Function Transformations - Math 2

Consider the piecewise function $f(x) = \begin{cases} x^2 & \text{if } x \leq 0 \\ 2x & \text{if } x > 0 \end{cases}$ . The function $g(x) = f(x-2) + 1$ shifts this function right 2 and up 1. At what value of $x$ does $g(x)$ change its defining rule?
Answer: $x = 2$ The original function f(x) changes its rule at x = 0. For g(x) = f(x-2) + 1, the input to f is (x-2). The function f changes behavior when its input equals 0, so we need x-2 = 0, which gives x = 2. At x = 2, g(x) transitions from using the x² rule to the 2x rule. Choice A uses the original transition point. Choice B uses x = 2-1. Choice D uses x = 0-2.
Consider the function $h(x) = 2f(\frac{1}{2}x - 1) + 3$ . If the domain of $f(x)$ is $[-4, 8]$ , what is the domain of $h(x)$ ?
Answer: $[-6, 18]$ When you encounter composite functions with transformations, you need to work from the inside out to find how the domain changes. The key is determining what input values for $x$ will make the inner function $f(\frac{1}{2}x - 1)$ stay within its original domain. Since $f(x)$ has domain $[-4, 8]$ , you need $\frac{1}{2}x - 1$ to produce values between $-4$ and $8$ . Set up the inequality: $-4 \leq \frac{1}{2}x - 1 \leq 8$ . Solving the left side: $-4 \leq \frac{1}{2}x - 1$ gives $-3 \leq \frac{1}{2}x$ , so $x \geq -6$ . Solving the right side: $\frac{1}{2}x - 1 \leq 8$ gives $\frac{1}{2}x \leq 9$ , so $x \leq 18$ . Therefore, the domain of $h(x)$ is $[-6, 18]$ . The "+3" outside doesn't affect the domain since it's just a vertical shift. Choice A $[-1, 7]$ represents a common error where students incorrectly apply transformations to the domain endpoints. Choice C $[-2, 4]$ might result from confusing horizontal and vertical shifts. Choice D $[-8, 16]$ could come from incorrectly handling the coefficient $\frac{1}{2}$ or mixing up the transformation order. Study tip: For composite function domains, always isolate the inner expression and set it equal to the bounds of the original domain. The outer transformations (like adding constants) don't change the domain—only the inner transformations matter.
The function $g(x) = -2f(\frac{x-3}{2}) + 1$ is derived from a parent function $f(x)$ . If the point $(4, 5)$ lies on the graph of $f(x)$ , what point lies on the graph of $g(x)$ ?
Answer: $(11, -9)$ To find the corresponding point on g(x), we work backwards from the transformations. If (4,5) is on f(x), then f(4) = 5. For g(x) = -2f((x-3)/2) + 1, we need to find x such that (x-3)/2 = 4, which gives x-3 = 8, so x = 11. Then g(11) = -2f(4) + 1 = -2(5) + 1 = -9. So (11, -9) is on g(x). Choice B incorrectly applies g(11) = -2(5) - 1 = -11. Choice C uses x = 5 from incorrect algebra. Choice D uses x = 8 from solving x-3 = 4 instead of (x-3)/2 = 4.
The function $f(x) = \sqrt{x}$ undergoes a sequence of transformations: first reflected across the line $y = x$ , then shifted left 3 units, then stretched vertically by a factor of 2. What is the resulting function?
Answer: $g(x) = 2(x+3)^2$ Starting with f(x) = √x: (1) Reflecting across y = x gives the inverse function: x = √y, so y = x². (2) Shifting left 3 units: y = (x+3)². (3) Stretching vertically by factor 2: y = 2(x+3)². Choice A applies transformations as if there was no reflection. Choice C incorrectly expands and modifies the expression. Choice D incorrectly applies the vertical stretch inside the square.
A function $f(x)$ is transformed to create $g(x)$ . The transformation includes a reflection across the x-axis, followed by a horizontal stretch by a factor of 3, and then a vertical shift up by 2 units. If the domain of $f(x)$ is $[-4, 8]$ and the range is $[-6, 12]$ , what is the domain of $g(x)$ ?
Answer: $[-12, 24]$ The transformations are applied as g(x) = -f(x/3) + 2. A horizontal stretch by factor 3 multiplies all x-values by 3, so the domain [-4, 8] becomes [-12, 24]. The reflection across x-axis and vertical shift affect the range, not the domain. Choice B incorrectly assumes horizontal transformations don't affect domain. Choice C applies compression instead of stretch. Choice D confuses the effect of vertical transformations on domain.
The graph of $y = f(x)$ has a maximum value of 8 at $x = -1$ . After applying the transformation $y = \frac{1}{2}f(2x+6) - 3$ , what are the coordinates of the corresponding maximum point?
Answer: $(-4, 1)$ The original maximum is at (-1, 8). For the transformation y = (1/2)f(2x+6) - 3: The horizontal transformation is f(2x+6) = f(2(x+3)), which compresses by 1/2 and shifts left 3 units. So x = -1 becomes x = -1 - 3 = -4. The vertical transformation (1/2)f(...) - 3 compresses by 1/2 and shifts down 3. So y = 8 becomes y = (1/2)(8) - 3 = 1. Choice B uses x = -3 - 3 = -6. Choice C uses y = 8 - 3 = 5, then (1/2)(5) = 2.5, getting close to -7. Choice D uses x = -1 - 2 = -3 from incorrect analysis.
A function $h(x)$ is obtained by applying the following transformations to $f(x)$ in order: reflect across the x-axis, stretch vertically by a factor of 3, shift right 2 units, then shift up 4 units. Which expression represents $h(x)$ ?
Answer: $h(x) = -3f(x-2) + 4$ Starting with f(x): (1) Reflect across x-axis: -f(x). (2) Stretch vertically by 3: -3f(x). (3) Shift right 2: -3f(x-2). (4) Shift up 4: -3f(x-2) + 4. Choice B omits the reflection and incorrectly shifts down. Choice C shifts left instead of right. Choice D omits the reflection across the x-axis.
Given that $f(x)$ has range $[-3, 5]$ , what is the range of $g(x) = -2f(\frac{x-4}{3}) + 1$ ?
Answer: $[-9, 7]$ When you see a function transformation problem involving ranges, you need to trace how each transformation affects the minimum and maximum values of the original function's range. Given that $f(x)$ has range $[-3, 5]$ , let's work through $g(x) = -2f(\frac{x-4}{3}) + 1$ step by step. The horizontal transformations ( $\frac{x-4}{3}$ ) don't affect the range, so we only need to consider the vertical transformations. Start with the range of $f(x)$ : $[-3, 5]$ . First, multiply by $-2$ . This scales the range by a factor of 2 and flips it because of the negative sign. The minimum $-3$ becomes $(-2)(-3) = 6$ , and the maximum $5$ becomes $(-2)(5) = -10$ . Since we flipped the function, our new range is $[-10, 6]$ . Next, add 1 to shift everything up: $[-10 + 1, 6 + 1] = [-9, 7]$ . Looking at the wrong answers: Choice A gives $[1, 13]$ , which suggests incorrectly applying $+2$ instead of $-2$ . Choice B shows $[-11, 5]$ , indicating the student likely forgot to flip the range when multiplying by $-2$ . Choice C gives $[-7, 9]$ , which appears to come from incorrect arithmetic in the scaling step. Study tip: When dealing with function transformations, handle vertical changes in order: first multiply/divide (remembering that negative coefficients flip the range), then add/subtract. Horizontal transformations inside the function don't affect the range at all.
The graph of $y = f(x)$ has been transformed to create $y = g(x)$ . The transformation can be described as a horizontal shift followed by a vertical scaling. If $f(x) = x^2$ and the vertex of $g(x)$ is at $(-3, 0)$ while $g(-1) = 8$ , what is the equation of $g(x)$ ?
Answer: $g(x) = 2(x + 3)^2$ When you encounter function transformations involving horizontal shifts and vertical scaling, you need to identify how each transformation affects the parent function's equation and key features. Starting with $f(x) = x^2$ , which has its vertex at $(0,0)$ , we know that $g(x)$ has vertex at $(-3,0)$ . A horizontal shift moves the vertex from $(0,0)$ to $(-3,0)$ , which means we shift 3 units left. This gives us the form $g(x) = a(x + 3)^2$ where $a$ represents the vertical scaling factor. To find $a$ , we use the condition $g(-1) = 8$ . Substituting: $g(-1) = a(-1 + 3)^2 = a(2)^2 = 4a = 8$ . Therefore, $a = 2$ . This confirms $g(x) = 2(x + 3)^2$ , which is choice B. Choice A, $g(x) = 2(x - 3)^2$ , represents a shift 3 units right (not left), placing the vertex at $(3,0)$ instead of $(-3,0)$ . Choice C, $g(x) = \frac{1}{2}(x + 3)^2$ , has the correct horizontal shift but wrong vertical scaling—when $x = -1$ , this gives $\frac{1}{2}(2)^2 = 2$ , not 8. Choice D, $g(x) = -2(x + 3)^2$ , has the correct shift and scaling magnitude but opens downward due to the negative coefficient, making the vertex a maximum point at $(-3,0)$ rather than a minimum. Remember: horizontal shifts inside parentheses work opposite to what you might expect— $(x + 3)$ means shift left 3 units, while $(x - 3)$ means shift right 3 units.
Consider the transformation $T(x) = -2f(\frac{x+4}{3}) + 5$ . If point $P(a,b)$ on the graph of $f(x)$ is transformed to point $Q(8,1)$ on the graph of $T(x)$ , what are the coordinates of point $P$ ?
Answer: $(4, 2)$ If Q(8,1) is on T(x), then T(8) = 1. So -2f((8+4)/3) + 5 = 1, which gives -2f(4) + 5 = 1, so -2f(4) = -4, and f(4) = 2. Therefore, point P has coordinates (4, 2). Choice A uses the x-coordinate from the transformed point. Choice B has the correct x-coordinate but wrong y-coordinate from incorrectly applying the vertical transformations. Choice D has the wrong x-coordinate from calculation errors.
If $g(x) = f(-x+4)$ and $f(x)$ is increasing on the interval $[1, 7]$ , on which interval is $g(x)$ decreasing?
Answer: $[-3, 3]$ The transformation g(x) = f(-x+4) = f(-(x-4)) reflects f(x) across the y-axis and shifts right 4 units. When a function is reflected across the y-axis, increasing intervals become decreasing intervals. If f(x) is increasing on [1,7], then after reflection it's decreasing on [-7,-1]. After shifting right 4 units, the interval becomes [-7+4, -1+4] = [-3,3]. Choice B keeps the original interval. Choice C gives the reflected interval before the horizontal shift. Choice D incorrectly shifts left instead of accounting for the reflection properly.
A function $f(x)$ is even, meaning $f(-x) = f(x)$ for all $x$ . If $h(x) = f(2x+4) - 3$ , which property does $h(x)$ satisfy?
Answer: $h(-x-4) = h(x)$ Since f(x) is even, f(-u) = f(u) for any u. For h(x) = f(2x+4) - 3, let's find h(-x-4): h(-x-4) = f(2(-x-4)+4) - 3 = f(-2x-8+4) - 3 = f(-2x-4) - 3. Since f is even: f(-2x-4) = f(2x+4). So h(-x-4) = f(2x+4) - 3 = h(x). Choice A: h(-x) = f(-2x+4) - 3 ≠ h(x) in general. Choice B: h(x) is not odd. Choice D: h(x+2) = f(2x+8) - 3 and h(-x-2) = f(-2x) - 3, and f(2x+8) ≠ f(-2x) in general.
Two functions are related by $g(x) = f(x + h) + k$ where $h$ and $k$ are constants. If $f(x)$ has a local minimum at $x = 5$ with value $-3$ , and $g(x)$ has its corresponding local minimum at $x = 2$ with value $1$ , which transformation maps the graph of $f(x)$ to the graph of $g(x)$ ?
Answer: Left 3 units, up 4 units When you see function transformations in the form $g(x) = f(x + h) + k$ , you're dealing with horizontal and vertical shifts. The key is understanding how the parameters $h$ and $k$ affect the graph's movement. Since $f(x)$ has a minimum at $x = 5$ and $g(x)$ has its corresponding minimum at $x = 2$ , the graph shifted 3 units to the left (from $x = 5$ to $x = 2$ ). For a leftward shift of 3 units, we need $x + h = x + 3$ , so $h = 3$ . The minimum value changed from $-3$ to $1$ , representing an upward shift of 4 units. This means $k = 4$ . Therefore, $g(x) = f(x + 3) + 4$ , which transforms $f(x)$ by moving it left 3 units and up 4 units. Answer D is correct: Left 3 units, up 4 units. Answer A suggests moving right 7 units and down 4 units, but the minimum point moved left (not right) and the value increased (not decreased). Answer B indicates rightward movement when the shift was actually leftward. Answer C correctly identifies the upward shift of 4 units but incorrectly suggests a leftward shift of 7 units instead of 3. Remember: In $f(x + h)$ , positive $h$ shifts the graph left, negative $h$ shifts right. This is counterintuitive but crucial. Always track a specific point (like the given minimum) to determine the actual direction and magnitude of the transformation.
The function $g(x) = -2f(\frac{1}{2}x + 3) - 4$ is a transformation of the parent function $f(x)$ . If the point $(6, 8)$ lies on the graph of $f(x)$ , which point lies on the graph of $g(x)$ ?
Answer: $(6, -20)$ To find the corresponding point on g(x), work backwards from the transformation. If g(x) = -2f(½x + 3) - 4 and we need f(6) = 8, solve ½x + 3 = 6 to get x = 6. Then g(6) = -2f(6) - 4 = -2(8) - 4 = -20. The point is (6, -20). Choice A reflects an error in horizontal shift direction. Choice B assumes the horizontal scaling affects the x-coordinate directly. Choice D incorrectly applies only the vertical scaling without the shift.
A function $f(x)$ has the property that $f(x+2) = f(x)$ for all $x$ in its domain. If $g(x) = f(2x-4)$ , what is the period of $g(x)$ ?
Answer: $1$ When you encounter questions about the period of composite functions, you need to understand how transformations affect periodicity. A function's period is the smallest positive value $p$ such that $f(x+p) = f(x)$ for all $x$ . Since $f(x+2) = f(x)$ , we know $f(x)$ has period 2. To find the period of $g(x) = f(2x-4)$ , we need to determine the smallest positive value $p$ such that $g(x+p) = g(x)$ . Let's work with the condition $g(x+p) = g(x)$ : $g(x+p) = f(2(x+p)-4) = f(2x+2p-4)$ For this to equal $g(x) = f(2x-4)$ , we need: $f(2x+2p-4) = f(2x-4)$ Since $f$ has period 2, this equation holds when $2p$ is a multiple of 2. The smallest positive value occurs when $2p = 2$ , giving us $p = 1$ . Looking at the wrong answers: Choice A ( $\frac{1}{2}$ ) results from incorrectly thinking the horizontal compression by factor 2 divides the period by 2. Choice B (2) assumes the period stays the same as the original function, ignoring the horizontal scaling. Choice C (4) might come from incorrectly adding the horizontal shift to the original period. The key insight is that horizontal compression by factor $a$ divides the period by $|a|$ , while horizontal shifts don't affect the period. When $f(x)$ has period 2, then $f(2x-4)$ has period $\frac{2}{2} = 1$ . The answer is D.

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