Progressive Graphs and functions practice questions
Use Clevolab for progressive 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 Progressive set covers coordinates, straight-line graphs, quadratic graphs, transformations, composite functions, and inverse functions. 234 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
If $f(x)=x+x$, what is $f(7)$?
- A7Answer option
- B12Answer option
- C14Correct answer
- D49Answer option
Why this answer is right
Add the input to itself: $7+7=14$.
The rule says to sum two copies of $x$. Substitute $x=7$. $$f(7)=7+7=14$$ This is not squaring; $7^2=49$ would be a different function.
Sample question
Find the equation of the line parallel to $2y-4x=7$ through $(1,3)$.
- A$y=-\tfrac{1}{2}x+1$Answer option
- B$y=2x-1$Answer option
- C$y=-2x+1$Answer option
- D$y=2x+1$Correct answer
Why this answer is right
Rewrite $2y-4x=7$ as $y=2x+\tfrac{7}{2}$, so slope is $2$. Through $(1,3)$: $3=2\cdot 1+b$ gives $b=1$, so $y=2x+1$.
Parallel lines share slope. From $2y-4x=7$, rearrange: $$2y=4x+7 \Rightarrow y=2x+\tfrac{7}{2}$$ So $m=2$. Use $y=2x+b$ with $(1,3)$: $$3=2\cdot 1+b \Rightarrow b=1$$ Therefore $$y=2x+1$$
Sample question
Find $f^{-1}(x)$ for $f(x)=3e^{2x}-5$. State it as a function of $x$.
- A$f^{-1}(x)=\tfrac{1}{2}\,\ln\!\big(\tfrac{x+5}{3}\big)$Correct answer
- B$f^{-1}(x)=\ln\!\big(\tfrac{x+5}{3}\big)$Answer option
- C$f^{-1}(x)=\tfrac{1}{2}\,\ln(x+5)-\ln 3$Answer option
- D$f^{-1}(x)=\tfrac{1}{2}\,\ln\!\big(\tfrac{x-5}{3}\big)$Answer option
Why this answer is right
Solve $y=3e^{2x}-5$: then $e^{2x}=(y+5)/3$, so $2x=\ln\!\big(\tfrac{y+5}{3}\big)$ and $x=\tfrac{1}{2}\,\ln\!\big(\tfrac{y+5}{3}\big)$. Replace $y$ by $x$.
Start from $y=3e^{2x}-5$. Rearrange: $e^{2x}=\dfrac{y+5}{3}$, then take natural logs: $2x=\ln\!\big(\dfrac{y+5}{3}\big)$. Divide by $2$ to isolate $x$: $$x=\tfrac{1}{2}\,\ln\!\big(\tfrac{y+5}{3}\big).$$ By definition, $f^{-1}(x)$ is this expression with $y$ replaced by $x$. Note the domain of $f^{-1}$ is $x>-5$, matching the range of $f$.
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.