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I know that is a simple question but I would like to know how to draw the graph of $|1-3x|$ by using transformations?

What do I do is simply draw $|x|$, then $|3x|$ and $|-3x|$ which are the same and then I shift the graph by $1$ unit to the left of $y-$axis which is wrong. We must shift $\cfrac{1}{3}$ to the right of $y-$axis but why?

I am using the following 'rule':

'For some $a \ge 0$ and a graph of a function $f$, if you want to draw f(x-a) then shift the graph to the right by $a$ units and if you want to draw $f(x+a)$, shift the graph to the left by $a$ units.'

3 Answers 3

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Your issue is that the order of applying transformations is not always commutative.

For example, let's carefully investigate your steps.

You start with the parent function $|x|$.

To apply the horizontal compression, we substitute $3x$ in place of $x$. So we have $|(3x)|=|3x|$.

Next you apply the horizontal reflection. We replace $x$ with $-x$, and this gives us $|3(-x)|=|-3x|$.

Notice so far, if we chose to apply these two transformations in the opposite order, we would get the same result.

Now, the trouble: you want to horizontally translate the graph one unit to the left. We must replace $x$ with $x+1$, which gives us $|-3(x+1)|=|-3x-3|$, and this is not what we want.

I recommend that you factor right at the beginning, getting $|-3(x-1/3)|$, and then apply the transformations.

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    What do you mean by "factor right at the beginning"? – 2017-02-17
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    Before you try to determine the horizontal translation, I recommend that you factor the leading coefficient. Once you factor, you shouldn't have trouble applying the transformations for graphing. – 2017-02-17
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Approach the problem from the other direction and you'll see your mistake. If you take your original function that you want to shift, $f(x) = | - 3x |$, then the function you get by shifting it left by one unit would be $f(x + 1) = | - 3(x + 1) | = | -3 -3x| \neq |1 - 3x|$ as you seem to think.

We can get the "shift" of the function you gave by seeing that $f(x - 1/3) = | - 3 (x - 1/3) | = | 1 - 3x |$. So what you're actually doing is plotting $f(x - 1/3)$, which by your rule mean shift the graph to the right by $\tfrac{1}{3}$.

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It is so because $1-3x=0$ when $x=1/3$.

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    I know, but there is a "rule" that we tell the students: 'For some $a \ge 0$ and a graph of a function $f$, if you want to draw f(x-a) then shift the graph to the right by $a$ units and if you want to draw $f(x+a)$, shift the graph to the left by $a$ units.' – 2017-02-16
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    The function from your problem is not of the form $f(x+a)$. It is $f(ax-b)$ Thus you take $ax-b=0$ and solve. Which in this particular instance, $-3x+1=0$ gives us $x=1/3$ – 2017-02-16
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    Has this cleared it up for you? – 2017-02-16