A-Level Graphs and functions exam-style questions
Use Clevolab for A-Level exam-style practice in graphs and functions. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.
About this topic
Clevolab treats graphs and functions as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.
The current A-Level set covers coordinates, straight-line graphs, quadratic graphs, transformations, composite functions, and inverse functions. 50 reviewed questions currently published for this page.
What you can practise
- Coordinates and graph interpretation
- Straight-line graphs and gradients
- Quadratic and non-linear graphs
- Function notation and evaluation
- Composite, inverse, and transformed functions
Real sample questions from the current set
These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.
Sample question
Define $f(x)=\begin{cases}kx+2,& x\le1\\ x^2+m,& x>1\end{cases}$. If $f$ is continuous and differentiable at $x=1$, find $(k,m)$.
- A$(k=1,\ m=2)$Answer option
- B$(k=3,\ m=4)$Answer option
- C$(k=2,\ m=3)$Correct answer
- D$(k=2,\ m=1)$Answer option
Why this answer is right
Continuity: $k\cdot1+2=1^2+m$ gives $m=k+1$. Differentiability: left gradient $k$ equals right gradient $2x$ at $x=1$, so $k=2$. Then $m=3$.
Continuity at $x=1$ requires $$k\cdot1+2=1^2+m\Rightarrow m=k+1.$$ Differentiability matches derivatives: left derivative is $k$; right derivative of $x^2+m$ at $1$ is $2$. Thus $$k=2,$$ and $$m=k+1=3.$$ Forgetting one of the two conditions gives the wrong pair.
Sample question
Given $A(1,2)$ and $B(9,6)$, find the intersection of the perpendicular bisector of $AB$ with the $y$-axis.
- A$(0,14)$Correct answer
- B$(0,10)$Answer option
- C$(0,4)$Answer option
- D$(0,-14)$Answer option
Why this answer is right
Midpoint is $(5,4)$. $m_{AB}=\tfrac{6-2}{9-1}=\tfrac{1}{2}$, so perpendicular slope is $-2$. Through $(5,4)$: $y-4=-2(x-5)$ gives $y=-2x+14$. At $x=0$, $y=14$.
Compute midpoint $$M=(5,4).$$ Gradient $$m_{AB}=\frac{6-2}{9-1}=\tfrac{1}{2},$$ so the perpendicular has slope $-2$. Equation: $$y-4=-2(x-5)\Rightarrow y=-2x+14.$$ Intersect the $y$-axis by setting $x=0$ to get $(0,14)$.
Sample question
A transformation $g(x)=f(x-a)+b$ maps the point $(4,-1)$ on $y=f(x)$ to $(7,5)$ on $y=g(x)$. Find $(a,b)$.
- A$(a=3,\ b=6)$Correct answer
- B$(a=-3,\ b=6)$Answer option
- C$(a=3,\ b=-6)$Answer option
- D$(a=-3,\ b=-6)$Answer option
Why this answer is right
For $y=f(x-a)+b$, the input to $f$ is $x-a$. To map input $4$, set $7-a=4$, so $a=3$. The output shifts by $+b$: $-1\mapsto5$ gives $b=6$.
Match the inside: $$x-a=4\text{ at }x=7\Rightarrow a=3.$$ The $y$-value is shifted by $+b$, so $$-1+b=5\Rightarrow b=6.$$ Confusing the sign on $a$ or shifting the wrong way yields the distractors.
How this page fits into Clevolab
Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.