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Quadratic functions questions
One parabola, three ways of writing it. Factorised shows the roots, completed-square shows the vertex, expanded shows the y-intercept. Underneath all three sits a single number, the discriminant, which knows how many times the curve meets the axis before you solve anything.
7 original questions · 23 marks · the quadratic functions notes · Algebra and functions
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Express x2 + 8x + 3 in the form (x + p)2 + q, and state the minimum point of the curve y = x2 + 8x + 3.
Worked answer
Half the 8, square it, correct for it: (x + 4)2 − 16 + 3 = (x + 4)2 − 13. The square is smallest when its bracket is zero, so the minimum point is (−4, −13). M1 for halving and squaring the 8, A1 for the completed square, B1 for the minimum point. Reading the minimum as (4, −13) is the sign slip to watch, since the bracket vanishes at x = −p.Use the discriminant to determine how many real roots 2x2 − 3x + 5 = 0 has.
Worked answer
b2 − 4ac = 9 − 40 = −31, which is negative, so there are no real roots. M1 for the discriminant, A1 for the conclusion. The parabola opens upward and its minimum sits above the axis, so it never touches down. Quote the value and then state the conclusion; a bare number scores half.Express 3x2 − 12x + 5 in the form a(x + p)2 + q, and state the minimum point.
Worked answer
Take the 3 out of the x terms first. 3(x2 − 4x) + 5 = 3[(x − 2)2 − 4] + 5, and distributing gives 3(x − 2)2 − 12 + 5 = 3(x − 2)2 − 7. The minimum point is (2, −7). M1 for taking the 3 out, A1 for the bracket, A1 for −7, B1 for the minimum point. That correction term must be multiplied by the 3 on the way back out, and answers that leave it as −4 + 5 = 1 lose two marks in one step.Find the values of k for which x2 + kx + 25 = 0 has a repeated root.
Worked answer
A repeated root means the discriminant is zero: k2 − 100 = 0, so k = 10 or k = −10. M1 for setting the discriminant to zero, A1 for both values. Two different parabolas, each just touching the axis, one touching to the left of the origin and one to the right. Giving only k = 10 halves the mark.The equation kx2 + 4x + 1 = 0 has two distinct real roots. Find the set of possible values of k.
Worked answer
Two distinct roots need 16 − 4k > 0, so k < 4. But k = 0 has to be excluded as well, because the equation is then linear and has a single root rather than two. So k < 4 and k ≠ 0. M1 for a positive discriminant, A1 for k < 4, B1 for excluding k = 0. The vanishing leading coefficient is the part most answers miss, and it is the mark that separates a 3 from a 2.Solve 9x − 10 × 3x + 9 = 0.
Worked answer
Since 9x = (3x)2, substitute u = 3x to get u2 − 10u + 9 = 0, which factorises as (u − 1)(u − 9) = 0. Then 3x = 1 gives x = 0, and 3x = 9 gives x = 2. M1 for the substitution, A1 for the factorised quadratic, A1 for x = 0, A1 for x = 2. Spot the disguise, substitute, and return; stopping at u = 1 and u = 9 answers a question nobody asked.Given that the equation (k + 1)x2 + 4kx + 9 = 0 has a repeated root, find the two possible values of k.
Worked answer
Set the discriminant to zero: (4k)2 − 4(k + 1)(9) = 0, so 16k2 − 36k − 36 = 0. Dividing by 4 gives 4k2 − 9k − 9 = 0, which factorises as (k − 3)(4k + 3) = 0, so k = 3 or k = −3/4. M1 for setting the discriminant to zero, A1 for the quadratic in k, dM1 for factorising it, A1 for k = 3, A1 for k = −3/4. Both are legitimate: k = 3 turns the equation into 4x2 + 12x + 9 = (2x + 3)2 = 0, and k = −3/4 into (x − 6)2/4 = 0. The unknown sits in a and in b at once, so the discriminant becomes a quadratic in k, and discarding the fractional root because it looks untidy throws away the final mark. Only k = −1 would need excluding, and neither answer is −1.
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