Transforming a function by a sequence geometric operations on its graph.

497 Views Asked by At

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?

3

There are 3 best solutions below

0
On BEST ANSWER

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:

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

3
On

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.

0
On

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.).