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Given that $f(x) = \ln x, \ x > 0$ sketch on separate axes the graphs of (i) $y = f(x)$, (ii) $y = |f(x)|$, (iii) $y = -f(x - 4)$ - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 7

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Given-that--$f(x)-=-\ln-x,-\-x->-0$--sketch-on-separate-axes-the-graphs-of--(i)-$y-=-f(x)$,--(ii)-$y-=-|f(x)|$,--(iii)-$y-=--f(x---4)$-Edexcel-A-Level Maths Pure-Question 4-2013-Paper 7.png

Given that $f(x) = \ln x, \ x > 0$ sketch on separate axes the graphs of (i) $y = f(x)$, (ii) $y = |f(x)|$, (iii) $y = -f(x - 4)$. Show, on each diagram, the p... show full transcript

Worked Solution & Example Answer:Given that $f(x) = \ln x, \ x > 0$ sketch on separate axes the graphs of (i) $y = f(x)$, (ii) $y = |f(x)|$, (iii) $y = -f(x - 4)$ - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 7

Step 1

(i) $y = f(x)$

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Answer

To sketch the graph of y=f(x)=lnxy = f(x) = \ln x:

  • The graph passes through the point (1, 0) because ln(1)=0\ln(1) = 0.
  • As xx approaches 0 from the right, lnx\ln x approaches -\infty, indicating a vertical asymptote at the line x=0x = 0.
  • The graph is increasing and concave down throughout its domain x>0x > 0.

Asymptote: The equation of the asymptote is x=0x = 0.

Step 2

(ii) $y = |f(x)|$

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Answer

To sketch the graph of y=f(x)=lnxy = |f(x)| = |\ln x|:

  • For x>1x > 1, the graph mirrors the original graph of y=lnxy = \ln x.
  • For 0<x<10 < x < 1, the graph will reflect the negative part of lnx\ln x, turning it positive.
  • The graph meets the point (1, 0) and has a vertical asymptote at x=0x = 0.

Asymptote: The equation of the asymptote is x=0x = 0.

Step 3

(iii) $y = -f(x - 4)$

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Answer

To sketch the graph of y=f(x4)=ln(x4)y = -f(x - 4) = -\ln(x - 4):

  • The graph shifts to the right by 4 units, changing the vertical asymptote to x=4x = 4.
  • The graph will now decrease from 00 at (5,0)(5, 0), heading towards -\infty as xx approaches 4 from the right.

Asymptote: The equation of the asymptote is x=4x = 4.

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