2(x − 3)² − 8 has its vertex at (3, −8). Read the sign the other way and you claim (−3, −8), where the curve is actually at 64. The claimed vertex misses the curve by 72.
The curve never moves. Each form puts a different feature of it in front of you and hides the others.
| Form | Written | Reads off immediately |
|---|---|---|
| General | 2x² − 12x + 10 | The y-intercept, (0, 10). It is the c. |
| Factorised | 2(x − 1)(x − 5) | The roots, 1 and 5. They are the p and the q. |
| Vertex | 2(x − 3)² − 8 | The vertex, (3, −8). It is the (h, k). |
Three sentences, one curve. A question that asks for the vertex is asking you to get into vertex form, and a question that asks for the roots is asking for the factorised one. Reading a feature off the wrong form is where the marks go, and it is avoidable: decide which form you need before you start rearranging.
In a(x − h)² + k the vertex is at (h, k), and the minus sign is already in the formula. So in
2(x − 3)² − 8
h is +3, not −3, and the vertex is at (3, −8). The k, by contrast, is read straight off with its own sign: −8 stays −8.
Why it has to be +3. The squared bracket is smallest when it is zero, and x − 3 is zero at x = 3. That is the whole argument, it takes four words, and it does not depend on remembering a convention.
The cost of getting it wrong is not subtle. At x = −3 the curve is at 64, so the claimed vertex at (−3, −8) sits 72 below the curve, and it is not even in the right half of the picture.
Two independent checks on a vertex, both free.
Either takes five seconds and both catch the sign error immediately.
General to factorised: factorise, or find the roots and build the brackets. 2x² − 12x + 10 = 2(x² − 6x + 5) = 2(x − 1)(x − 5). Take the a out first; it makes the inside factorise with small numbers.
General to vertex: complete the square, and take the a out first here too.
2(x² − 6x + 5) = 2[(x − 3)² − 9 + 5] = 2[(x − 3)² − 4] = 2(x − 3)² − 8
Vertex to general: expand. 2(x² − 6x + 9) − 8 = 2x² − 12x + 18 − 8 = 2x² − 12x + 10, which is where we started.
The trap in completing the square is the a. Half of −6 is −3 only after the 2 has been taken outside. Halving the −12 instead gives (x − 6)², a vertex at x = 6, and an axis that is not between the roots.
Not every quadratic factorises, and all of them have a vertex. 2x² − 12x + 20 has no real roots, so there is no factorised form over the reals, and completing the square still gives 2(x − 3)² + 2 with vertex (3, 2). The vertex form always exists, which is why −b/(2a) is the check to trust.
The machine will find a vertex and a root without being told which form you are in, which makes it the fastest way to check an algebraic rearrangement you have just done by hand.
When you may use it. Analysis Paper 1 is non-calculator, and completing the square is Paper 1 work. Use the machine to check your three forms agree, not to produce them.
The mark people lose. Writing the vertex of a(x − h)² + k as (−h, k). The minus is already printed in the formula, so 2(x − 3)² − 8 has h = 3. The check that kills it: the axis of symmetry is the average of the roots, (1 + 5)/2 = 3, and also −b/(2a) = 12/4 = 3. Two independent routes to the same 3, and neither of them is −3.
Throughout: f(x) = 2x² − 12x + 10.
1. What is the y-coordinate of the y-intercept?
2. Where is the axis of symmetry? Give the x value.
3. What is the y-coordinate of the vertex?
4. What is the curve's value at x = −3?
5. In a(x − h)² + k, why is the vertex at x = +h?
1 markThe correct form reached, or the correct feature identified.
1 markThe algebra, with the a handled before completing the square.
1 markThe coordinates, as a pair, with the signs right.
A vertex given as a single number loses the third mark. It is a point, so write (3, −8), and check the 3 against the average of the roots before you do.
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