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Graphs, proportion and transformations questions
Curve sketching starts by reading the equation rather than plotting points. Factors give the roots, a squared factor touches the axis instead of crossing it, and a reciprocal curve approaches its asymptotes without meeting them. Four standard transformations then turn one known graph into a whole family.
7 original questions · 28 marks · the graphs, proportion and transformations notes · Algebra and functions
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The curve C has equation y = (x − 1)(x + 2)2. State the coordinates of the points where C crosses or touches the x-axis, and the coordinates of its y-intercept.
Worked answer
The simple factor gives a crossing at (1, 0) and the squared factor gives a touch at (−2, 0). Putting x = 0 gives y = (0 − 1)(0 + 2)2 = −4, so the y-intercept is (0, −4). B1 B1 B1 for the three points. Squared factor means touch, simple factor means crossing, and the factorised form hands over the whole sketch without any expanding.y is directly proportional to x2, and y = 12 when x = 2. Find y when x = 5.
Worked answer
Write y = kx2. Then 12 = 4k gives k = 3, so y = 3x2 and at x = 5, y = 75. M1 for the statement with a constant in it, A1 for k = 3, A1 for 75. Writing y = 12x2 straight from the given pair is the standard error, and it comes from skipping the step where k is pinned down.y is inversely proportional to x, and y = 8 when x = 3. Find y when x = 6, and state the equations of the asymptotes of the graph of y against x.
Worked answer
Write y = k/x, so k = 8 × 3 = 24 and at x = 6, y = 4. Doubling x halves y, which is the check worth doing in your head. The graph is the reciprocal curve, with asymptotes x = 0 and y = 0, the two axes themselves. M1 A1 for the model and k, A1 for y = 4, B1 for both asymptotes. Both asymptotes belong in a complete answer.The curve y = f(x) passes through the point (4, 3). Find the coordinates of the corresponding point on y = 2f(x), and on y = f(2x).
Worked answer
Outside the bracket, the output doubles, so (4, 3) becomes (4, 6). Inside the bracket, 2x must equal 4, so x = 2 and the point becomes (2, 3). B1, then M1 A1. Outside stretches vertically by scale factor 2; inside squashes horizontally by scale factor ½. The inside move always runs backwards, which is the whole difficulty of the topic.The curve y = f(x) has a maximum point at (2, 5) and crosses the x-axis at (−1, 0) and (4, 0). Find the coordinates of the maximum point and of the x-axis crossings of the curve y = 2f(x + 3).
Worked answer
The x + 3 inside translates the graph 3 to the left, and the factor 2 outside stretches it vertically by scale factor 2. Every point (a, b) therefore moves to (a − 3, 2b). The maximum (2, 5) goes to (−1, 10). The crossings have b = 0, and doubling 0 changes nothing, so (−1, 0) goes to (−4, 0) and (4, 0) goes to (1, 0). B1 for the translation, B1 for the stretch, A1 for the maximum, A1 for both crossings. Shifting right instead of left is the error the mark scheme is watching for here.The curve C1 has equation y = 4/x. The curve C2 has equation y = 4/(x − 2) + 3. Describe fully the single transformation that maps C1 onto C2, write down the equations of the asymptotes of C2, and find the exact coordinates of the points where C2 crosses the coordinate axes.
Worked answer
Replacing x by x − 2 shifts the picture 2 to the right, and adding 3 lifts it 3 up, so the transformation is a translation by the vector (2, 3). The asymptotes travel with it, moving from x = 0 and y = 0 to x = 2 and y = 3. For the y-axis crossing put x = 0: y = 4/(−2) + 3 = 1, giving (0, 1). For the x-axis crossing put y = 0: 4/(x − 2) = −3, so x − 2 = −4/3 and x = 2/3, giving (2/3, 0). B1 for the translation, B1 B1 for the asymptotes, B1 for (0, 1), M1 A1 for (2/3, 0). The original curve never meets either axis at all; both crossings are created by the shift, so a candidate who answers 'none' has stopped thinking one step too early.The period T seconds of a pendulum is directly proportional to the square root of its length L cm. A pendulum of length 25 cm has period 1 second. Find the period of a pendulum of length 81 cm, and find the percentage increase in length needed to double the period of any pendulum.
Worked answer
Write T = k√L. Then 1 = k√25 = 5k, so k = 1/5 and T = √L/5. At L = 81, T = 9/5 = 1.8 seconds. To double T you must double √L, which means multiplying L by 4, an increase of 300%. M1 A1 for the model and k, A1 for 1.8, M1 for doubling the square root, A1 for 300%. Two traps sit in the last part. Doubling L only multiplies T by √2, and the answer wanted is the increase, 300%, not the final multiplier of 400%.
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