex's $x = -\frac{b}{2a}$ , which doesn't involve the unknown
Then
put the vertex into the fact: one equation in the unknown

You are given	Start from	Then
both x-intercepts $p, q$	$y = a(x-p)(x-q)$	one more point finds $a$
the vertex $(h, k)$	$y = a(x-h)^2 + k$	one more point finds $a$
the axis of symmetry + one root	reflect the root in the axis to get the other root	then row 1
two points with equal y-values	the axis is at the midpoint of their x-values, so $-\frac{b}{2a} =$ that value	one more fact finds the rest
the y-intercept and two other points	$y = ax^2 + bx + c$ , with $c$ = the y-intercept	substitute the other two points: 2 simultaneous equations in $a$ and $b$ . GDC on : solve simultaneous linear equations
$a$ and $b$ , one unknown constant, and a fact about the vertex	the vertex's $x = -\frac{b}{2a}$ , which doesn't involve the unknown	put the vertex into the fact: one equation in the unknown

The y-intercept gives $c$ immediately in general form, and fits any row as the "one more point".

Three points with nothing special about them would need three equations in three unknowns. That is HL content (AHL 1.16), so an SL question always gives you something that removes an unknown: a coefficient already given (e.g. $y = x^2 + bx + c$ ), the y-intercept (gives $c$ ), both roots (factored form), the vertex (vertex form), or two points with equal y-values (the axis). Look for it.

Steps

List what you are given, and count: $y = ax^2 + bx + c$ has three unknowns, so you need three facts.
Choose the form that uses the most facts at once (the table above).
Write one equation per remaining fact, substituting each point.
Solve: by hand on ; on , two linear equations can go to the GDC: solve simultaneous linear equations (write them down first).
Give the answer in the form asked (expand if it says $ax^2 + bx + c$ ), and check it against a fact you haven't used.

Common mistakes

Forgetting $a$
writing $(x+1)(x-5)$ assumes $a = 1$ . Always write $a(\ldots)$ and find it.
Root signs
a root at $x = -1$ gives the factor $(x+1)$ .
Leaving the wrong form
if they ask for $ax^2 + bx + c$ , expand.

More mistakes (3)

Not checking the sign of $a$ against the graph
a graph that opens downward needs $a < 0$ . If your $a$ has the wrong sign, look for an arithmetic slip.
Substituting the wrong point
when the fact is about the vertex, find the vertex's x-coordinate first and substitute there. Any other point of the line proves nothing: it usually gives a wrong value, and even the other intersection (here $x = 4.5$ ), which happens to give $p = 21$ , scores no marks.
Keeping the known root
"meets the graph again" asks for the other intersection. In Example 2.4, $x = 3$ is the vertex, not $B$ , so leave it out of the final answer.

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