Map three points through three transformations

The diagram above shows the graph of y=f(x)y=f(x).

Q1 original y=f(x)xyOx=2y=1A(4,3)B(1,−1)C(3,0)
Question diagram 2
Q1 original y=f(x)xyOx=2y=1A(4,3)B(1,−1)C(3,0)

Source graph: A(4,3), B(1,-1), C(3,0), asymptotes x=2, y=1

On separate diagrams, sketch the graphs of

(i) y=2f(x)+1y=2f(x)+1, marks

(ii) y=f(∣x∣)y=f(|x|), marks

(iii) y=f(1−2x)y=f(1-2x). marks

showing in each case, the coordinates of the points corresponding to A,B,CA,B,C and the equations of the asymptotes.

[2010/SRJC/Prelim/II/3]

Total marks: 8

Solution

(i)

Scaling and translation

Sketch the graph of y=2f(x)+1y=2f(x)+1, showing the coordinates of the points corresponding to AA, BB and CC and the equations of the asymptotes.

Explore this idea

Hint

  1. Which operation acts on yy first: the “2×” or the “+1”?

  2. Apply the same two operations to the yy-coordinate of every point, and to the equation of every horizontal asymptote.

  3. What happens to a vertical asymptote when only yy changes?

Q1(i) y=2f(x)+1
Q1(i) y=2f(x)+1Scroll across to see the full diagram.

Answer · (i)

A′=(4,7),B′=(1,−1),C′=(3,1)A'=(4,7),B'=(1,-1),C'=(3,1); asymptotes x=2,y=3x=2,y=3.

Explore the idea

Scale, then translate

Step through the two transformations. The faint curve is where the graph was before the step.

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

Start: y = f(x)

The printed graph, with A(4,3)A(4,3), B(1,−1)B(1,-1), C(3,0)C(3,0) and asymptotes x=2x=2, y=1y=1.

  • Graph after the step you have selected
  • Images of the key points A, B and C
  • Vertical and horizontal asymptotes of the current graph
Ideas to try
  1. Look at point C. It lies on the xx-axis, so doubling does nothing to it. Then +1 lifts it to y=1y=1.
  2. Which coordinates change at each step?
(ii)

Modulus of the input

Sketch the graph of y=f(∣x∣)y=f(|x|), showing the coordinates of the points corresponding to AA, BB and CC and the equations of the asymptotes.

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Hint

  1. f(∣x∣)f(|x|) only ever feeds non-negative numbers into ff. What does that say about the graph for x<0x<0?

  2. Keep the part of the graph with x≥0x\ge0, then mirror it in the yy-axis.

  3. Each asymptote with x>0x>0 gets a mirror image.

Q1(ii) y=f(|x|)
Q1(ii) y=f(|x|)Scroll across to see the full diagram.

Answer · (ii)

A′=(±4,3),B′=(±1,−1),C′=(±3,0)A'=(\pm4,3),B'=(\pm1,-1),C'=(\pm3,0); asymptotes x=±2,y=1x=\pm2,y=1.

Sharp point at the origin.

Explore the idea

Keep the right, mirror it onto the left

Replacing x by |x| makes the graph symmetric about the y-axis.

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

Start: y = f(x)

The printed graph.

  • Graph after the step you have selected
  • Images of the key points A, B and C
  • Vertical and horizontal asymptotes of the current graph
Ideas to try
  1. Compare the two graphs for x>0x>0. Are they the same?
  2. Look near the origin. Why is there a sharp corner?
(iii)

Combined transformations

Sketch the graph of y=f(1−2x)y=f(1-2x), showing the coordinates of the points corresponding to AA, BB and CC and the equations of the asymptotes.

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Hint

  1. Rewrite 1−2x1-2x as −(2x−1)-(2x-1) or find a chain of three simple steps: translate, scale, reflect.

  2. For transformations of xx, the effect is the inverse of what the sign suggests. Work with a point to check.

  3. Track AA, BB, CC and the vertical asymptote through each step.

Q1(iii) y=f(1−2x)
Q1(iii) y=f(1−2x)Scroll across to see the full diagram.

Answer · (iii)

A′=(−3/2,3),B′=(0,−1),C′=(−1,0)A'=(-3/2,3),B'=(0,-1),C'=(-1,0); asymptotes x=−1/2,y=1x=-1/2,y=1.

The short-key positive B ordinate is a typo.

Explore the idea

Translate, squash, flip

Three steps turn f(x) into f(1 − 2x). Use the arrows to move one step at a time and watch point A.

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

Start: y = f(x)

The printed graph.

  • Graph after the step you have selected
  • Images of the key points A, B and C
  • Vertical and horizontal asymptotes of the current graph
Ideas to try
  1. Follow point A through all four stages: (4,3), (3,3), (3/2,3), (−3/2,3).
  2. Which step does not change the yy-coordinates at all?
Takeaway

The one-line check

For y=f(1−2x)y=f(1-2x), a point (a,b)(a,b) on the original graph moves to (1−a2, b)\left(\frac{1-a}{2},\,b\right). For AA: 1−42=−32\frac{1-4}{2}=-\frac32 ✓. If your sketch disagrees with this formula, find the step where it went wrong.