Sec 4 A Math: Quadratic Functions & Inequalities, practice questions & worked solutions

Quadratic Functions & Inequalities practice questions selected from 2025 Singapore school A Math prelim papers, each with a full worked solution.

About this topic & key methods

These questions support Secondary 4 and O-Level Additional Mathematics revision.

Attempt each question on paper before opening its worked solution. Keep the source question number when checking against the original paper.

Key methods

Complete the square to find a turning point and a maximum or minimum.
Use the discriminant to determine the number of real roots or intersections.
Solve quadratic inequalities and check the relevant intervals.
Form and solve simultaneous equations involving a line and a curve.

For guided practice, see our Sec 4 A Maths tuition programme.

Questions & worked solutions

Unless the question specifies otherwise, give numerical answers to 3 significant figures and angles in degrees to 1 decimal place. Angles in radians are stated explicitly.

Q1St Joseph’s Institution2025Prelim P1 Q4●○○ Foundational4 marks

Quadratic Functions

Express $-9 + 6x - 2x^2$ in the form $a(x + b)^2 + c$ and hence explain why $-9 + 6x - 2x^2$ is negative for all real values of $x$ .

Show worked solution▾

$\begin{aligned} -2x^2 + 6x - 9 &= -2\left(x^2 - 3x + \tfrac{9}{2}\right)\\ &= -2\left(x^2 - 3x + \left(\tfrac{3}{2}\right)^2 - \left(\tfrac{3}{2}\right)^2 + \tfrac{9}{2}\right)\\ &= -2\left(\left(x - \tfrac{3}{2}\right)^2 + \tfrac{9}{4}\right)\\ &= -2\left(x - \tfrac{3}{2}\right)^2 - \left(\tfrac{9}{2}\right) \end{aligned}$ Since $\left(x - \dfrac{3}{2}\right)^2 \geq 0$ , $-2\left(x - \tfrac{3}{2}\right)^2 \leq 0 \textbf{and} -2\left(x - \tfrac{3}{2}\right)^2 - \left(\tfrac{9}{2}\right) \leq -\left(\tfrac{9}{2}\right)$ for all real values of $x$ .

Therefore $-9 + 6x - 2x^2$ is negative for all real values of $x$ .

Q2Bedok South Secondary School2025Prelim P1 Q5●●○ Standard5 marks

Simultaneous Equations

The line $3x + 4y = 13$ intersects the curve $y = \dfrac{6x - 10}{3x - 1}$ at points $P$ and $Q$ . Calculate the exact length of the line segment $PQ$ .

Show worked solution▾

$\begin{aligned} 4x - y &= 5 \text{------[1]}\\ y &= 2x^2 - 6x + 7 \text{------[2]} \end{aligned}$ Subst (2) into (1) $\begin{aligned} 4x - (2x^2 - 6x + 7) &= 5\\ 4x - 2x^2 + 6x - 7 &= 5\\ 2x^2 - 10x + 12 &= 0\\ x^2 - 5x + 6 &= 0\\ (x - 2)(x - 3) &= 0\\ x = 2 \ &\text{or}\ x = 3 \end{aligned}$ Subst respective $x$ values into [1], $\begin{array}{c|c} y = 4(2) - 5 & y = 4(3) - 5\\ y = 3 & y = 7 \end{array}$ $P(2, 3) \ \text{and}\ Q(3, 7)$ $\begin{aligned} PQ &= \sqrt{(3 - 2)^2 + (7 - 3)^2}\\ PQ &= \sqrt{17} \end{aligned}$

Q3Catholic High School2025Prelim P1 Q1●●○ Standard5 marks

Quadratic Functions

A quadratic function is given by $f(x) = 5 - 4x - 2x^2$ .

Express $f(x)$ in the form $c - a(x+b)^2$ , where $a$ , $b$ and $c$ are constants and $a > 0$ .
Sketch the graph of $y = f(x)$ , indicating clearly he turning point and the $y$ -intercept.
Hence find the range of values of $k$ for which the equation $5 - 4x - 2x^2 = k$ has at most one root.

Show worked solution▾

(a)

$\begin{aligned} 5-4x-2x^2 &= -2\!\left[x^2+2x\right]+5\\ &= -2(x+1)^2+7 \end{aligned}$

(b)

Turning point $(-1,\,7)$ ; $y$ -intercept $(0,\,5)$ .

Catholic High School 2025 P1 Q1: worked solution diagram

(c)

$k \geq 7$

Q4Chung Cheng High (Yishun)2025Prelim P2 Q2●●○ Standard5 marks

Quadratic Functions

The equation of a curve is $y = px^2 - (p + 2)x + 1$ , where $p$ is a constant.

Given that $p > 0$ , by completing the square, find the minimum value of $y$ . Express your answer in the form $\dfrac{a - p^2}{bp}$ , where $a$ and $b$ are integers.
Given instead that $p < 0$ , find the coordinates of the maximum point in terms of $p$ .

Show worked solution▾

(a)

$\begin{aligned} y &= px^2 - (p + 2)x + 1\\ &= p\left(x^2 - \frac{p + 2}{p}x\right) + 1\\ &= p\left[x^2 - \frac{p + 2}{p}x + \left(\frac{p + 2}{2p}\right)^2 - \left(\frac{p + 2}{2p}\right)^2\right] + 1\\ &= p\left[\left(x - \frac{p + 2}{2p}\right)^2 - \left(\frac{p + 2}{2p}\right)^2\right] + 1\\ &= p\left(x - \frac{p + 2}{2p}\right)^2 + 1 - \frac{p^2 + 4p + 4}{4p^2} \times p\\ &= p\left(x - \frac{p + 2}{2p}\right)^2 + \frac{4p - p^2 - 4p - 4}{4p}\\ &= p\left(x - \frac{p + 2}{2p}\right)^2 + \frac{-p^2 - 4}{4p} \end{aligned}$ Hence, the minimum value of $y$ is $\dfrac{-p^2 - 4}{4p}$ .

(b)

$\left(\frac{p + 2}{2p},\ \frac{-p^2 - 4}{4p}\right)$

Q5Cedar Girls’ Secondary School2025Prelim P1 Q2●●○ Standard5 marks

Quadratic Functions

Find the range of values of the constant $m$ for which the curve $y = (m-3)x^2 + mx + m$ lies completely above the $x$ -axis.

Show worked solution▾

$\begin{aligned} y &= (m-3)x^2+mx+m\\ b^2-4ac &< 0\\ m^2-4(m-3)(m) &< 0\\ m^2-4m^2+12m &< 0\\ -3m^2+12m &< 0\\ -3m(m-4) &< 0\\ m<0 \;&\text{or}\; m>4 \end{aligned}$ $\begin{aligned} m-3 &> 0\\ m &> 3 \end{aligned}$ Therefore $m>4$ .

Q6Jurongville Secondary School2025Prelim P1 Q2●●○ Standard5 marks

Quadratic Functions

Find the set of values of the constant $p$ for which the curve $y = px^2 + (p - 2)x + 5$ lies entirely above the $x$ -axis.

Show worked solution▾

$\begin{aligned} p &> 0\\ (p - 2)^2 - 4(p)(5) &< 0\\ p^2 - 24p + 4 &< 0 \end{aligned}$ $\begin{aligned} \text{Critical Points: } p &= 12 \pm 2\sqrt{35}\\ 12 - 2\sqrt{35} < p &< 12 + 2\sqrt{35}\\ 0.168 < p &< 23.8 \end{aligned}$

Q7Maris Stella High School2025Prelim P1 Q3●●○ Standard5 marks

Quadratic Functions

Find the range of values of the constant $p$ such that $y = px^2 - 4x + p - 3$ is always positive for all real values of $x$ .

Show worked solution▾

for $y=px^2-4x+p-3$ to be always positive, $\begin{aligned} p&>0 \text{ and discriminant} <0,\\ (-4)^2-4(p)(p-3) &< 0\\ 16-4p^2+12p &< 0\\ p^2-3p-4 &> 0\\ (p+1)(p-4) &> 0\\ p<-1 \;&\text{or}\; p>4\\ \text{since } p>0,\ &p>4 \end{aligned}$

Q8Ngee Ann Secondary School2025Prelim P1 Q3●●○ Standard5 marks

Simultaneous Equations

A line that is not parallel to the $x$ -axis has the equation $a^2 x + y = 3a - b - 3$ , where $a$ and $b$ are constants. A curve has the equation $2x^2 + y^2 = b + 3$ . Given that the line and the curve intersect at the point $(0, a)$ , find the value of $a$ and of $b$ .

Show worked solution▾

$\begin{aligned} a^2x+y &= 3a-b-3\\ y &= -a^2x+3a-b-3 \text{-----(1)}\\ 2x^2+y^2 &= b+3 \text{-----(2)} \end{aligned}$ When $x=0$ , $y=a$ ,

From (1), $a=3a-b-3$ , $b=2a-3$ :–(3).

From (2), $a^2=b+3$ :–(4).

Substitute (3) into (4), $\begin{aligned} a^2 &= 2a-3+3\\ a^2-2a &= 0\\ a(a-2) &= 0\\ a=0\ (\text{rej, since not parallel}) \;&\text{or}\; a=2\\ b &= 1 \end{aligned}$

Q9School of Science and Technology2025Prelim P1 Q3●●○ Standard5 marks

Quadratic Functions

The equation of a curve is $y = (4 + p)x^2 - 2x + p$ , where $p$ is a constant. The curve has a maximum point and intersects the line $y = 2x - 1$ . Find the range of values of $p$ .

Show worked solution▾

$\begin{aligned} (4 + p)x^2 - 2x + p &= 2x - 1\\ (4 + p)x^2 - 4x + p + 1 &= 0 \end{aligned}$ $a = 4 + p, b = -4, c = p + 1$ Since the curve intersects the line, $b^2 - 4ac \geq 0$ $\begin{aligned} 16 - 4(4 + p)(p + 1) &\geq 0\\ 4 - \left(p^2 + 5p + 4\right) &\geq 0\\ -p^2 - 5p &\geq 0\\ p^2 + 5p &\leq 0\\ p(p + 5) &\leq 0\\ -5 \leq p &\leq 0 \end{aligned}$ Since curve has a maximum point, $p + 4 < 0$ $p < -4$ Hence $-5 \leq p < -4$

Q10Tanjong Katong Secondary School2025Prelim P1 Q4●●○ Standard5 marks

Quadratic Functions

The height, $y$ m, above the ground of a pellet shot by a shooting device can be modelled by the equation $y = -2x^2 + 20x + 12$ , where $x$ m is the horizontal distance travelled by the pellet.

Express $y = -2x^2 + 20x + 12$ in the form $y = a(x - b)^2 + c$ , where $a$ , $b$ and $c$ are constants.
Explain why the maximum height reached by the pellet is 62 metres.
A two-metre-tall pole is placed on the ground, at a horizontal distance of 10 metres from the shooting device. Explain if the pellet will hit the pole.

Show worked solution▾

(a)

$\begin{aligned} -2x^2 + 20x + 12 &= -2\left(x^2 - 10x\right) + 12\\ &= -2\left[x^2 - 10x + \left(\frac{10}{2}\right)^2 - \left(\frac{10}{2}\right)^2\right] + 12\\ &= -2\left[(x - 5)^2 - 25\right] + 12\\ &= -2(x - 5)^2 + 62 \end{aligned}$

(b)

$\begin{aligned} \text{Height} &= -2(x - 5)^2 + 62\\ (x - 5)^2 &\geq 0\\ -2(x - 5)^2 &\leq 0\\ -2(x - 5)^2 + 62 &\leq 62 \end{aligned}$ Height $\leq 62$ m

Since height is $\leq 62$ m, the maximum height reach by the ball is 62 metres.

(c)

Height $= -2(x - 5)^2 + 62$

When $x = 10$ , $y = 12 > 2$ The pellet will not hit the object.

Need help with quadratic functions & inequalities? See our Sec 4 A Maths tuition programme.

Frequently asked questions

Do these A Math questions include worked solutions?▾

Yes. Every question includes the worked solution for all the parts shown.

Which year are the questions from?▾

All the selected questions are from 2025 Singapore school Preliminary Examination papers.

How should I use this topic page?▾

Attempt a question fully on paper, then open the worked solution and compare your method and answer.

Sec 4 A Math: Quadratic Functions & Inequalities, practice questions & worked solutions

Key methods

Questions & worked solutions

Frequently asked questions

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