Questions on graphs, proportion and transformations

Question 1

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.

Accepted 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.

Question 2

y is directly proportional to x², and y = 12 when x = 2. Find y when x = 5.

Accepted 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.

Question 3

On separate axes, sketch y = 6/x and y = −6/x². For each curve state the equations of its asymptotes and which quadrants the curve occupies.

Accepted Answer

For y = 6/x the sign of y is the sign of x, so one branch lies in the first quadrant and the other in the third, each approaching both axes. B1 for the shape in the right two quadrants. For y = −6/x² the square makes the denominator positive, so y takes the sign of −6 alone: both branches lie below the x-axis, in the third and fourth quadrants, symmetric about the y-axis. B1 for both branches below the axis. In each case the asymptotes are x = 0 and y = 0. B1 for stating them as equations for both curves. The curves approach the axes without meeting them, and the value at x = 0 does not exist, so neither curve has a y-intercept; a sketch that passes through the origin has no asymptote there.

Question 4

On the same axes, sketch y = 4/x and y = x² + 1. Use your sketch to state the number of real solutions of x³ + x = 4.

Accepted Answer

Dividing x³ + x = 4 by x, which is not zero at any solution, gives x² + 1 = 4/x, so the solutions are the intersections of the two curves. B1 for both curves sketched, the reciprocal in the first and third quadrants and the parabola with its vertex at (0, 1). In the first quadrant the parabola rises from 1 while the reciprocal falls from very large values towards zero, so they cross exactly once; in the third quadrant 4/x is negative and the parabola is positive, so there is no meeting. The equation has one real solution. B1. It lies between x = 1, where 4/x = 4 is above the parabola's 2, and x = 2, where 4/x = 2 is below the parabola's 5.

Question 5

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.

Accepted 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.

Question 6

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).

Accepted 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.

Question 7

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).

Accepted 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.

Question 8

The curve y = f(x), where f(x) = (x − 1)(x − 5), is a parabola with minimum point (3, −4). Sketch the curve y = 1/f(x), stating the equations of its asymptotes and the coordinates of its turning point.

Accepted Answer

The zeros of f become vertical asymptotes, so x = 1 and x = 5. B1. As |x| grows f is unbounded, so its reciprocal flattens onto y = 0, the horizontal asymptote. B1. The minimum (3, −4) of f, on a stretch where f is negative, becomes a maximum at the reciprocal height: (3, −¼). B1 for the turning point with its type. The sign is kept throughout: between x = 1 and x = 5 the parabola is negative, so the middle part of the reciprocal lies below the axis, rising to −¼ and dropping without bound towards each asymptote; outside that interval both parts are positive and tend to zero. B1 for a sketch with the three parts on the correct sides of the axis. The two curves meet where f is 1 or −1, which fixes the middle of the sketch. Placing the turning point at (3, 4), by inverting the height and dropping the sign, puts the middle part on the wrong side of the axis.

Question 9

The curve C has equation y = (2x + 3)/(x − 1). State the equations of the asymptotes of C, find the coordinates of the points where C meets the coordinate axes, and write the equation of C in the form y = a + b/(x − 1).

Accepted Answer

The denominator is zero at x = 1, so x = 1 is the vertical asymptote, and the leading coefficients are 2 and 1, so as |x| grows y tends to 2: the horizontal asymptote is y = 2. B1 for both. Setting y = 0 gives 2x + 3 = 0, so C crosses the x-axis at (−3/2, 0); setting x = 0 gives y = 3/(−1) = −3, so it crosses the y-axis at (0, −3). B1 for both crossings. Dividing out, 2x + 3 = 2(x − 1) + 5, so y = 2 + 5/(x − 1). M1 for the division, A1 for a = 2 and b = 5. The last form is y = 5/x translated 1 to the right and 2 up, and its two constants are the two asymptotes over again, so the division checks the diagnosis.

Question 10

The curve y = f(x) has a minimum point at (2, −3). Find the coordinates of the minimum point of the curve y = 3f(x − 1) + 2, and of the curve y = f(2x + 4).

Accepted Answer

For y = 3f(x − 1) + 2 the bracket moves the curve 1 to the right, then the 3 stretches heights by scale factor 3 and the 2 lifts it. The minimum goes to (2 + 1, 3(−3) + 2) = (3, −7). B1. For y = f(2x + 4) the minimum occurs where the inside equals 2, the original x-coordinate: 2x + 4 = 2 gives x = −1, and the height is unchanged, so the point is (−1, −3). M1 for setting 2x + 4 equal to 2, A1 for the point. Read f(2x + 4) as f(2(x + 2)): a horizontal squash by ½ then a shift 2 left lands on −1, and so does a shift 4 left followed by the squash, but squashing first and then shifting 4 left lands on −3, which is wrong. Tracking the one point through the bracket avoids the ordering question altogether.

Question 11

Sketch the curve y = x²(x − 2)(x + 2), showing where it meets the x-axis and the shape of the curve at each such point. Find the number of real solutions of x²(x − 2)(x + 2) = −1, justifying your answer.

Accepted Answer

The curve is a positive quartic, so it runs from top-left to top-right. The simple factors give crossings at x = −2 and x = 2, and the squared factor gives a touch at x = 0. B1 for the ends and the three axis meetings. Just either side of 0 the curve is about −4x², negative, so the touch at the origin is from below and the origin is a local maximum, with a dip on each side reaching down between the crossings. B1 for that shape. Expanding, y = x⁴ − 4x² = (x² − 2)² − 4, a square minus 4, so the lowest height on the curve is −4, reached at x = ±√2. M1 for completing the square in x². The line y = −1 lies between the floor of the two dips at −4 and the local maximum at 0, so it cuts each dip twice: the equation has four real solutions. A1. Solving directly confirms it, since x⁴ − 4x² + 1 = 0 gives x² = 2 ± √3, both positive, each giving two values of x. Reading the touch at 0 as a touch from above, a local minimum, would put the line y = −1 through only two points.

Question 12

The diagram shows the curve y = f(x), where f(x) is a cubic. Use the graph to write down the roots of f(x) = 0, stating which root is repeated and how the graph shows it. Use the graph to state the number of real solutions of f(x) = 2. Hence write down the coordinates of the maximum point of the curve y = f(x + 1) − 3.

Accepted Answer

The curve crosses the x-axis at x = −1, so x = −1 is a root. B1 for that root. It meets the x-axis again at x = 2 but turns there instead of crossing, and a touch means the factor is squared, so x = 2 is the repeated root. B1 for x = 2 named as the repeated one. The graph turns at (0, 4) and again at (2, 0), so f runs up to 4, back down to 0 and up again. The line y = 2 lies between those two turning values, so it cuts the rising branch to the left of the maximum, the falling branch between the turning points and the rising branch beyond the minimum: f(x) = 2 has 3 real solutions. B1 for 3. The curve y = f(x + 1) − 3 is y = f(x) translated 1 to the left and 3 down, so the maximum read off the graph, (0, 4), moves to (0 − 1, 4 − 3). M1 for the maximum point taken from the graph with both translations applied. The maximum point is (−1, 1). A1 for the coordinates.

Question 13

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.

Accepted 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 the answer is two, not none.

Question 14

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.

Accepted 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%.

Question 15

The curve C has equation y = (x + 4)/(x − 2), and the line L has equation y = 2x + 1. Find the coordinates of the points where L meets C. Find also the set of values of x for which (x + 4)/(x − 2) > 2x + 1.

Accepted Answer

At a meeting the two heights agree, and x ≠ 2 there, so multiply up: x + 4 = (2x + 1)(x − 2) = 2x² − 3x − 2. M1 for equating and clearing the fraction. That rearranges to 2x² − 4x − 6 = 0, or x² − 2x − 3 = (x − 3)(x + 1) = 0. A1 for the quadratic. So x = −1 with y = −1, and x = 3 with y = 7: the points are (−1, −1) and (3, 7). A1 for both. For the inequality, multiplying by x − 2 is unsafe because its sign is unknown, and the curve's vertical asymptote x = 2 is a third critical value alongside the two crossings. B1 for x = 2 as a critical value. Test each interval: at x = −2 the curve is −½ and the line is −3, so the curve is above; at x = 0 the curve is −2 and the line is 1, so it is below; at x = 2.5 the curve is 13 and the line is 6, above; at x = 4 the curve is 4 and the line is 9, below. M1 for testing the four intervals or reading them from a sketch. So x < −1 or 2 < x < 3. A1. On a sketch the curve is y = 6/x moved 2 right and 1 up, with asymptotes x = 2 and y = 1, and the line cuts the left branch once and the right branch once; the curve is above the line to the left of the first crossing, and again on the right branch between the asymptote and the second crossing. Treating the crossings as the only boundaries gives −1 < x < 3, which includes x = 0 where the curve is plainly below the line.

Question 16

The curve C has equation y = (x² + 3)/(x − 1). Without differentiating, find the set of values that y can take, and find the coordinates of the two stationary points of C.

Accepted Answer

A height k is reached exactly when the line y = k meets the curve, so set the expression equal to k and clear the fraction: x² + 3 = k(x − 1), which is x² − kx + (k + 3) = 0. M1 for setting up the quadratic in x. A real x needs a non-negative discriminant: k² − 4(k + 3) = k² − 4k − 12 = (k − 6)(k + 2) ≥ 0. A1 for the discriminant in factorised form. M1 for solving the quadratic inequality in k: an upward parabola in k, non-negative outside its roots, so y ≤ −2 or y ≥ 6. A1. No height strictly between −2 and 6 is reached, which is the gap between the two branches either side of x = 1. At each end of the range the discriminant is zero, so the line y = k is a tangent and the repeated root is the x-coordinate of a stationary point. At k = 6 the quadratic is x² − 6x + 9 = (x − 3)², giving (3, 6); at k = −2 it is x² + 2x + 1 = (x + 1)², giving (−1, −2). A1 for both points. Substituting back checks them: (9 + 3)/2 = 6 and (1 + 3)/(−2) = −2. The maximum on the left branch is lower than the minimum on the right, because the vertical asymptote separates the two branches.