Function Transformations - Algebra 3

1. Starting with the graph of $y=f(x)$, compress it horizontally by a factor of $\frac{1}{2}$, then shift the result 2 units to the right. Which equation describes the resulting graph?
   Answer: $y=f(2x-4)$ Whenever you see horizontal compressions or shifts, remember that all horizontal changes happen inside $f$'s parentheses—and they work in the opposite direction from what you might expect: compress by $\frac12$ means multiply $x$ by $2$, and shift right $2$ means replace $x$ by $x-2$. Start with $y=f(x)$. Compress horizontally by a factor of $\frac12$: inside becomes $2x$, so you get $y=f(2x)$. Now shift that graph 2 units to the right: replace every $x$ in $f(2x)$ with $x-2$. That gives $$y=f(2(x-2))=f(2x-4).$$ That is the correct choice. The choice $y=f(2x-2)$ is a common trap: it factors as $f(2(x-1))$, so it compresses by $\frac12$ but then shifts right only 1 unit, not 2. The choice $y=f(2x)-2$ correctly compresses, but the $-2$ is outside the function, making it a vertical shift down 2, not a horizontal shift right. The choice $y=f(x-2)$ shifts the graph 2 units right but never applies the horizontal compression. A good check: after compressing, factor out the coefficient before reading the shift. In $f(2x-4)=f(2(x-2))$, the shift right is clearly 2. Horizontal transformations are inside the parentheses; vertical shifts are outside. Keep those two worlds separate.
2. Suppose $f(3)=7$. Let $g(x)=-2f(x-1)+5$. Based only on this information, which point must lie on the graph of $g$?
   Answer: $(4,-9)$ This question tests function transformations: horizontal shifts happen inside the parentheses, while vertical stretches and shifts happen outside. When you see a "must lie on the graph" question with a known point like $f(3)=7$, your goal is to find the $x$-value that makes the inside expression equal to $3$. Here, set $x-1=3$, so $x=4$. Then $g(4)=-2f(3)+5=-2(7)+5=-14+5=-9$. Therefore, the point $(4,-9)$ must lie on the graph of $g$. The choice $(2,-9)$ comes from incorrectly treating $x-1$ as the input value itself, or shifting in the wrong direction. The choice $(3,-9)$ uses $3$ as the new input, but $g(3)$ would require $f(2)$, which is unknown. The choice $(4,19)$ correctly finds $x=4$ but makes a sign error: it computes $+14+5$ instead of $-14+5$. In general, be careful that multiplying by $-2$ changes both the magnitude and the sign. A strong strategy is to start from the known point $(3,7)$ and apply the transformations in order: shift right by $1$ to get $(4,7)$, reflect and stretch vertically by $-2$ to get $(4,-14)$, then shift up $5$ to get $(4,-9)$. Matching "inside equals known input" first will always locate the guaranteed point.
3. A student claims that the graph of $g(x)=f(3-x)$ can be obtained by shifting the graph of $f$ right 3 units and then reflecting across the $y$-axis. Which statement is true?
   Answer: Incorrect; the correct sequence is to reflect across the $y$-axis, then shift right 3. Since $3-x=-(x-3)$, $f(3-x)$ is obtained by reflecting $f$ across the $y$-axis and then shifting right 3. Shifting right 3 first and then reflecting gives $f(-x-3)$, not $f(3-x)$. Reflecting across the $x$-axis changes outputs, not inputs.
4. The zeros of a function $f$ are $-3$ and $5$. What are the zeros of $g(x)=f(2x-4)$?
   Answer: $\{0.5,4.5\}$ To keep a zero, the input to $f$ must be $-3$ or $5$. Solve $2x-4=-3$ for $x=0.5$, and $2x-4=5$ for $x=4.5$. Dividing the zeros by 2 without adding 4 gives $\{-1.5,2.5\}$; using $2x+4$ or multiplying by 2 and adding 4 gives the other lists.
5. The point $(4,-2)$ lies on the graph of $y=f(x)$. After the transformation $g(x)=-3f(2x+6)+5$, which point on the graph of $g$ corresponds to this point?
   Answer: $(-1,11)$ Set $2x+6=4$, so $x=-1$. Then $g(-1)=-3f(4)+5=-3(-2)+5=11$. The corresponding point is $(-1,11)$. $(-5,11)$ comes from solving $2x+6=-4$, $(1,11)$ comes from treating the inner shift as $x+3=4$, and $(-1,-1)$ comes from dropping the reflection sign.
6. The graph of $f(x)=\frac{1}{x}$ is transformed into $g(x)=\frac{2}{x-3}+1$. Which sequence of transformations produces this graph?
   Answer: Vertical stretch by factor 2, shift right 3, shift up 1. $g(x)=2f(x-3)+1$. Applying a vertical stretch by 2, then a shift right 3, then a shift up 1 gives $2f(x-3)+1$. Shifting up before the vertical stretch stretches the 1-unit shift into a 2-unit shift, yielding a different vertical placement.
7. The function $g$ is formed by taking $f(x)=\frac{1}{x}$, shifting its graph 2 units left, reflecting it across the $x$-axis, and shifting it 1 unit down. Which statement about the graph of $g$ is true?
   Answer: The vertical asymptote is $x=-2$, and the horizontal asymptote is $y=-1$. When you see a transformation question involving a rational function like $f(x)=\frac1x$, track each shift and reflection one step at a time. The original vertical asymptote is $x=0$, and the horizontal asymptote is $y=0$. Shifting the graph 2 units left replaces $x$ with $x+2$, giving $y=\frac1{x+2}$, so the vertical asymptote moves to $x=-2$. Reflecting across the $x$-axis gives $y=-\frac1{x+2}$. Finally, shifting 1 unit down gives $g(x)=-\frac1{x+2}-1$. Because $-\frac1{x+2}\to 0$ as $x\to\pm\infty$, the horizontal asymptote is $y=-1$. So the true statement is: vertical asymptote $x=-2$, horizontal asymptote $y=-1$. The choice saying the vertical asymptote is $x=2$ reverses the direction of the shift — moving left means subtracting from $x$, which lands on $-2$, not $2$. The choice saying the horizontal asymptote is $y=1$ misses the combined effect of the reflection and downward shift: the reflection flips the sign, and the shift moves it down, so the asymptote goes to $-1$, not $1$. The choice with both $x=2$ and $y=1$ combines both errors, and the choice with $x=-2$ but $y=1$ correctly identifies the vertical shift direction but not the final vertical position. On exam day, write the transformed equation first. The asymptotes come directly from the denominator's zero and the end behavior — don't rely on memory of the graph alone.
8. Which transformation of $f(x)=2^x$ produces the same graph as $g(x)=2^{x+3}$?
   Answer: A vertical stretch by a factor of 8 Whenever you see a transformation question, rewrite the target in terms of $f(x)$ and decide whether the change is inside the exponent or outside it. Here, $$g(x)=2^{x+3}=2^x\cdot 2^3=8\cdot 2^x=8f(x).$$ That means every output value of $f$ is multiplied by $8$, so the graph is exactly the vertical stretch of $f$ by a factor of $8$. The other choices confuse where the $+3$ acts. A reflection across the $x$-axis followed by a vertical shift up $3$ would produce $-2^x+3$, not $8\cdot2^x$. A vertical shift up $3$ would produce $2^x+3$, which adds $3$ to each output instead of multiplying it by $8$. A horizontal shift right $3$ would produce $$f(x-3)=2^{x-3}=2^x\cdot2^{-3}=\frac{2^x}{8},$$ which compresses the graph vertically by $8$, not stretches it. The equivalent horizontal move would actually be a shift left $3$, since $f(x+3)=2^{x+3}$, but that choice is not listed. The key habit: distinguish input changes (inside the exponent, horizontal shifts) from output changes (outside the exponent, vertical shifts and stretches). Then use exponent laws to convert forms—you will quickly see that $g$ is just $f$ stretched vertically by $8$.
9. Which sequence of transformations maps $f(x)=x^3$ to $g(x)=(2x-6)^3$?
   Answer: Horizontal compression by factor $\frac12$, then shift right 3. Rewrite $2x-6=2(x-3)$. Start with $f(2x)$ (horizontal compression by factor $\frac12$), then replace $x$ by $x-3$: $f(2(x-3))=(2x-6)^3$. The distractor with shift right 3 before compression gives $f(2x-3)$, not $f(2x-6)$.
10. The graph of $y=f(x)$ is transformed by first a horizontal stretch by a factor of 3, then a shift left 2 units. Which equation represents the transformed graph?
    Answer: $y=f\left(\frac{x+2}{3}\right)$ A horizontal stretch by factor 3 replaces $x$ by $x/3$, giving $f(x/3)$. A subsequent shift left 2 replaces $x$ by $x+2$, giving $f((x+2)/3)$. The form $f(x/3+2)$ corresponds to shifting left before stretching; $f(3(x+2))$ is a compression, not a stretch.
11. The function $f$ has domain $[-2,6]$ and range $[1,5]$. If $g(x)=2f(x-4)-3$, what are the domain and range of $g$?
    Answer: Domain $[2,10]$, range $[-1,7]$ The inside transformation $x-4$ shifts the domain right 4: $[-2+4,6+4]=[2,10]$. The outside transformation $2(\cdot)-3$ maps range values $y$ to $2y-3$, so $[2(1)-3,2(5)-3]=[-1,7]$. The incorrect options use a left shift or add/subtract 3 before stretching.
12. Starting with the graph of $y=f(x)$, shift the graph 3 units to the right, then reflect it across the $y$-axis. Which equation describes the resulting graph?
    Answer: $y=f(-x-3)$ Whenever a transformations question appears, focus on what happens inside the function versus outside. Inside changes affect $x$ horizontally; outside changes affect $y$ vertically. Also, horizontal shifts are counterintuitive: shifting right 3 means replacing $x$ with $x-3$, so the shifted graph is $y=f(x-3)$. Now reflect across the $y$-axis: reflection replaces every $x$ with $-x$. Apply that to the already-shifted expression: $y=f((-x)-3)=f(-x-3)$. So the resulting graph is indeed $y=f(-x-3)$. Now see why the others miss the mark. $y=f(x-3)$ only describes the shift right; it forgets the reflection entirely. $y=f(-x)+3$ is a vertical shift upward by 3, not a horizontal shift or a true input reflection; it confuses inside and outside transformations. The tempting choice $y=f(-x+3)$ often comes from reflecting first and then shifting right, or shifting left and then reflecting—either way, the order of operations is wrong. Since the problem first shifts right, then reflects, the $-3$ must stay inside after the reflection, giving $-x-3$, not $-x+3$. Study tip: track transformations step by step, rewriting the function after each move. For horizontal changes, ask yourself exactly what $x$ is replaced by before moving on. This prevents the classic "reflection order" trap.
13. The graph of $g$ is obtained from $f(x)=\sqrt{x}$ by shifting 3 units left, vertically stretching by a factor of 2, and then reflecting across the $x$-axis. Which of the following is $g(x)$?
    Answer: $g(x)=-2\sqrt{x+3}$ Whenever you see a function transformation problem, separate horizontal changes inside the function from vertical changes outside. For $f(x)=\sqrt{x}$, shifting 3 units left means replacing $x$ by $x+3$, giving $\sqrt{x+3}$. A vertical stretch by factor 2 multiplies the output by 2: $2\sqrt{x+3}$. Then reflecting across the $x$-axis multiplies the output by $-1$, so $g(x)=-2\sqrt{x+3}$. The choice $2\sqrt{-x+3}$ has a different horizontal structure: $-x+3=3-x$, which reflects the graph across the $y$-axis and shifts it right, not left; it also lacks the $x$-axis reflection. The choice $-2\sqrt{x-3}$ has the vertical stretch and $x$-axis reflection, but $x-3$ shifts the graph right 3 units, not left. The choice $2\sqrt{x+3}$ has the correct left shift and vertical stretch, but multiplying by 2 instead of $-2$ means the $x$-axis reflection is missing. A useful habit is to build the transformations step-by-step and check the sign inside the radical: $x+$ constant means shift left, $x-$ constant means shift right. A negative sign outside the function is your signal that the final reflection across the $x$-axis has been applied.
14. Starting from $f(x)=x^2$, which sequence of transformations, applied in the order listed, produces $h(x)=-\frac{1}{2}(x-4)^2+3$?
    Answer: Reflect across the $x$-axis, compress vertically by a factor of $\frac{1}{2}$, shift 4 units right, shift 3 units up. Whenever you see a transformed quadratic like $h(x)=-\frac12(x-4)^2+3$, start from $f(x)=x^2$ and compare to the form $y=a(x-h)^2+k$. Here $a=-\frac12$, $h=4$, and $k=3$. The negative sign means reflect across the $x$-axis, the factor $\frac12$ means compress vertically, $(x-4)$ means shift 4 units right, and $+3$ means shift 3 units up. Applying the vertical changes first, then the shifts, gives exactly: reflect to $-x^2$, compress to $-\frac12 x^2$, shift right to $-\frac12(x-4)^2$, then shift up to $-\frac12(x-4)^2+3$. The distractor that says "stretch vertically by a factor of 2, shift 4 units left" misreads the reciprocal and confuses the horizontal direction. The sequence "compress, shift right, shift up, then reflect" would produce $-\left(\frac12(x-4)^2+3\right)=-\frac12(x-4)^2-3$, turning the final shift downward. The sequence "shift right, shift up, reflect, then compress" gives $-\frac12(x-4)^2-\frac32$, because the reflection and compression are applied to the vertical shift too. Order truly matters. A good strategy: first identify the reflection/compression coefficient and the shifts from the form $y=a(x-h)^2+k$, then apply vertical changes before shifts unless you are careful about how later transformations affect earlier shifts.
15. Starting with $f(x)=\sqrt{x}$, reflect its graph across the $x$-axis and then shift it 4 units up. What are the domain and range of the transformed function?
    Answer: Domain: $x\ge 0$; range: $y\le 4$ When you see a transformation question, identify each operation in order and ask: does this change the inputs ($x$) or the outputs ($y$)? Reflecting across the $x$-axis and shifting vertically affect only outputs, so the domain of the square root function stays $x \ge 0$. Start with $f(x)=\sqrt{x}$, where $\sqrt{x}\ge 0$. Reflect across the $x$-axis: $y=-\sqrt{x}$, so outputs become $\le 0$. Then shift 4 units up: $y=-\sqrt{x}+4$. Now every output is $4$ minus a nonnegative number, so the largest possible output is $4$, and outputs decrease without bound below it. Thus the range is $y \le 4$. The correct answer is domain $x\ge 0$; range $y\le 4$. Each wrong choice traps a specific misunderstanding. "Domain: $x\ge 0$; range: $y\ge 4$" treats the upward shift as raising the lower bound, but a negative square root flips the inequality direction. "Domain: $x\le 0$; range: $y\le 4$" incorrectly assumes reflecting across the $x$-axis somehow changes the domain; reflections across the $x$-axis do not affect input values. "Domain: $x\ge 0$; range: $y\le -4$" confuses the 4-unit shift with moving the range below $-4$; the shift is upward, so the ceiling should be $+4$, not $-4$. Study tip: for transformations, remember that vertical stretches/reflections/shifts change the range, while horizontal ones change the domain. Track the graph of $y=\sqrt{x}$ through each step, and write the final equation before finding domain and range.
16. The graph of $f(x)=\sin x$ is transformed to $g(x)=-2\sin(3x-\pi)+1$. Which statement about $g$ is correct?
    Answer: The period is $\frac{2\pi}{3}$, the phase shift is $\frac{\pi}{3}$ to the right, and the range is $[-1,3]$. Rewrite $3x-\pi=3(x-\pi/3)$, so the period is $2\pi/3$ and the phase shift is $\pi/3$ to the right. The amplitude is 2 and the graph is reflected and shifted up 1, so the range is $[-1,3]$. The $\pi$ phase shift comes from not factoring out 3; the left shift reverses the sign; the $2\pi$ period ignores the horizontal compression.
17. A student was asked to transform $f(x)$ by stretching vertically by a factor of 2, reflecting across the $x$-axis, and then shifting up 3 units. The student wrote $g(x)=-2(f(x)+3)$. Compared with the correct graph, the student's graph is:
    Answer: 9 units lower The intended graph is $-2f(x)+3$. The student's expression $-2(f(x)+3)$ simplifies to $-2f(x)-6$. For every $x$, the student's y-value is $(-2f(x)-6)-(-2f(x)+3)=-9$, so the graph is 9 units lower.
18. Which sequence of transformations of $f(x)=x^3$ produces $g(x)=(2-x)^3$?
    Answer: Reflect across the $y$-axis, then shift right 2. Because $2-x=-(x-2)$, $g(x)=f(-(x-2))$. Reflect $f(x)=x^3$ across the $y$-axis to get $f(-x)$, then shift right 2 to get $f(-(x-2))=(2-x)^3$. Shifting first and then reflecting gives $f(-x-2)$, and shifting left after reflection gives $f(-x-2)$ as well.
19. The graph of $f(x)=\sqrt{x}$ is reflected across the $y$-axis, then shifted left 3 units and down 2 units. Which equation defines the resulting function $g$?
    Answer: $g(x)=\sqrt{-x-3}-2$ Reflecting $f(x)=\sqrt{x}$ across the $y$-axis gives $\sqrt{-x}$. A shift left 3 replaces $x$ by $x+3$, producing $\sqrt{-(x+3)}=\sqrt{-x-3}$, and a shift down 2 gives $\sqrt{-x-3}-2$. The option $\sqrt{-x+3}-2$ corresponds to shifting before reflecting; $\sqrt{-x+3}+2$ and $\sqrt{-x-3}+2$ have the vertical shift in the wrong direction.
20. The graph of $f$ contains the point $(1,4)$. The graph of $g(x)=f(2x-6)$ is a transformation of the graph of $f$. Which point must be on the graph of $g$?
    Answer: $(3.5,4)$ Whenever you see a transformation question like $g(x)=f(2x-6)$, focus on what happens to the input of $f$. Since $(1,4)$ is on $f$, you know $f(1)=4$. For $g$ to output $4$, you need the inside expression to equal $1$: set $2x-6=1$. Solving gives $2x=7$, so $x=3.5$. Therefore $(3.5,4)$ must be on $g$. This is the correct point. The other choices show common missteps. The point $(5,4)$ comes from solving $x-6=1$ after forgetting the coefficient $2$, or from shifting right by $6$ instead of applying the stretch. The point $(0.5,4)$ might come from solving $2x=1$ and ignoring the $-6$, treating the transformation as just a horizontal compression. The point $(1,4)$ assumes that the original point passes through unchanged, which would only be true if $g(x)=f(x)$ or if $2x-6=1$ at $x=1$, but $2(1)-6=-4$, not $1$. A reliable strategy: rewrite $2x-6$ as $2(x-3)$, but then set the whole inside equal to the known input, not just part of it. For any point $(a,b)$ on $f$, find $x$ such that the expression inside $f$ equals $a$. This turns transformations into a simple equation and avoids visual guessing. On the algebra exam, this input-equation method works for all horizontal stretches, compressions, and shifts.

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