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I am solving the following problem:

Let $f(x) =\sqrt{x}$. Find a formula for a function $g$ whose graph is obtained from $f$ from the given sequence of transformations:

  • shift right $3$ units
  • horizontal shrink by a factor of $2$
  • shift up $1$ unit

I think that$ g(x) = f(2(x-3)) + 1 = \sqrt{2x-6} + 1$, but in the answers it says $\sqrt{2x-3} + 1$, so i assume $g(x) = f(2x-3) +1$, but wouldn't that mean that the horizontal shrink was done first and afterwards the right horizontal shift?

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    $\begingroup$ With these kinds of problems, in general, an effective way of verifying which answer is correct is to simply consider a specific ordered pair (as I do in my answer). Doing this will give you assurance in regards to which function is correct--then the matter becomes pinpointing your flaw in reasoning (which is what OriginalOldMan did). $\endgroup$ – Daniel W. Farlow Mar 31 '15 at 23:25
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When we do a horizontal shrink by a factor of $b$ we replace $x$ with $bx$, rather than multiplying the whole expression by b. So:

$$g(x) = f(2x-3) + 1$$

not:

$$g(x) = f(2(x-3)) + 1$$

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The ambiguity is in the shrink step. The origin of the shrink has to be specified - points to the right of this origin are moved to the left towards the origin and points to the left of the origin are moved to the right towards the origin.

I believe that you are shrinking around 3, and the answer is shrinking around zero.

Since the shrinking origin is not specified, I believe that your answer would be acceptable.

But then, so would the other.

Ask your teacher.

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  • $\begingroup$ Does the order of the sequence of transformations give any insight into where the origin of the shrink would make the most sense? $\endgroup$ – Jonny Mar 31 '15 at 23:31
  • $\begingroup$ @Jonny I don't think there is any ambiguity, especially when you consider the transformations of a single ordered pair as I do in my answer. Probably the clearest or least ambiguous way of looking at it. $\endgroup$ – Daniel W. Farlow Mar 31 '15 at 23:37
  • $\begingroup$ @crash The fact that you apply the transformations in the same order as the sequence results in an unambiguous determination of the origin. However, couldn't "shift right by 3 units" also include shifting the origin right by three units? $\endgroup$ – Jonny Mar 31 '15 at 23:47
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There is actually a really easy way to test which function is correct. First, for $f(x)=\sqrt{x}$, you know you have the ordered pair $(4,2)$.

Now consider the transformations to obtain $g(x)$:

  1. Shift right by $3$ units.
  2. Horizontal shrink by a factor of $2$.
  3. Shift up by $1$ unit.

Now consider what happens to the ordered pair $(4,2)$ when you apply 1-3 above: $$ (4,2)\overset{\text{(1)}}{\implies} (7,2)\overset{\text{(2)}}{\implies} (7/2,2)\overset{\text{(3)}}{\implies} (7/2,3). $$ Thus, we know $g(x)$ must have the ordered pair $(7/2,3)$.

Trying out $g(x) = \sqrt{2x-3}+1$ when $x=7/2$ yields $(7/2,3)$, as desired.

Trying out $g(x) = \sqrt{2x-6}+1$ when $x=7/2$ yields $(7/2,2)\neq (7/2,3)$.

Thus, we clearly must have that $g(x) = \sqrt{2x-3}+1$.

Note: The answer provided by OriginalOldMan highlights the important point here, but I provided my answer for concrete verification, and this way is probably the easiest way to go when in doubt concerning the validity of your transformations (i.e., consider a particular ordered pair, etc.).

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