Find the vertex

Vertex form, range, max/min value

In 14 of 20 recent exams
2% of the marks
Paper 1 or 2
3 to 6 marks

How it is set in the exam

The most common quadratics question. Paper 1 (no calculator), 3–6 marks, often part (a) of a longer question, followed by "hence…". Sometimes the roots come first, then the axis, the vertex and a sketch. On Paper 2 (calculator) it appears inside contexts (maximum height, maximum profit).

Spot it in the exam

Exam phrases to spot (10)

“express in the form $a(x-h)^2 + k$ ”
“ $f(x)$ can be written in the form $a(x-h)^2 + k$ . Find the value of $h$ and the value of $k$ ”
“write down the value of $a$ ”
“hence write down the coordinates of the vertex”
“write down the equation of the axis of symmetry”
“find the x-intercepts … hence write down the equation of the axis of symmetry”
“find the range of $f$ ”
“find the minimum (maximum) value of $f$ ”
“sketch the graph of $f$ , labelling the intercepts and the vertex”
“describe the transformations that map $y = x^2$ onto $y = f(x)$ ”

In disguise (5)

"find the maximum height" of a projectile whose height is a quadratic in $t$ .
" $f(x) = p$ has one / two / no solutions", with the unknown $p$ only in the constant term: $y = p$ is a horizontal line, so compare $p$ with the vertex's y-value $k$ . For $a > 0$ : one solution if $p = k$ , two if $p > k$ , none if $p < k$ (the other way round for $a < 0$ ). $\Delta$ of $f(x) - p = 0$ is an accepted alternative, but slower. If the unknown is anywhere else, it is Find k from the number of roots. (The same horizontal line against a cubic meets its turning points:.)
$f(x) = p$ with two positive solutions, $p$ again only in the constant term: the line $y = p$ must cut the graph twice to the right of the y-axis. That needs the vertex on the right ( $h > 0$ ), and then $p$ lies strictly between the vertex's y-value $k$ and the y-intercept $f(0)$ (at $p = f(0)$ one solution is $x = 0$ , which is not positive).
"hence show that $f(x) > 0$ for all $x$ ": from vertex form, $(x-h)^2 \ge 0$ , so $f(x) \ge k > 0$ when $a > 0$ and $k > 0$ . With no vertex form to start from (a fixed quadratic, often a denominator or a factor in a calculus question), complete the square first, or show $\Delta < 0$ with $a > 0$ : the graph opens upward and never meets the x-axis.
the greatest value of $\ln$ or $\log_k$ of a quadratic that opens downward, or the least value of the square root of one that opens upward: the log and the root are increasing, so the extreme is at the vertex, and the answer is the log (or root) of the vertex's y-value, not the y-value itself (Logarithm graphs)

Looks like it, but isn't

an identity with no unknown constants to find ("show that $x^2 - 6x + 13 \equiv (x - 3)^2 + 4$ "): Prove it: from one side to the other (identities and integer claims), with the same completing-the-square working.

Worked example

Complete the square, then "hence"

Paper 1 (no calculator)

Question

Let $f(x) = 2x^2 - 12x + 23$ .
(a) Express $f(x)$ in the form $a(x-h)^2 + k$ . [3]
(b) Write down the coordinates of the vertex of the graph of $f$ . [1]
(c) Describe the transformations that map $y = x^2$ onto $y = f(x)$ . [2]

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Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

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See it move: Move the parabola into place

a

h

k

$y = x^{2}$
Inside goes the other way: $x^{2}$ alone: no move sideways.
Outside goes as it says: no move up or down.
This is $y = x^{2}$ itself: no stretch and no move yet.
Land the vertex on $(3, 5)$ to match $y = 2x^{2} - 12x + 23$ , the curve in yellow.

Complete the square:

f(x) = 2x^{2} - 12x + 23

Which method?

The question	Use
"Express in the form $a(x-h)^2 + k$ "	Complete the square
Only the vertex (or the max/min value, or the range) is needed	$x = -\frac{b}{2a}$ , then substitute to get the y-value
It factorises, or the roots are already known (often from an earlier part)	Midpoint of the roots gives $x$ , then substitute
You know calculus	$f'(x) = 0$ gives the vertex's $x$ (stationary points); markschemes accept it

Steps

Find the vertex $(h, k)$ by the method in the table. Watch the sign of $h$ .
Axis of symmetry: write it as an equation, $x = h$ . A bare number scores . The vertex is a minimum if $a > 0$ , a maximum if $a < 0$ , and $k$ is that minimum or maximum value. Give the vertex as coordinates, $(h, k)$ : the y-value alone loses the mark.
Range (domain all real numbers): $f(x) \geq k$ if $a > 0$ , $f(x) \leq k$ if $a < 0$ . If the domain is restricted, also work out the values at its endpoints.
Sketch: the shape from the sign of $a$ , then label the x-intercepts, the y-intercept $(0, c)$ and the vertex, each in its place.

Common mistakes

Sign of $h$
$2(x-3)^2 + 5$ has vertex $(3, 5)$ , not $(-3, 5)$ .
Forgetting to multiply by $a$
$2\left[(x-3)^2 - 9\right]$ gives $-18$ , not $-9$ .
Sign slips when $a < 0$
$-x^2 + 4x + 1 = -(x^2 - 4x) + 1$ , not $-(x^2 + 4x) + 1$ . Then $-\left[(x-2)^2 - 4\right] = -(x-2)^2 + 4$ . Check by expanding your answer.

More mistakes (5)

The axis as a number
the axis of symmetry is the line $x = -1$ . Writing " $-1$ " or " $-\frac{b}{2a} = -1$ " scores .
The vertex as one number
the vertex is the point $(-1, -18)$ . Writing " $f(-1) = -18$ " or " $-18$ " loses the final mark (" $x = -1$ , $y = -18$ " is accepted).
Unlabelled sketches
the right shape earns only the first mark. Each intercept and the vertex earns its own mark, so label each one with its value.
Transformation order
translating by $\binom{3}{5}$ first and then stretching gives $2\left[(x-3)^2 + 5\right] = 2(x-3)^2 + 10$ , not $2(x-3)^2 + 5$ . Stretch first, then translate.
Restricted domains
if the domain is, say, $0 \leq x \leq 5$ , the range may end at an endpoint rather than at the vertex. Check both ends.

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