Why is the turning point of a parabola with roots 2 and 8 at x=5, not x=2 or x=8?

Because the parabola is symmetric about its axis of symmetry, which sits exactly halfway between the two roots. The midpoint of 2 and 8 is (2+8)/2 = 5. At x=2 and x=8, the parabola crosses the x-axis (y=0); those are roots, not the turning point.

What is the formula for the x-coordinate of the turning point from two roots?

For a parabola with roots a and b, the x-coordinate of the turning point is (a+b)/2, which is the average (midpoint) of the two roots. This must be computed with the correct signs: for roots -3 and 7, the answer is (-3+7)/2 = 2, not (3+7)/2 = 5.

Why does averaging the magnitudes give the wrong answer when one root is negative?

Because the midpoint formula requires signed arithmetic. For roots -3 and 7, the correct calculation is (-3+7)/2 = 4/2 = 2. Using magnitudes gives (3+7)/2 = 5, which is not the midpoint: x=5 is 8 units from -3 but only 2 units from 7 (not equidistant). Only x=2 is equidistant from both roots.

GCSE Maths Higher · AQA · Completing the square and turning points

Turning point x-coordinate is the midpoint of the roots, not a single root

When a parabola has roots at $x=2$ and $x=8$ , the most common error is to write the turning point x-coordinate as $2$ or $8$ . Those are roots: points where $y=0$ . The turning point is a different point. The parabola is symmetric about its axis of symmetry, and that axis sits exactly halfway between the two roots: $\frac{2+8}{2} = 5$ .

In symbolic form: for a parabola with roots $a$ and $b$ , the x-coordinate of the turning point is $\frac{a+b}{2}$ , the average of the roots. Writing $a$ , $b$ , or $a+b$ is wrong.

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How to spot it in your own work

You wrote $x=2$ (or $x=8$ ) as the turning point x-coordinate for roots $x=2$ and $x=8$ , using one root directly instead of the midpoint $\frac{2+8}{2}=5$ .
You wrote $x=10$ (the sum $2+8$ ) instead of the midpoint $\frac{2+8}{2}=5$ , forgetting to divide by 2.
You wrote $x=5$ for roots $-3$ and $7$ by averaging magnitudes $\frac{3+7}{2}=5$ instead of using signed values: $\frac{-3+7}{2}=2$ .
On a symbolic question you wrote $a+b$ or just $a$ for the turning point x-coordinate of a parabola with roots $a$ and $b$ , rather than $\frac{a+b}{2}$ .

An exam question that triggers it

Here is the structure of a typical AQA Higher question on this misconception:

A parabola has roots at $x=a$ and $x=b$ .

Write an expression for the x-coordinate of the turning point.

The misconception writes $a$ , $b$ , or $a+b$ . The fix: the turning point is the midpoint of the roots, so the expression is $\frac{a+b}{2}$ .

Why students fall for this

Students know that the turning point is connected to the roots of a quadratic. The roots appear explicitly as numbers, so using one of them as the turning point feels natural. The error is conflating two different features of the curve: a root is where $y=0$ , while the turning point is where the gradient is zero. They are not the same point.

A second pattern appears with the sum trap $a+b$ : students recall that the sum of the roots appears in the quadratic formula or in Vieta's formulas, and write $a+b$ without dividing by 2. The midpoint requires dividing the sum by 2.

A third pattern is the magnitude-average trap for mixed-sign roots: for roots $-3$ and $7$ , students compute $\frac{3+7}{2}=5$ by dropping the sign on $-3$ . The correct answer is $\frac{-3+7}{2}=2$ .

The fix: Midpoint of the roots: x-coordinate of the turning point is (a+b)/2

Step 1: identify the two roots. For a parabola with roots at $x=2$ and $x=8$ , let $a=2$ and $b=8$ .

Step 2: compute the midpoint. $\frac{a+b}{2} = \frac{2+8}{2} = \frac{10}{2} = 5$ .

Step 3: verify by equidistance. $5 - 2 = 3$ and $8 - 5 = 3$ . Both distances are equal, confirming $x=5$ is the midpoint.

Step 4: for mixed-sign roots, keep the signs. For roots $-3$ and $7$ : $\frac{-3+7}{2} = \frac{4}{2} = 2$ . Check: $2-(-3)=5$ and $7-2=5$ . Equidistant. Trap: $\frac{3+7}{2}=5$ gives $x=5$ , which is 8 units from $-3$ but only 2 units from $7$ , so it is not the midpoint.

Worked example

A parabola has roots at $x=2$ and $x=8$ . Find the x-coordinate of its turning point.

Midpoint formula. $\frac{2+8}{2} = \frac{10}{2} = 5$ .
Turning point x-coordinate.
$x = 5$
Traps: $x=2$ (root, not TP), $x=8$ (root, not TP), $x=10$ (sum without /2).

A parabola has roots at $x=-3$ and $x=7$ . Find the x-coordinate of its turning point.

Signed midpoint. $\frac{-3+7}{2} = \frac{4}{2} = 2$ .
Turning point x-coordinate.
$x = 2$
Trap: $x=5$ from $\frac{3+7}{2}$ drops the sign. Verify: $2-(-3)=5$ and $7-2=5$ ✓

A parabola has roots at $x=a$ and $x=b$ . Write an expression for the x-coordinate of the turning point.

Apply the midpoint formula.
$\frac{a+b}{2}$
Also written $\frac{b+a}{2}$ . Both are correct by commutativity.
Traps. $a$ or $b$ : single roots, not the midpoint. $a+b$ : sum without dividing by 2.

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Common questions

Why is the turning point at $x=5$ and not at $x=2$ or $x=8$ for a parabola with those roots?: Because at $x=2$ and $x=8$ the curve crosses the x-axis (y=0). Those are roots, not the turning point. The turning point is where the gradient is zero, which is the axis of symmetry at $x=\frac{2+8}{2}=5$ .
Why does the formula require signed arithmetic for roots like $-3$ and $7$ ?: Because the midpoint of two numbers on a number line is always $\frac{a+b}{2}$ using their signed values. For $a=-3$ and $b=7$ : $\frac{-3+7}{2}=2$ . Using magnitudes gives $\frac{3+7}{2}=5$ , which is 8 units from $-3$ and 2 units from $7$ , not equidistant.
Is $\frac{b+a}{2}$ also a correct answer?: Yes. Addition is commutative, so $\frac{b+a}{2} = \frac{a+b}{2}$ . Both expressions are correct answers to the symbolic question.

Related misconceptions

Turning point sign from vertex form: reading x from the bracketWriting the turning point of y=(x-7)^2+8 as (-7,8) by copying the printed sign instead of solving x-7=0 to get x=7, a companion misconception on the same Higher topic.
Completing the square: forgetting to subtract the adjustment constantKeeping the original constant unchanged when completing the square, writing (x+4)^2-5 instead of (x+4)^2-21, the first misconception in the same Higher completing-the-square topic.

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