Read a derivative and a reciprocal from a shifted graph

(a) A curve y=f(x)y=f(x) is transformed by a stretch with scale factor 2 parallel to the yy-axis, followed by a reflection about the xx-axis. The equation of the resultant curve is y=ln⁡[1(x+a)2]y=\ln[\frac1{(x+a)^2}], where x>−ax>-a and a>1a>1. Find f(x)f(x) and sketch the graph of y=f(x)y=f(x). marks

(b) The diagram shows the graph of y=g(x+1)y=g(x+1). The graph passes through the origin and has a turning point A(3,4)A(3,4). The asymptotes of the graph are x=2x=2 and y=2x−4y=2x-4.

Q6(b) source y=g(x+1)xyOx=2y=2x−4OA(3,4)
Question diagram 2
Q6(b) source y=g(x+1)xyOx=2y=2x−4OA(3,4)

Source graph y=g(x+1)

On separate diagram, sketch the following graphs indicating the points corresponding to the axial intercepts, turning point and asymptotes, where necessary.

(i) y=g(x)y=g(x) marks

(ii) y=g′(x)y=g'(x) marks

(iii) y=1g(x)y=\frac1{g(x)} marks

[modified 2015/RVHS/Prelim/I/10]

Total marks: 12

Solution

(a)

Inverting transformations

A curve y=f(x)y=f(x) is transformed by a stretch with scale factor 2 parallel to the yy-axis, followed by a reflection in the xx-axis. The resulting curve has equation y=ln⁡[1(x+a)2]y=\ln\left[\dfrac{1}{(x+a)^2}\right], where x>−ax>-a and a>1a>1. Find f(x)f(x) and sketch y=f(x)y=f(x).

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Hint

  1. Apply the two transformations to y=f(x)y=f(x) in order and equate with the given curve.

  2. Simplify the logarithm: ln⁡1u2=−2ln⁡u\ln\frac1{u^2}=-2\ln u.

  3. Sketch ln⁡(x+a)\ln(x+a): asymptote at x=−ax=-a, intercepts at x=1−ax=1-a and y=ln⁡ay=\ln a.

Q6(a) logarithm
Q6(a) logarithmScroll across to see the full diagram.

Answer · (a)

f(x)=ln⁡(x+a)f(x)=\ln(x+a); intercepts (1−a,0),(0,ln⁡a)(1-a,0),(0,\ln a); asymptote x=−ax=-a.

x>−ax>-a, a>1.

Explore the idea

The logarithm and its key points

Drawn with a = 2: f(x) = ln(x + 2).

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Y = ln(x + a)
  • Asymptote x = −a
  • Intercepts
Ideas to try
Change the point of view: how would the graph look for a=5a=5? Which features move?
(b)i

Translations

The graph of y=g(x+1)y=g(x+1) passes through the origin and has a turning point A(3,4)A(3,4). Its asymptotes are x=2x=2 and y=2x−4y=2x-4. Sketch y=g(x)y=g(x), indicating the points corresponding to the axial intercepts and the turning point, and the asymptotes.

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Hint

  1. g(x+1)g(x+1) is g(x)g(x) shifted left by 1. To recover gg, shift the given graph right by 1.

  2. Apply the shift to every point and every asymptote.

Q6(b)(i) translated g
Q6(b)(i) translated gScroll across to see the full diagram.

Answer · (b)i

Intercept (1,0)(1,0), minimum (4,4)(4,4); asymptotes x=3,y=2x−6x=3,y=2x-6.

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Slide the given graph one unit right

The given graph is g(x+1). Sliding it right by 1 recovers g(x).

Teal: y = g(x + 1) (given). No comparison overlay.

Start: y = g(x + 1) (given)

The printed graph: through OO, turning point A(3,4)A(3,4), asymptotes x=2x=2 and y=2x−4y=2x-4.

  • Graph at the selected step
Ideas to try
Where does the origin go? Where does the vertical asymptote go?
(b)ii

Sketching the derivative

Sketch the graph of y=g′(x)y=g'(x), indicating the points corresponding to the axial intercepts and the turning point, and the asymptotes.

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Hint

  1. The derivative is the gradient: trace the gradient of gg as xx increases.

  2. A turning point of gg becomes a zero of g′g'.

  3. A vertical asymptote stays; a slanted asymptote of gradient 2 becomes a horizontal asymptote y=2y=2.

Q6(b)(ii) gradient source sketch
Q6(b)(ii) gradient source sketchScroll across to see the full diagram.

Answer · (b)ii

Zero (4,0)(4,0); asymptotes x=3,y=2x=3,y=2.

Positive for x<3 and x>4; negative for 3<x<4. Numerical y-intercept unspecified.

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From a graph to its gradient

The faint curve is g(x). The solid curve is its gradient g′(x).

Teal: y = g(x). No comparison overlay.

Start: y = g(x)

The graph from part (i).

  • Graph at the selected step
Ideas to try
  1. Where is gg decreasing? Is g′g' below the axis there?
  2. Why does g′g' approach 2 on both sides?
(b)iii

Reciprocal graphs

Sketch the graph of y=1g(x)y=\dfrac{1}{g(x)}, indicating the points corresponding to the turning point and the asymptotes.

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Hint

  1. Zeros of gg become vertical asymptotes of 1g\frac1g.

  2. The minimum (4,4)(4,4) becomes a maximum (4,14)(4,\frac14).

  3. Large ∣g∣|g| gives 1g→0\frac1g\to0, so y=0y=0 is a horizontal asymptote.

Q6(b)(iii) reciprocal g
Q6(b)(iii) reciprocal gScroll across to see the full diagram.

Answer · (b)iii

Maximum (4,1/4)(4,1/4); pole x=1x=1; hole (3,0)(3,0); horizontal asymptote y=0y=0.

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Take the reciprocal of g

Zeros of g become asymptotes; the minimum of g becomes a maximum.

Teal: y = g(x). No comparison overlay.

Start: y = g(x)

The graph from part (i).

  • Graph at the selected step
Ideas to try
Which branch of gg is negative? What does that say about the matching branch of 1/g1/g?