Pure: Functions, Graphs and Transformations
Functions, graphs and transformations covers sketching polynomial and reciprocal graphs from their equation, and the effect of each single transformation (a translation, a stretch or a reflection, in either direction) on a graph's equation, its key points, its intercepts and its asymptotes. It also covers the modulus graphs y = |f(x)| and y = f(|x|), which are each built from y = f(x) by reflecting only part of it. Every one of these questions can be answered from a sketch or a short list of given features, without ever being shown the full equation of the curve.
Method
- To sketch a polynomial given in factorised form, such as y = (x-a)(x-b)(x-c), read off the x-intercepts directly as x = a, b and c, find the y-intercept by substituting x = 0, and determine the end behaviour (as x tends to +infinity and -infinity) from the sign and the parity (odd or even) of the highest power of x.
- The reciprocal graphs y = 1/x^n both have asymptotes x = 0 and y = 0; for odd n (such as y = 1/x) the two branches sit in opposite quadrants with rotational symmetry about the origin, while for even n (such as y = 1/x^2) both branches lie above the x-axis with reflective symmetry in the y-axis.
- For a direct or inverse proportion relationship (y = kx^n or y = k/x^n), find the constant k by substituting one known pair of values, then use that equation for any further calculation; the graph is a straight line through the origin only for y = kx (n=1).
- Learn what each single transformation does to the GRAPH of y = f(x): y = f(x)+a translates it by (0, a); y = f(x+a) translates it by (-a, 0); y = af(x) stretches it vertically by scale factor a; y = f(ax) stretches it horizontally by scale factor 1/a; y = -f(x) reflects it in the x-axis; y = f(-x) reflects it in the y-axis.
- To find where a specific point (p, q) on y = f(x) moves to under a transformation, apply the same rule to p that the transformation applies to x: for y = cf(ax+b)+d, the point maps to ((p-b)/a, cq+d).
- To combine several transformations given in words (such as a stretch, scale factor 1/2, parallel to the x-axis, followed by a translation of 3 units in the positive y-direction), apply them to the equation, and to each given point, in the ORDER stated, one at a time.
- A stationary point's nature (maximum or minimum) is unchanged by every translation and by a stretch in any direction; only a transformation that flips the graph vertically (y = -f(x)) swaps every maximum for a minimum and every minimum for a maximum.
- To sketch y = |f(x)|, keep any part of y = f(x) already on or above the x-axis unchanged, and reflect in the x-axis any part that is below it, so the whole graph becomes non-negative and touches (rather than crosses) the x-axis at each original x-intercept.
- To sketch y = f(|x|), keep only the part of y = f(x) for x >= 0, discard whatever the original graph did for x < 0, and reflect the kept part in the y-axis to create a new part for x < 0; the result is always symmetric about the y-axis, whatever the original function.
Worked example
The curve y = f(x) has a maximum point at (2, 5), and passes through the point (0, 1). It also has a horizontal asymptote y = -3. Find (a) the coordinates of the maximum point, (b) the coordinates of the image of (0, 1), and (c) the equation of the horizontal asymptote, on the curve y = 2f(x - 4) + 1.
- The transformation y = 2f(x-4)+1 combines a horizontal translation of +4 (replacing x with x-4 shifts the graph 4 units to the right), a vertical stretch of scale factor 2, and a vertical translation of +1; a point (p, q) on y=f(x) maps to (p+4, 2q+1).
- Maximum point (2, 5) maps to (2+4, 2(5)+1) = (6, 11); the point remains a maximum, since a positive vertical stretch and a translation both preserve the nature of a stationary point.
- Point (0, 1) maps to (0+4, 2(1)+1) = (4, 3).
- A horizontal asymptote is a y-value the curve approaches as x tends to +-infinity; the horizontal translation does not change WHICH y-value is approached, only where it is approached, so only the vertical stretch and translation affect it: y = -3 maps to y = 2(-3)+1 = -5.
- Final answer: maximum point (6, 11); image of (0,1) is (4, 3); new asymptote y = -5.
Practice questions
Try each question, then tap to reveal the answer.
Q1The graph of y = f(x) is translated by the vector (0, -6) to give the graph of y = g(x). Write down the equation of g(x) in terms of f(x).Show answer
Answer: g(x) = f(x) - 6.
Q2The graph of y = f(x) has a minimum point at (3, -2). State the coordinates of the minimum point on the graph of y = f(x + 5).Show answer
Answer: (-2, -2) (a translation of (-5, 0) moves the point 5 units to the left: 3-5=-2; the y-coordinate is unchanged).
Q3The graph of y = f(x) passes through the point (6, 10). Find the coordinates of the corresponding point on the graph of y = f(2x).Show answer
Answer: (3, 10) (for y=f(ax), a point (p,q) maps to (p/a, q); here (6,10) maps to (6/2, 10) = (3,10)).
Q4Describe fully the single transformation that maps the graph of y = x^3 onto the graph of y = -x^3.Show answer
Answer: A reflection in the x-axis.
Q5The graph of y = f(x) has range -5 <= y <= 7. State the range of y = 2f(x) - 1.Show answer
Answer: -11 <= y <= 13 (each y-value maps to 2y-1: 2(-5)-1=-11 and 2(7)-1=13, and the order is preserved since the stretch factor 2 is positive).
Q6Sketch y = |2x - 6| by stating its minimum point and its y-intercept.Show answer
Answer: Minimum point (3, 0), where 2x-6=0 (the graph touches the x-axis here, since |2x-6| >= 0 everywhere); y-intercept at x=0: |2(0)-6| = |-6| = 6, so (0, 6).
Q7The curve y = (x+2)(x-1)(x-5) crosses the x-axis at three points and the y-axis at one point. State the coordinates of all four points.Show answer
Answer: (-2, 0), (1, 0), (5, 0) and (0, 10) (the y-intercept is (2)(-1)(-5) = 10 at x=0).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
The curve y = f(x) has x-intercepts at x = -2 and x = 5 only, and a maximum point at (1.5, 12). (a) State the x-intercepts and the coordinates of the maximum point on the curve y = 3f(x). (3) (b) State the x-intercepts of the curve y = f(x - 4). (2) (c) Explain why the x-intercepts of the curve y = f(x) + 6 cannot be found from the information given. (1)
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The graph of y = f(x), where f(x) = 2x - 4, is defined for -1 <= x <= 5. It is a straight line from (-1, -6) to (5, 6), crossing the x-axis at (2, 0). (a) State the coordinates of the point where the graph of y = |f(x)| meets the x-axis, and explain why the graph touches the x-axis there rather than crossing it. (2) (b) Find the coordinates of the y-intercept of the graph of y = |f(x)|. (2) (c) State the range of y = |f(x)| on this domain. (2)
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The graph of y = f(x) for x >= 0 passes through (0, 3), has a minimum point at (2, -1), and continues increasing for x > 2, passing through (5, 4). No information is given about f(x) for x < 0. (a) State the coordinates of the minimum point (or points) on the graph of y = f(|x|). (2) (b) State the coordinates of the point where y = f(|x|) crosses the y-axis. (1)
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