Quadratic Functions

• Anchor for questions like 6. A “find the values of \(k\)” question of this shape is three nested skills in order: combine the line and curve into one quadratic, form the discriminant condition, then solve the resulting inequality using section 2.5. Write those three headings on your page before you start, and each one tells you what the next line of working has to be.
• The five-second self-test, any time. Say these three sentences. “\(a(x-h)^2+k\) turns at \((h,k)\).” “Combine first, then \(b^2-4ac\).” “Upward parabolas are negative between the roots.” If any one of them is slow to arrive, that is the section to revisit.

• Complete the square when \(a=1\), when \(a\neq1\), when \(a\) is negative and when \(b\) is odd.
• State a turning point as a coordinate pair, not just as a value.
• Decide maximum or minimum from the sign of \(a\), or from the second derivative.
• Sketch a quadratic with opening direction, axis of symmetry, \(y\)-intercept, roots and turning point all marked.
• Determine a range for an unrestricted domain and for a restricted domain.
• Recognise when the vertex lies outside a stated domain and use the endpoint instead.
• Compute \(b^2-4ac\) and state the correct root condition.
• Combine a line and a curve into one quadratic before testing tangency.
• Find an unknown constant from a tangency or root condition.
• Solve a quadratic equation by factorisation, by completing the square and by formula.
• Leave exact roots in surd form and only round when asked.
• Solve a quadratic inequality and write the solution set with correct endpoints.

Completing the square returns in Chapter 8, coordinate geometry of the circle, when an equation given as \(x^2+y^2+2gx+2fy+c=0\) must be put into centre-radius form. The discriminant returns in that same chapter, deciding whether a straight line is a tangent to a circle, a chord, or does not meet it. The sign diagram of section 2.5 is reused unchanged in Chapter 14 to decide where a derivative is positive, and therefore where a function is increasing. Time spent making these secure now is repaid later in the course.

Completing the Square

An algebraic rearrangement that rewrites a quadratic ax^2 + bx + c in the equivalent form a(x - h)^2 + k. Because a squared bracket is never negative, the constant k is the minimum value of the function when a is positive and the maximum value when a is negative, attained at x = h; the point (h, k) is therefore the turning point of the parabola. The rearrangement changes only how the expression is written, never its value at any x.

Quadratic Function

A function of a single variable whose highest power is two, written f(x) = ax^2 + bx + c with a not equal to zero. Its graph is a parabola, symmetric about a vertical axis through a single turning point. The same function can be written in expanded form, in factorised form a(x - p)(x - q) which displays its roots, or in completed-square form a(x - h)^2 + k which displays its turning point (h, k); all three describe one curve and are equal for every value of x.

Range of a Quadratic Function

The set of output values a quadratic function actually takes over its stated domain. For an upward parabola with turning point (h, k) and no domain restriction the range is f(x) greater than or equal to k; for a downward parabola it is f(x) less than or equal to k. When the domain is restricted the range must be read from the part of the curve that survives: if the turning point lies inside the domain it still supplies one boundary, but if it lies outside, the function is monotonic there and every boundary the range has comes from an endpoint of the domain instead. A closed interval supplies two endpoints and so gives a range bounded on both sides, while a one-sided domain such as x greater than or equal to 3 supplies one endpoint and gives a range bounded on one side only.

Quadratic Formula

The general solution of ax^2 + bx + c = 0, giving x equal to negative b plus or minus the square root of b squared minus 4ac, all over 2a. It is supplied in the Cambridge 0606 List of formulas, so it need not be memorised, but it must be applied correctly: every coefficient carries its own sign, the whole numerator is divided by 2a, and the quantity under the root is the discriminant, so a negative value there means the equation has no real roots and the formula returns nothing.

Discriminant

The quantity b^2 - 4ac formed from the coefficients of a quadratic equation ax^2 + bx + c = 0, usually written as the Greek capital delta. Its sign alone determines the number of real roots without the equation being solved: positive gives two distinct real roots, zero gives two equal real roots, and negative gives no real roots. Applied to the single quadratic obtained by equating a line and a curve, the same three cases correspond to the line cutting the curve twice, being tangent to it, and not meeting it at all.

Quadratic Inequality

An inequality in which a quadratic expression is compared with zero, such as ax^2 + bx + c less than or equal to zero. Its answer is a solution set, not a pair of numbers: the set of all x for which the statement is true. The roots of the corresponding equation are the critical values where the expression changes sign, and the opening direction of the parabola then decides whether the wanted region lies between those roots or outside them. Endpoints belong to the set only when the inequality is non-strict.

• “\(2x^2-8x+3=2(x-2)^2-4+3=2(x-2)^2-1\)” Why it fails The \(-4\) sits inside a bracket that is multiplied by \(2\), so it contributes \(-8\), not \(-4\). Test Expand the wrong answer: \(2(x-2)^2-1=2x^2-8x+8-1=2x^2-8x+7\neq2x^2-8x+3\).
• “The turning point of \(2(x-2)^2-5\) is \((-2,-5)\).” Why it fails The form is \(a(x-h)^2+k\), so \(x-2\) means \(h=+2\). The bracket is zero when \(x=2\). Test Substitute: \(f(-2)=2(16)-5=27\), not \(-5\). The point \((-2,-5)\) is not on the curve.
• “\(\dfrac{dy}{dx}=0\) gives \(x=2\), so the minimum value is 2.” Why it fails \(x=2\) is where the minimum occurs, not the minimum value. The value is \(f(2)=-5\). Test Ask which axis the answer belongs to. A “value of the function” is always a \(y\)-coordinate.
• “A stationary point was found, so it is a minimum.” Why it fails Nothing in \(\dfrac{dy}{dx}=0\) says which. For \(-3x^2+12x-5\) the stationary point is a maximum. Test Check the sign of \(a\), or of \(\dfrac{d^2y}{dx^2}=2a\), before naming it.
• “The range of \(2(x-2)^2-5\) is always \(f(x)\ge-5\).” Why it fails Only when the domain contains \(x=2\). On \(x\ge3\) the value \(-5\) is never produced by any permitted input. Test Ask: can I name an \(x\) in the domain with \(f(x)=-5\)? If not, \(-5\) is not in the range.
• “For \(x\ge3\) the range is \(-3\le f(x)\le\) something.” Why it fails The domain \(x\ge3\) is unbounded above, and the curve rises without limit, so there is no upper boundary to find. Test An unbounded domain on the rising side of an upward parabola always gives an unbounded range above.
• “\(x\ge3\) so I substitute \(x=3\) and \(x=\infty\).” Why it fails \(\infty\) is not a number and cannot be substituted. Describe the behaviour instead: as \(x\) increases beyond \(3\), \(f(x)\) increases without bound. Test Write the range as a one-sided inequality, \(f(x)\ge-3\), which says exactly that.
• “On a closed interval the maximum is always at an endpoint.” Why it fails True only for an upward parabola. For \(g(x)=-x^2+4x+5\) on \(0\le x\le5\) the maximum is \(9\), at the vertex \(x=2\); both endpoints give less (\(g(0)=5\), \(g(5)=0\)). Test Ask which boundary the vertex is supplying before you go looking for the other one. \(a>0\) means the vertex gives the minimum; \(a<0\) means it gives the maximum.
• “The range is \(-1\le x\le4\).” Why it fails That is the domain restated. The range is a statement about \(f(x)\), not about \(x\). Test A range should have \(f(x)\) or \(y\) in it, never \(x\) alone.
• “For \(y=x^2-4x+7\) and \(y=mx+1\), \(\Delta=(-4)^2-4(1)(7)\).” Why it fails Those are the coefficients of the curve, not of the combined equation. The discriminant is only defined once the two have been merged into \(x^2-(m+4)x+6=0\). Test Ask: which single equation, equal to zero, am I taking the discriminant of? If you cannot write it down, you are not ready to use \(\Delta\).
• “\(\Delta>0\), so the roots are equal.” Why it fails Equal roots need \(\Delta=0\) exactly. A positive discriminant gives two different roots. Test \(\pm\sqrt\Delta\) gives two different numbers unless \(\Delta\) is zero.
• “\((m+4)^2=24\), so \(m+4=2\sqrt6\) and \(m=-4+2\sqrt6\).” Why it fails The negative square root has been dropped. There are two tangent lines, so there must be two values of \(m\). Test Whenever you take a square root of both sides, write the \(\pm\) before you simplify.
• “\(\Delta<0\), so the roots are imaginary.” Why it fails Complex numbers are outside 0606. The expected statement is “no real roots”. Test Answer in the vocabulary of the syllabus you are sitting.
• “\(kx^2+4x+k=0\) has equal roots, so \(k=\pm2\) or \(k=0\).” Why it fails \(k=0\) destroys the \(x^2\) term, so the equation is no longer quadratic and has no discriminant. Test Whenever the coefficient of \(x^2\) contains the unknown, state the condition that it is non-zero.
• “\(x(x-3)=4\), so \(x=4\) or \(x-3=4\).” Why it fails The null factor law works only against zero. Two numbers with product \(4\) can be anything. Fix Expand and rearrange: \(x^2-3x-4=0\), so \((x-4)(x+1)=0\) and \(x=4\) or \(x=-1\). The second root would have been lost.
• “\(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\), and with \(b=-7\) that is \(\dfrac{-7\pm\dots}{4}\).” Why it fails \(-b\) is \(-(-7)=+7\). The formula negates whatever \(b\) is, sign included. Test Check the sum of your roots against \(-\dfrac{b}{a}\); a sign error shows up immediately.
• “\(x=\dfrac{7\pm\sqrt{17}}{4}\), so \(x=2.78\) or \(x=0.72\).” Why it fails Nothing, unless the question asked for exact values — in which case the surd form is the answer and the decimals are a different, approximate one. Rule “Exact”, “in surd form” and “leave your answer in terms of…” all forbid rounding.
• “\((x-3)^2=5\), so \(x-3=\sqrt5\) and \(x=3+\sqrt5\).” Why it fails The \(\pm\) has been dropped, losing the root \(3-\sqrt5\). Test For \(x^2-6x+4\), \(\Delta=(-6)^2-4(1)(4)=36-16=20>0\) — two roots are expected, so a single answer cannot be complete.
• “My calculator gave the roots, so I do not need working.” Why it fails A calculator returns decimals, not surds, and cannot answer a “find \(k\), then use \(k\)” question at all. Use it properly Solve algebraically, then use the calculator to verify the decimal value of your exact answer.
• “\((2x+1)(x-3)\le0\), so \(2x+1\le0\) and \(x-3\le0\).” Why it fails A product is negative when the factors have opposite signs, not when both are negative. Both negative would make the product positive. Fix Use the sign diagram in Figure 2.8, or the shape of the graph. Never split a product inequality into two independent inequalities.
• “The answer is \(x=-\frac12\) and \(x=3\).” Why it fails Those are the critical values — the solution of the equation. The inequality's answer is a whole interval of values. Test Your final line should contain an inequality sign, not an equals sign.
• “\(2x^2-5x-3\le0\) gives \(-\frac12<x<3\).” Why it fails The endpoints have been dropped. At \(x=3\) the expression is \(0\), and \(0\le0\) is true, so \(3\) belongs to the solution set. Rule The endpoint convention copies the original sign: \(\le\) and \(\ge\) include, \(<\) and \(>\) exclude.
• “\(x^2-x-6>0\) gives \(-2<x<3\).” Why it fails That is the region where the expression is negative. The upward parabola is positive outside its roots. Test Substitute \(x=0\), which lies in the claimed set: \(-6>0\) is false, so the set is wrong.
• “\(-x^2+4x-3\ge0\) becomes \(x^2-4x+3\ge0\).” Why it fails Multiplying by \(-1\) reverses the inequality. The correct statement is \(x^2-4x+3\le0\). Test Try a number: \(x=2\) satisfies the original, and \(4-8+3=-1\), which satisfies \(\le0\) but not \(\ge0\).

• Cambridge IGCSE Additional Mathematics 0606 is assessed by two written papers of equal weight, each 2 hours, each 80 marks, each 50% of the qualification. Every candidate takes both — 0606 has no Core or Extended tier, so nothing in this chapter is optional for anyone. Paper 1 is a non-calculator paper — a calculator is not allowed at all. Paper 2 requires a scientific calculator. Either paper can examine any part of the content, so every method in this chapter has to work on paper as well as on a machine. Quadratic reasoning appears in Topic 2 questions directly and then reappears inside later topics, so the habits below matter beyond this chapter.
• Write the combined equation before the discriminant. If a question involves a line and a curve, the first line of your working should be the single quadratic equal to zero. Everything else follows from it, and the method stays visible even if the arithmetic slips later.
• Check a completed square by expanding it back. It costs one line and it catches two easy slips: the wrong sign inside the bracket and a forgotten multiplication by \(a\).
• Do not decimalise early. If the roots are \(3\pm\sqrt5\), write \(3\pm\sqrt5\). Converting to \(5.236\) and \(0.764\) throws away exactness and cannot be undone.
• The accuracy rules, once, for the whole chapter. Answers are expected in their simplest form unless the question says otherwise. Where a question asks for exact values, the answer may need surds, a fraction, \(\pi\), \(e\) or logarithms — in Topic 2 it is almost always a surd or a fraction. Where an answer is not exact, give at least 3 significant figures, or at least 1 decimal place for an angle in degrees, unless the question sets a different accuracy; carry more accuracy than that inside your working. Do not round an answer that is already exact to four or five significant figures. And never mix a fraction and a decimal inside a single numerical value: the syllabus gives \(\dfrac{1}{0.2}\) as an example of what is not acceptable, so write \(5\), or keep the whole value as a fraction.
• On the calculator. Paper 1 does not allow one, so on that paper every root in this chapter has to come out of factorisation, completing the square or the formula by hand.