Quadratics from Their Turning Point

Write y = a(x − p)² + q from the turning point, then use a root or point to find a

In the exam for Edexcel International AS Maths · 6 real past-paper questions below

Turning point, then find a

You can already read the turning point from completed square form, and write the equation when nothing sits in front of the bracket (Completed Square Form and Turning Points). AS questions add one thing: the number in front is unknown, and you must find it.

Formula

y = a (x - p)^{2} + q

$a$ — the stretch. Unknown at first. Positive for a minimum, negative for a maximum.
$p$ — the x of the turning point. It goes in the bracket with its sign changed.
$q$ — the y of the turning point. It stays outside.

The turning point gives you $p$ and $q$ . It cannot give $a$ : every curve with that turning point, steep or flat, passes through it. So you need one more point on the curve.

Worked example

1
A quadratic curve has a minimum at (−1, 3) and passes through (1, 11). Find its equation.
2
$y = a (x + 1)^{2} + 3$ — Write the form. x = −1 makes the bracket (x + 1).
3
$11 = a (1 + 1)^{2} + 3$ — Put in the other point: x = 1, y = 11.
4
$11 = 4 a + 3$ — (1 + 1)² = 2² = 4.
5
$a = 2$ — Take 3 from both sides, then divide by 4. Positive, as a minimum needs.
6
$y = 2 (x + 1)^{2} + 3$ — The equation. Expanded, it is y = 2x² + 4x + 5. Either form scores.

Rule

Marks: M1 for writing the form $a (x - p)^{2} + q$ , M1 for putting in another point to find $a$ , A1 for the answer. If you write the form with $a = 1$ you still earn the first M1, but nothing after it.

Common mistake

Stopping at $y = (x + 1)^{2} + 3$ . Check it: at x = 1 it gives 7, not 11. Always use the extra point.

Key idea

Turning point gives the bracket and the number outside. One more point gives $a$ .

The mirror line

A parabola is symmetric about the vertical line through its turning point. So if you know one place where it crosses the x-axis, you know the other.

Memory trick

The turning point sits exactly halfway between the two roots, like the middle of a seesaw. Step from the known root to the middle, then take the same step again.

Worked example

1
A quadratic curve has a minimum at (5, −12) and crosses the x-axis at x = 3 and x = k. Find k, then the equation.
2
$5 - 3 = 2$ — The known root is 2 to the left of the mirror line x = 5.
3
$k = 5 + 2 = 7$ — So the other root is 2 to the right of it.
4
$y = a (x - 3) (x - 7)$ — Now you know both roots, so use the factor form.
5
$- 12 = a (5 - 3) (5 - 7)$ — Put in the turning point: x = 5, y = −12.
6
$- 12 = - 4 a, a = 3$ — 2 × (−2) = −4.
7
$y = 3 (x - 3) (x - 7)$ — Expanded: y = 3x² − 30x + 63.

The other route works too: $y = a (x - 5)^{2} - 12$ , then put in (3, 0) to get $4 a - 12 = 0$ , so $a = 3$ . Both give the same curve. Use whichever needs fewer steps.

Rule

“Deduce k” is worth one mark and needs no working: the word deduce means it follows from what you were told. Write k = 7 and move on.

Common mistake

Doubling the known root: k = 2 × 3 = 6 is wrong. Measure from the root to the mirror line x = 5, not from 0.

Key idea

Roots are mirror images in x = p. Two roots let you use $a (x - r_{1}) (x - r_{2})$ .

Sketch from a maximum

If the turning point is a maximum, the curve is an upside-down U, so $a$ is negative. A sketch question wants the shape and every point where the curve meets the axes, with coordinates.

Worked example

1
A quadratic curve has a maximum at (3, 18) and passes through the origin. Sketch it, then find its equation.
2
$\cap shape$ — A maximum means the curve opens downwards.
3
$(0, 0)$ — One crossing is given: the origin. It is 3 to the left of x = 3.
4
$(6, 0)$ — Mirror it: 3 to the right of x = 3. Label it as 6 on the x-axis.
5
$peak at (3, 18)$ — Draw the ∩ through (0, 0) and (6, 0), with its top between them, to the right of the y-axis.
6
$y = a x (x - 6)$ — Roots 0 and 6 give factors x and (x − 6).
7
$18 = a (3) (3 - 6) = - 9 a$ — Put in the maximum: x = 3, y = 18.
8
$a = - 2$ — Negative, as a maximum needs: a good check.
9
$y = - 2 x (x - 6) = 12 x - 2 x^{2}$ — Same as 18 − 2(x − 3)². Either form scores.

Rule

Sketch marks: one for a ∩ shape through the origin with its top to the right of the y-axis, one for the other crossing labelled. Write 6 on the x-axis, or (6, 0); writing (0, 6) loses the mark.

Common mistake

Drawing a U shape. A maximum always means ∩, whatever the numbers look like.

Key idea

A thrown ball follows a ∩ path. Galileo showed this in the early 1600s by rolling balls off a table and marking where they landed.

Key idea

Maximum: ∩ and $a < 0$ . Mirror the known crossing, label every axis point, then find $a$ .

Quadratic models

Some questions use a quadratic in completed square form as a model: an equation that describes something real over a time t. Three parts come up again and again.

Worked example

1
The number of fish in a lake, N hundred, t years after it was stocked, is modelled by N = 45 − 0.5(t − 6)², for 0 ≤ t ≤ 12.
2
$t = 0 : N = 45 - 0.5 (0 - 6)^{2}$ — “At the start” means t = 0, not t = 1.
3
$= 45 - 0.5 \times 36 = 27$ — So 2700 fish at the start. Give the units of the context.
4
$(t - 6)^{2} \geq 0$ — The squared bracket is never negative, so it can only take away from 45.
5
$N_{m a x} = 45$ — Greatest value 45, so 4500 fish, when t = 6. Read it straight off: no working needed.
6
$t = 16 : N = 45 - 0.5 (10)^{2} = - 5$ — Try the model far outside its range.
7
$N < 0$ — A negative number of fish is impossible, so the model fails for t = 16.

Rule

When asked why a model should not be used, give a reason in the context: “the number of fish would be negative”, or “the model is only for 0 ≤ t ≤ 12, and 16 is outside that”. Writing just “−5 < 0” scores nothing, because it does not mention the fish.

Common mistake

Answering the maximum with the time, 6. The maximum is the value of N, 45. The time is when it happens.

Key idea

t = 0 for the start. The number outside the bracket for the maximum. A negative or out-of-range value to say why the model fails.

Which method? Read the question

The wording tells you which step to take.

You see	You do
Turning point and one other point	Write $a (x - p)^{2} + q$ , put in the point, find $a$
Turning point and one root; “deduce” the other	Reflect the root in x = p
Both roots known	$a (x - r_{1}) (x - r_{2})$ , then put in the turning point
“Sketch”, given a maximum	∩ shape, label every axis crossing
Model: value at the start	Put t = 0
Model: maximum value	The number outside the bracket
Model: why not use it here	A value that is impossible in the context, or outside the range

Key idea

Turning point gives $p$ and $q$ ; symmetry gives the other root; one more point gives $a$ .

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Real past-paper questions on quadratics from their turning point

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