STEP functions and curve sketching: guide and practice problems
Many STEP questions ask you to sketch a curve and then use the sketch to count solutions. A good sketch is a proof tool, not decoration.
Key ideas
- Check symmetry first: odd and even functions halve the work.
- For a rational function, find the asymptotes, the intercepts and the range (treat y = f(x) as a quadratic in x and use the discriminant).
- To count the solutions of f(x) = k, draw y = f(x) and slide the horizontal line y = k.
- Compositions: work out f(f(x)) explicitly; a short cycle (f applied n times gives x) makes large powers easy.
Common traps
- Forgetting the case where the 'quadratic in x' has leading coefficient 0.
- Sketches with no coordinates of turning points or intercepts.
- Assuming a function has an inverse without checking it is one-to-one.
Practice problems
Original problems in the style of STEP. Try each one for at least 20 minutes before you open a hint; write a full solution, then compare.
Problem 1: Range of a rational function
Let \(y=\dfrac{(x-1)(x-2)}{x^2+x+1}\).
- Show that the denominator is positive for all real \(x\), and state the horizontal asymptote.
- By treating the equation as a quadratic in \(x\), show that \(y\) can take exactly the values in the interval \(\alpha\le y\le\beta\), and find \(\alpha\) and \(\beta\) in surd form.
- Find where the curve crosses its asymptote, and sketch it.
Hint 1
Complete the square: \(x^2+x+1=(x+\tfrac12)^2+\tfrac34\).
Hint 2
Rearrange to \((y-1)x^2+(y+3)x+(y-2)=0\). For a real \(x\) to exist the discriminant must be non-negative (take care with \(y=1\)).
Full solution
(i) \(x^2+x+1=(x+\tfrac12)^2+\tfrac34>0\). As \(x\to\pm\infty\), \(y\to1\): the asymptote is \(y=1\).
(ii) \(y(x^2+x+1)=x^2-3x+2\) gives \((y-1)x^2+(y+3)x+(y-2)=0\). If \(y\ne1\) a real \(x\) exists iff \((y+3)^2-4(y-1)(y-2)\ge0\), that is \(-3y^2+18y+1\ge0\), or \(3y^2-18y-1\le0\). The roots of \(3y^2-18y-1=0\) are \(y=3\pm\tfrac{2\sqrt{21}}3\). When \(y=1\) the equation is \(4x-1=0\), which has a solution, and \(1\) lies in the interval anyway. So \(\alpha=3-\tfrac{2\sqrt{21}}3\approx-0.055\) and \(\beta=3+\tfrac{2\sqrt{21}}3\approx6.055\).
(iii) \(y=1\) at \(x=\tfrac14\) only. The curve crosses the \(x\)-axis at \(1\) and \(2\), has its minimum \(\alpha\) between them and its maximum \(\beta\) at negative \(x\), and approaches \(y=1\) at both ends.
Results: \(\alpha=3-\frac{2\sqrt{21}}{3}\), \(\beta=3+\frac{2\sqrt{21}}{3}\); crosses \(y=1\) at \(x=\frac14\).
2 more functions and curve sketching problems
How many roots, for each k; A function of order four. Each with two hints and a full solution.
Next steps
Related topics: Integration · Algebra and inequalities. Stuck? Read the problem-solving strategies. Ready for the real thing? Official STEP past papers.
STEP is run by OCR. A Level Math Revision is independent: it is not affiliated with or endorsed by OCR, the University of Cambridge or any university. Every problem here is our own, written in the style of STEP; for real papers use OCR's official past papers.