GCSE Graphs and functions exam-style questions
Use Clevolab for GCSE 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 GCSE 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
What is the relationship between the lines $y=2x+3$ and $2y-4x=1$?
- AParallelCorrect answer
- BPerpendicularAnswer option
- CThe same lineAnswer option
- DNeither parallel nor perpendicularAnswer option
Why this answer is right
Rewrite $2y-4x=1$ as $y=2x+\tfrac{1}{2}$. Both have gradient $2$ but different intercepts, so they are parallel.
Compare gradients. Rearrange $2y-4x=1$ to slope–intercept form: $$2y=4x+1 \quad\Rightarrow\quad y=2x+\tfrac{1}{2}$$ Both lines have gradient $2$. Different intercepts mean distinct, so they are parallel and do not coincide.
Sample question
What is the minimum value of $y=(x-4)^2+1$?
- A$0$Answer option
- B$1$Correct answer
- C$-1$Answer option
- D$4$Answer option
Why this answer is right
A square is always non-negative. $(x-4)^2\ge 0$, so $y\ge 1$, with equality at $x=4$. Minimum is $1$.
Since $(x-4)^2\ge 0$ for all $x$, adding $1$ gives $$y=(x-4)^2+1\ge 1$$ The minimum occurs when the square is $0$, at $x=4$. Therefore the minimum value is $1$.
Sample question
Find the equation of the line perpendicular to $4y+x=7$ that passes through $(0,-1)$.
- A$y=-\tfrac{1}{4}x-1$Answer option
- B$y=\tfrac{1}{4}x-1$Answer option
- C$y=4x-1$Correct answer
- D$y=-4x-1$Answer option
Why this answer is right
Rewrite $4y+x=7$ as $y=-\tfrac{1}{4}x+\tfrac{7}{4}$, so perpendicular gradient is $4$. Through $(0,-1)$ gives $y=4x-1$.
First find the gradient of the given line. $$4y+x=7\Rightarrow y=-\tfrac{1}{4}x+\tfrac{7}{4}$$ Perpendicular lines have gradients whose product is $-1$, so the perpendicular gradient is $4$. Through $(0,-1)$, the equation is $$y=4x-1$$
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.