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Suppose I wanted to sketch the graph $y=\sqrt{5x-10}$ for $5x-10 \ge0$

Is there a direct method?

I know that I can define first $g(x)=\sqrt{x}$ and consider $y=g(5x)$ or $y=\sqrt{5x}$ This is a stretch scale factor $\frac{1}{5}$ parallel to the $x$ axis.

I define a new function $h(x)=\sqrt{5x}$ then consider $y=h(x-2)$, a translation by the vector $[2,0]^T$

But is there a direct method? I'm thinking along the lines of $y=f(5(x-2))$ but I couldn't work it.

I'd like to do it more directly because my students are struggling when they do it in steps. The topic arose when finding the range of a function and sketching all the intermediate graphs confuses them.

I had searched the site and the web but couldn't find anything. Hopefully I've not missed a duplicate question as I'm surprised it hadn't been asked previously.

Thanks for your help in advance.

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  • $\begingroup$ Don't you mean $5x-10=5(x-2)$ rather than $5x-10=5(x-\tfrac 25)$? $\endgroup$ – String Sep 22 '15 at 15:14
  • $\begingroup$ Yes, sorry edited, $\endgroup$ – Karl Sep 22 '15 at 15:18
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We have $$ f(ax+b)=f(a(x+\tfrac ba)) $$ corresponding to a stretch by a factor $\frac 1a$ followed by a horisontal translation by $-\frac ba$. You simply invert the operations, so multiplied by $a$ becomes stretched by $\frac 1a$ and adding $\frac ba$ becomes shifted by $-\frac ba$. So in steps:

  1. Given $f(ax+b)$ write down $a,\frac ba$.
  2. Invert both in turn to have $\frac 1a,-\frac ba$.
  3. Conclude that the graph of $f$ have been stretched by a factor $\frac 1a$ followed by a horisontal shift by $-\frac ba$.
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  • $\begingroup$ Ok thanks. I think I may have been doing the transformations in step 3 in the reverse order to yours. I guess they're not commutative and hence my error? $\endgroup$ – Karl Sep 22 '15 at 15:29
  • $\begingroup$ @Karl: No they are certainly not commutative, since the stretch is symmetrical around the $y$-axis, so if the graph have been shifted horisontally the axis of symmetry of the stretch has so too :o) $\endgroup$ – String Sep 22 '15 at 15:32
  • $\begingroup$ That sorts it then. Thanks! $\endgroup$ – Karl Sep 22 '15 at 15:33
  • $\begingroup$ @Karl: No problem. All the best! $\endgroup$ – String Sep 22 '15 at 15:35

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