Add Maths · 2 Quadratic functions

2 Quadratic functions

Before you start

Expanding a perfect square: ${(x + p)}^{2} = x^{2} + 2 p x + p^{2}$ .
Factorising quadratics, including those with a leading coefficient, e.g. $2 x^{2} + 5 x - 3 = (2 x - 1) (x + 3)$ .
Simplifying surds without a calculator, e.g. $\sqrt{40} = 2 \sqrt{10}$ and $\sqrt{\frac{5}{3}} = \frac{\sqrt{15}}{3}$ .
Range notation from the Functions lesson: a range is an inequality in $f (x)$ , not in $x$ .

Syllabus objectives covered (5)

Find the maximum or minimum value of the quadratic function f : x ↦ ax² + bx + c by completing the square or by differentiation.
Use the maximum or minimum value of f(x) to sketch the graph of y = f(x) or determine the range for a given domain.
Know the conditions for f(x) = 0 to have: (i) two real roots (ii) two equal roots (iii) no real roots and the related conditions for a given line to: (i) intersect a given curve (ii) be a tangent to a given curve (iii) not intersect a given curve.
Solve quadratic equations for real roots.
Find the solution set for quadratic inequalities either graphically or algebraically.

Part 1 of 8

Completing the square: where it comes from

Every quadratic $a x^{2} + b x + c$ can be rewritten as $a {(x + p)}^{2} + q$ . The effort is worth it because that form shows the turning point at a glance.
Start from what a perfect square looks like when you expand it.
${(x + p)}^{2} = x^{2} + 2 p x + p^{2}$
The coefficient of $x$ is $2 p$ . To match $x^{2} - 6 x$ you need $2 p = - 6$ , so $p = - 3$ . In words: **halve the coefficient of $x$ **.
But ${(x - 3)}^{2} = x^{2} - 6 x + 9$ has an extra $+ 9$ that $x^{2} - 6 x$ did not have. Take it straight back off.
$x^{2} - 6 x = {(x - 3)}^{2} - 9$
Then bring in the constant term.
$x^{2} - 6 x + 11 = {(x - 3)}^{2} - 9 + 11 = {(x - 3)}^{2} + 2$
Check by expanding: ${(x - 3)}^{2} + 2 = x^{2} - 6 x + 9 + 2 = x^{2} - 6 x + 11$ . Do this every time. It takes ten seconds and catches sign slips.
An odd coefficient of $x$ just gives fractions. Keep them exact.
$x^{2} + 5 x + 1 = {(x + \frac{5}{2})}^{2} - \frac{25}{4} + 1 = {(x + \frac{5}{2})}^{2} - \frac{21}{4}$

Part 2 of 8

Reading off the maximum or minimum

A square is never negative: ${(x - 3)}^{2} \geq 0$ , and it equals 0 only when $x = 3$ .
So ${(x - 3)}^{2} + 2$ is never less than 2, and it equals 2 exactly when $x = 3$ . The minimum value is 2, and it occurs at $x = 3$ .
In general, $a {(x + p)}^{2} + q$ turns where the bracket is zero, at $x = - p$ , and the value there is $q$ . Watch the sign: ${(x - 3)}^{2}$ is zero at $x = + 3$ .
If $a > 0$ , the turning point is a minimum. If $a < 0$ , the square is multiplied by a negative number, so $a {(x + p)}^{2} \leq 0$ and the turning point is a maximum.
When $a \neq 1$ , take $a$ out of the $x^{2}$ and $x$ terms only, complete the square inside the bracket, then multiply back out.
$2 x^{2} + 12 x + 7 = 2 (x^{2} + 6 x) + 7 = 2 [{(x + 3)}^{2} - 9] + 7 = 2 {(x + 3)}^{2} - 11$
The trap is the $- 9$ . It is inside the bracket, so it is multiplied by 2 and becomes $- 18$ . Forget that and you get $- 2$ instead of $- 11$ .
A negative $a$ works the same way. Take out $- 2$ , and check the sign of the $x$ term inside: $- 2 \times (- 4 x) = + 8 x$ .
$5 + 8 x - 2 x^{2} = - 2 (x^{2} - 4 x) + 5 = - 2 [{(x - 2)}^{2} - 4] + 5 = 13 - 2 {(x - 2)}^{2}$
So $5 + 8 x - 2 x^{2}$ has maximum value 13, at $x = 2$ .
The objective also mentions differentiation. That route comes with calculus in topic 14. For now, completing the square is your method, and on a non-calculator paper it gives exact values directly.

Part 3 of 8

Sketching a parabola from the completed square

A sketch needs four things: the right way up, the vertex, the $y$ -intercept, and the $x$ -intercepts if there are any. Label each with its coordinates.
Take $y = x^{2} - 2 x - 8$ . Completing the square gives ${(x - 1)}^{2} - 9$ , so the vertex is the minimum point $(1, - 9)$ .
The $y$ -intercept comes from $x = 0$ , which leaves just the constant: $(0, - 8)$ .
The $x$ -intercepts come from solving $x^{2} - 2 x - 8 = 0$ : $(x + 2) (x - 4) = 0$ , so $(- 2, 0)$ and $(4, 0)$ .
The vertex comes from the completed square, the intercepts from $x = 0$ and $y = 0$ , and the roots sit symmetrically about $x = 1$ .
The curve is symmetric about the vertical line through its vertex, $x = 1$ . The roots $- 2$ and $4$ are each 3 units from that line, which is a free check on your arithmetic.
Shape: $a > 0$ gives a U with a minimum; $a < 0$ gives an upside-down U with a maximum. If a U has its vertex above the $x$ -axis, there are no $x$ -intercepts to find.

Part 4 of 8

The range for a given domain

The range is every output the function produces over its domain. For a quadratic, the only candidates for the smallest and largest values are the vertex and the ends of the domain.
Step 1: is the $x$ -coordinate of the vertex inside the domain? If it is, the vertex value is one end of the range.
Step 2: work out $f$ at each end of the domain. The other end of the range is the largest of these (for a U) or the smallest (for an upside-down U).
For $f (x) = x^{2} - 2 x - 8$ with $- 1 \leq x \leq 5$ : the vertex $x = 1$ is inside, giving $- 9$ . The ends give $f (- 1) = - 5$ and $f (5) = 7$ . The largest is 7.
$- 9 \leq f (x) \leq 7$
Now change the domain to $x \geq 3$ . The vertex is outside it, and the curve only rises from there. So the least value is at the end: $f (3) = - 5$ , and the range is $f (x) \geq - 5$ .
Write the range with $f (x)$ . If an end of the domain is excluded (a strict $<$ ) and that end gives the extreme value, that value is excluded from the range too.

Part 5 of 8

Solving quadratic equations exactly

Get zero on one side first. Every method below depends on it.
Factorise when the numbers allow. It is fastest and exact. If a product is zero, one of its factors must be zero.
The quadratic formula always works. It is printed on the List of formulas, so you don't need to memorise it.
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
On a non-calculator paper, simplify the surd fully. For $2 x^{2} - 4 x - 3 = 0$ , $a = 2$ , $b = - 4$ , $c = - 3$ , and $\sqrt{40} = 2 \sqrt{10}$ .
$x = \frac{4 \pm \sqrt{16 + 24}}{4} = \frac{4 \pm 2 \sqrt{10}}{4} = \frac{2 \pm \sqrt{10}}{2}$
Every term on top must be divided. $\frac{4 \pm 2 \sqrt{10}}{4}$ is not $1 \pm 2 \sqrt{10}$ . Dividing the 4 but not the $2 \sqrt{10}$ is a very common slip.
Completing the square is the third route, and the syllabus expects you to be able to use it. Get the square on its own, square-root both sides, and remember the $\pm$ .
Never divide both sides by $x$ . From $3 x^{2} = 5 x$ , dividing by $x$ loses the solution $x = 0$ . Factorise instead: $x (3 x - 5) = 0$ , so $x = 0$ or $x = \frac{5}{3}$ .
'Solve for real roots' means real numbers only. If $b^{2} - 4 a c$ is negative, there is no real square root to take, and the correct answer is that there are no real roots.

Part 6 of 8

The discriminant: how many roots, without solving

The expression under the square root in the formula, $b^{2} - 4 a c$ , is the discriminant. It tells you how many real roots there are before you find any. The formula is given in the exam, but these conditions are not, so learn them.
$b^{2} - 4 a c > 0$ : you add and subtract a positive square root, giving two different real roots. The graph crosses the $x$ -axis twice.
$b^{2} - 4 a c = 0$ : adding and subtracting zero gives the same value twice, so there are two equal roots (a repeated root). The graph touches the $x$ -axis at its vertex.
$b^{2} - 4 a c < 0$ : a negative number has no real square root, so there are no real roots. The graph never meets the $x$ -axis.
Same shape, shifted up: the sign of $b^{2} - 4 a c$ decides whether the curve crosses, touches or misses the $x$ -axis.
'Has real roots', with no other words, allows either two different or two equal roots, so the condition is $b^{2} - 4 a c \geq 0$ .
A typical use: find $k$ so that $k x^{2} + 4 x + 1 = 0$ has two equal roots. Here $a = k$ , $b = 4$ , $c = 1$ , so $16 - 4 k = 0$ and $k = 4$ . Check: $4 x^{2} + 4 x + 1 = {(2 x + 1)}^{2}$ , a perfect square.
Write $a$ , $b$ and $c$ down before you substitute. When $b$ is an expression, bracket it: ${(k - 3)}^{2}$ , not $k - 3^{2}$ .

Part 7 of 8

A line and a curve: two points, tangent, or no meeting

Where a line meets a curve, their $y$ -values are equal. So set the two expressions for $y$ equal and rearrange to the form $a x^{2} + b x + c = 0$ .
Each real root of that equation is the $x$ -coordinate of a meeting point. So the discriminant of this equation counts the meeting points.
$b^{2} - 4 a c > 0$ : the line meets the curve at two distinct points. $b^{2} - 4 a c = 0$ : the line is a tangent to the curve. $b^{2} - 4 a c < 0$ : the line does not meet the curve.
Example: $y = x + k$ and $y = x^{2} - 3 x + 5$ . Equate and rearrange.
$x^{2} - 3 x + 5 = x + k \Rightarrow x^{2} - 4 x + (5 - k) = 0$
For a tangent, the discriminant is zero, with $a = 1$ , $b = - 4$ and $c = 5 - k$ .
$16 - 4 (5 - k) = 0 \Rightarrow 4 k - 4 = 0 \Rightarrow k = 1$
Find the point of contact by putting $k = 1$ back in: $x^{2} - 4 x + 4 = {(x - 2)}^{2} = 0$ , so $x = 2$ and $y = 3$ . The curve agrees: $4 - 6 + 5 = 3$ .
Sliding the line $y = x + k$ down changes two meeting points to one (a tangent) to none, and the discriminant tracks each change.
The constant here is $c = 5 - k$ , not 5. Moving everything to one side before reading off $a$ , $b$ and $c$ is where most errors happen.
Read the wording. 'Two distinct points' needs $> 0$ . If a question only says the line meets the curve, a tangent counts as meeting, so the condition is $\geq 0$ .

Part 8 of 8

Quadratic inequalities

Solve $x^{2} - 3 x - 10 < 0$ . First solve the equation $x^{2} - 3 x - 10 = 0$ to find the critical values: $(x + 2) (x - 5) = 0$ , so $x = - 2$ and $x = 5$ .
Now sketch $y = x^{2} - 3 x - 10$ : a U crossing the axis at $- 2$ and $5$ . The inequality asks where $y < 0$ , which is where the curve is below the $x$ -axis.
Below the axis between the roots gives one combined inequality; above the axis gives two separate pieces joined by 'or'.
That is the stretch between the roots.
$- 2 < x < 5$
For $x^{2} - 3 x - 10 > 0$ you want where the curve is above the axis: the two outer pieces. Two separate regions need two inequalities joined by 'or'.
$x < - 2 or x > 5$
The syllabus says solutions must be written in the correct form. Write $- 2 < x < 5$ as one combined inequality, and $x < - 2$ or $x > 5$ as two. Never write $5 < x < - 2$ : no number is above 5 and below $- 2$ at once.
Use $\leq$ or $\geq$ in your answer exactly when the question does. ' $\leq 0$ ' includes the roots themselves.
If $a$ is negative, either sketch the upside-down U, or multiply the whole inequality by $- 1$ first and reverse the sign. So $6 + x - x^{2} \geq 0$ becomes $x^{2} - x - 6 \leq 0$ .
Never solve $(x + 2) (x - 4) > 7$ by making each bracket greater than 7. Expand, move the 7 across so one side is zero, then factorise again. The last worked example does this.
Inequalities also finish discriminant questions. ' $2 x^{2} + k x + 8 = 0$ has no real roots' means $k^{2} - 64 < 0$ . The critical values are $\pm 8$ . As a function of $k$ this is a U, which is below the axis between its roots, so $- 8 < k < 8$ .

Worked examples

Example 1. $f (x) = 7 + 12 x - 3 x^{2}$ . (a) Write $f (x)$ in the form $a - b {(x + c)}^{2}$ . (b) State the maximum value of $f (x)$ and the value of $x$ at which it occurs. (c) Find the range of $f$ for the domain $0 \leq x \leq 5$ .

Show the working

(a) Take $- 3$ out of the $x^{2}$ and $x$ terms. Check the sign inside: $- 3 \times (- 4 x) = + 12 x$ .
$7 + 12 x - 3 x^{2} = - 3 (x^{2} - 4 x) + 7$
Complete the square inside the bracket: half of $- 4$ is $- 2$ , and ${(x - 2)}^{2} = x^{2} - 4 x + 4$ , so subtract 4.
$- 3 [{(x - 2)}^{2} - 4] + 7$
Multiply the $- 3$ back through both terms in the bracket. The $- 4$ becomes $+ 12$ .
$- 3 {(x - 2)}^{2} + 12 + 7 = 19 - 3 {(x - 2)}^{2}$
Check by expanding: $19 - 3 (x^{2} - 4 x + 4) = 19 - 3 x^{2} + 12 x - 12 = 7 + 12 x - 3 x^{2}$ . Correct. So $a = 19$ , $b = 3$ , $c = - 2$ .
(b) $3 {(x - 2)}^{2} \geq 0$ , and it is subtracted from 19. So $f (x)$ can never exceed 19, and it equals 19 when ${(x - 2)}^{2} = 0$ , that is, when $x = 2$ .
(c) The vertex $x = 2$ lies inside $0 \leq x \leq 5$ , so the maximum 19 belongs to the range.
Now the ends: $f (0) = 7$ and $f (5) = 7 + 60 - 75 = - 8$ . For an upside-down U the least value is at an end, here $- 8$ .
Both ends are included, so both signs are $\leq$ .
$- 8 \leq f (x) \leq 19$

Answer: (a) $19 - 3 {(x - 2)}^{2}$ (b) maximum value $19$ , when $x = 2$ (c) $- 8 \leq f (x) \leq 19$

Example 2. Find the values of $k$ for which the line $y = k x - 2$ is a tangent to the curve $y = x^{2} + 2 x + 2$ . Hence find the set of values of $k$ for which the line does not meet the curve.

Show the working

Set the two expressions for $y$ equal, and rearrange so one side is zero.
$x^{2} + 2 x + 2 = k x - 2 \Rightarrow x^{2} + (2 - k) x + 4 = 0$
Group the $x$ terms into one coefficient so you can read off $b$ : $a = 1$ , $b = 2 - k$ , $c = 4$ .
A tangent gives a repeated root, so the discriminant is zero.
${(2 - k)}^{2} - 16 = 0$
${(2 - k)}^{2} = 16$ , so $2 - k = 4$ or $2 - k = - 4$ . That gives $k = - 2$ or $k = 6$ .
Check $k = 6$ : the equation becomes $x^{2} - 4 x + 4 = {(x - 2)}^{2} = 0$ , a repeated root at $x = 2$ . The line gives $y = 6 (2) - 2 = 10$ and the curve gives $4 + 4 + 2 = 10$ . They touch at $(2, 10)$ .
The line does not meet the curve when the discriminant is negative.
${(2 - k)}^{2} - 16 < 0$
This is a quadratic inequality in $k$ , with critical values $- 2$ and $6$ from above. The $k^{2}$ coefficient is $+ 1$ , so as a function of $k$ it is a U, which is negative between its roots.
Write the answer as one combined inequality.
$- 2 < k < 6$

Answer: Tangent when $k = - 2$ or $k = 6$ . The line does not meet the curve when $- 2 < k < 6$ .

Example 3. Solve $3 x^{2} - 6 x - 2 = 0$ , giving your answers in exact form. Use completing the square, then check with the quadratic formula.

Show the working

Take 3 out of the $x$ terms and complete the square.
$3 (x^{2} - 2 x) - 2 = 3 [{(x - 1)}^{2} - 1] - 2 = 3 {(x - 1)}^{2} - 5$
Set this equal to zero and get the square on its own.
$3 {(x - 1)}^{2} = 5 \Rightarrow {(x - 1)}^{2} = \frac{5}{3}$
Square-root both sides, with $\pm$ .
$x - 1 = \pm \sqrt{\frac{5}{3}}$
Simplify the surd by rationalising: $\sqrt{\frac{5}{3}} = \frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{5} \times \sqrt{3}}{3} = \frac{\sqrt{15}}{3}$ .
$x = 1 \pm \frac{\sqrt{15}}{3}$
Now the formula, with $a = 3$ , $b = - 6$ , $c = - 2$ . The discriminant is $36 + 24 = 60$ , and $\sqrt{60} = \sqrt{4 \times 15} = 2 \sqrt{15}$ .
$x = \frac{6 \pm 2 \sqrt{15}}{6} = 1 \pm \frac{\sqrt{15}}{3}$
Both routes agree. In the exam one clean method is enough: use the formula when fractions would make completing the square messy, and completing the square when the question asks for it or you already have the completed form.

Answer: $x = 1 + \frac{\sqrt{15}}{3}$ or $x = 1 - \frac{\sqrt{15}}{3}$ (equivalently $x = \frac{3 \pm \sqrt{15}}{3}$ )

Example 4. Find the set of values of $x$ for which $(x + 2) (x - 4) > 7$ .

Show the working

The 7 on the right is the trap. Knowing a product is greater than 7 tells you nothing about each bracket on its own, so you cannot split it. Get zero on one side first.
Expand, then subtract 7.
$x^{2} - 2 x - 8 - 7 > 0 \Rightarrow x^{2} - 2 x - 15 > 0$
Find the critical values by solving the equation.
$(x - 5) (x + 3) = 0 \Rightarrow x = - 3 or x = 5$
Sketch $y = x^{2} - 2 x - 15$ : a U through $- 3$ and $5$ . You want $y > 0$ , the parts above the axis, which are the two outer pieces.
Test a value to be sure: $x = 0$ gives $(2) (- 4) = - 8$ , which is not greater than 7. So the middle region is correctly left out.
Two separate regions, so two inequalities joined by 'or'.
$x < - 3 or x > 5$

Answer: $x < - 3$ or $x > 5$

Where people go wrong

Mistake: Forgetting to multiply the subtracted constant by $a$ : writing $2 (x^{2} + 6 x) + 7$ as $2 {(x + 3)}^{2} - 9 + 7$ .

Instead: $2 [{(x + 3)}^{2} - 9] + 7 = 2 {(x + 3)}^{2} - 18 + 7 = 2 {(x + 3)}^{2} - 11$ .

The $- 9$ is inside the bracket being multiplied by 2. Expanding your answer back out catches this at once.

Mistake: Reading the vertex of ${(x - 3)}^{2} + 2$ as $(- 3, 2)$ .

Instead: The vertex is $(3, 2)$ .

The turning point is where the bracket is zero. Solve $x - 3 = 0$ rather than copying the sign you can see.

Mistake: Using $b^{2} - 4 a c > 0$ for 'the equation has real roots'.

Instead: Use $b^{2} - 4 a c \geq 0$ .

Two equal roots are still real roots. Only 'two distinct' or 'two different' real roots needs the strict $> 0$ .

Mistake: Writing the solution of $x^{2} - 2 x - 15 > 0$ as $- 3 > x > 5$ or $5 < x < - 3$ .

Instead: Write $x < - 3$ or $x > 5$ .

A combined inequality $p < x < q$ means $x$ lies between $p$ and $q$ . The outer regions are two separate pieces and need 'or'.

Mistake: Multiplying an inequality by $- 1$ to make $a$ positive, without reversing the sign.

Instead: $6 + x - x^{2} \geq 0$ becomes $x^{2} - x - 6 \leq 0$ .

Multiplying by a negative number reverses order: $2 < 3$ , but $- 2 > - 3$ .

Check yourself

Write $x^{2} + 10 x + 18$ in the form ${(x + p)}^{2} + q$ , and state its minimum value.
Answer
${(x + 5)}^{2} - 25 + 18 = {(x + 5)}^{2} - 7$ . Minimum value $- 7$ , when $x = - 5$ .
Write $2 x^{2} - 8 x + 3$ in the form $a {(x + p)}^{2} + q$ . Is its turning point a maximum or a minimum?
Answer
$2 [{(x - 2)}^{2} - 4] + 3 = 2 {(x - 2)}^{2} - 5$ . A minimum, at $(2, - 5)$ , because $a = 2 > 0$ .
State the range of $f (x) = {(x - 1)}^{2} + 4$ for the domain $x \geq 3$ .
Answer
The vertex $x = 1$ is outside the domain, and the curve rises for $x \geq 3$ , so the least value is $f (3) = 8$ . Range: $f (x) \geq 8$ .
Find the values of $k$ for which $x^{2} + k x + 9 = 0$ has two equal roots.
Answer
$b^{2} - 4 a c = k^{2} - 36 = 0$ , so $k = 6$ or $k = - 6$ .
Show that the line $y = 2 x - 5$ does not meet the curve $y = x^{2} + 1$ .
Answer
$x^{2} + 1 = 2 x - 5$ gives $x^{2} - 2 x + 6 = 0$ . $b^{2} - 4 a c = 4 - 24 = - 20 < 0$ , so there are no real roots and the line does not meet the curve.
Solve $2 x^{2} + 5 x - 3 = 0$ .
Answer
$(2 x - 1) (x + 3) = 0$ , so $x = \frac{1}{2}$ or $x = - 3$ .
Solve $x^{2} + 4 x - 1 = 0$ , giving exact answers.
Answer
${(x + 2)}^{2} - 5 = 0$ , so $x + 2 = \pm \sqrt{5}$ and $x = - 2 + \sqrt{5}$ or $x = - 2 - \sqrt{5}$ .
Solve $6 + x - x^{2} \geq 0$ .
Answer
Multiply by $- 1$ and reverse the sign: $x^{2} - x - 6 \leq 0$ , so $(x - 3) (x + 2) \leq 0$ . The U is on or below the axis between its roots: $- 2 \leq x \leq 3$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. $f (x) = 2 x^{2} - 12 x + 7$ for $x \in ℝ$ .

(a) Write $f (x)$ in the form $a {(x + b)}^{2} + c$ , where $a$ , $b$ and $c$ are constants. 3 marks No calculator

Mark scheme

B1 a = 2
B1 b = −3
B1 c = −11

(b) Write down the coordinates of the minimum point of the curve $y = f (x)$ . 1 mark No calculator

Mark scheme

B1 (3, −11); FT their b and c from (a)

(c) Find the range of $f$ when the domain is $1 \leq x \leq 7$ . 2 marks No calculator

Mark scheme

M1 uses the minimum −11 (vertex inside the domain) and evaluates f(1) = −3 and/or f(7) = 21
A1 −11 ≤ f(x) ≤ 21 (accept f or y in place of f(x))

Question 2. A line has equation $y = k x + 1$ , where $k$ is a constant. A curve has equation $y = x^{2} + 3 x + 5$ .

(a) Show that the $x$ -coordinates of any points where the line meets the curve satisfy $x^{2} + (3 - k) x + 4 = 0$ . 2 marks No calculator

Mark scheme

M1 x² + 3x + 5 = kx + 1 seen
A1 rearranged to the given equation, with the x terms grouped, no errors seen

(b) Find the values of $k$ for which the line is a tangent to the curve. 3 marks No calculator

Mark scheme

M1 (3 − k)² − 4(1)(4) seen
M1 sets the discriminant equal to 0 and solves for k
A1 k = −1 and k = 7

(c) Find the set of values of $k$ for which the line meets the curve at two distinct points. 2 marks No calculator

Mark scheme

M1 (3 − k)² − 16 > 0 used, choosing the outer region of their critical values
A1 k < −1 or k > 7

Question 3.

(a) Solve $4 x^{2} - 4 x - 7 = 0$ , giving your answers in the form $p \pm q \sqrt{2}$ , where $p$ and $q$ are rational numbers. 3 marks No calculator

Mark scheme

M1 correct substitution into the quadratic formula, or 4(x − ½)² − 8 = 0 oe
M1 √128 simplified to 8√2, or (x − ½)² = 2 reached
A1 x = ½ ± √2

(b) Find the set of values of $x$ for which $(x + 1) (x - 5) \leq 2 x - 13$ . 3 marks No calculator

Mark scheme

M1 expands and rearranges to x² − 6x + 8 ≤ 0 oe
A1 critical values 2 and 4
A1 2 ≤ x ≤ 4

Question 4. The curve $y = 5 + 4 x - x^{2}$ .

(a) Sketch the curve, stating the coordinates of its turning point and of the points where it meets the axes. 3 marks No calculator

Mark scheme

B1 upside-down U shape
B1 (−1, 0), (5, 0) and (0, 5) all stated
B1 turning point (2, 9) stated

(b) Write down the set of values of $k$ for which the equation $5 + 4 x - x^{2} = k$ has two distinct real roots. 1 mark No calculator

Mark scheme

B1 k < 9; FT their maximum value from (a)

(c) Write down the set of values of $x$ for which $5 + 4 x - x^{2} > 0$ . 1 mark No calculator

Mark scheme

B1 −1 < x < 5

How this is examined

A line meeting a curve: the discriminant with an unknown constant: Equate a line and a curve, then find the values of $k$ for a tangent, or the set of values of $k$ for two points or no intersection; 4–6 marks across the parts. A three-term quadratic with every term on one side, $b^{2} - 4 a c$ written with the correct $a$ , $b$ and $c$ and the correct condition ( $= 0$ , $> 0$ or $< 0$ ), both critical values of $k$ , and the final set in correct inequality notation.
Completing the square and reading off the turning point: Write $f (x)$ in the form $a {(x + b)}^{2} + c$ , then state the maximum or minimum value or the coordinates of the vertex; 3–4 marks. All three constants correct (check by expanding), then the turning point at $x = - b$ with value $c$ , called a maximum or minimum according to the sign of $a$ .
Writing the solution set of a quadratic inequality: Find the set of values of $x$ for which a quadratic inequality holds; about 3 marks, sometimes as the last step of a discriminant question. A rearrangement to a quadratic compared with 0, both critical values, and the correct region written as $p < x < q$ , or as $x < p$ or $x > q$ . The right critical values with the wrong region is not a complete answer.
Exact roots without a calculator: Solve a quadratic, giving answers in the form $p \pm q \sqrt{r}$ ; 2–3 marks. A correct substitution into the formula or a correct completed square, then the surd simplified fully with every term of the numerator divided.
Sketches and ranges from the turning point: Sketch $y = f (x)$ showing the vertex and axis intercepts, or state the range of $f$ for a restricted domain; 1–3 marks. The right orientation with the vertex and both axis intercepts labelled with coordinates; for a range, a check that the vertex is inside the domain, the end values worked out, and an inequality in $f (x)$ .

Words to know

Completed square form: A quadratic written as $a {(x + p)}^{2} + q$ . Its turning point is $(- p, q)$ .
Vertex (turning point): The minimum point of a U-shaped parabola ( $a > 0$ ) or the maximum point of an upside-down one ( $a < 0$ ).
Discriminant: $b^{2} - 4 a c$ . Positive: two distinct real roots. Zero: two equal roots. Negative: no real roots.
Tangent: A line that touches a curve at one point without crossing it there. For a line and a parabola, the equation you get by equating them has discriminant $0$ .
Critical values: The roots of the equation you get by replacing the inequality sign with $=$ . They mark the ends of the solution regions.
Solution set: All the values that satisfy an inequality, written e.g. as $- 2 < x < 5$ , or $x < - 2$ or $x > 5$ .

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