Function Graph Transformations - Math 3

1. The rational function $r(x) = \frac{x^2 - 4}{x - 2}$ is transformed by shifting left 3 units and up 5 units. What is the equation of the hole in the transformed function?
   Answer: $(-1, 9)$ When you encounter rational functions with holes that undergo transformations, you need to first identify the hole in the original function, then apply the transformations to that point. The function $r(x) = \frac{x^2 - 4}{x - 2}$ has a hole because the numerator and denominator share a common factor. Since $x^2 - 4 = (x-2)(x+2)$, you can simplify: $r(x) = \frac{(x-2)(x+2)}{x-2} = x+2$ (for $x \neq 2$). The hole occurs where the original denominator equals zero: at $x = 2$. At this point, the simplified function gives $y = 2 + 2 = 4$, so the original hole is at $(2, 4)$. Now apply the transformations to this hole. Shifting left 3 units means subtracting 3 from the x-coordinate: $2 - 3 = -1$. Shifting up 5 units means adding 5 to the y-coordinate: $4 + 5 = 9$. The transformed hole is at $(-1, 9)$, which is answer D. Looking at the wrong answers: A) $(1, 9)$ correctly applies the vertical shift but adds 3 to the x-coordinate instead of subtracting (confusing left with right shift). B) $(2, 4)$ is the original hole position without any transformations applied. C) $(-1, 4)$ correctly applies the horizontal shift but ignores the vertical shift entirely. Remember: transformations affect the coordinates of all key points (holes, intercepts, asymptotes) in the same way they affect the entire graph. Always identify special points first, then transform them systematically.
2. The function $g(x) = 2^x$ is transformed to create $h(x) = -3 \cdot 2^{-2x+4} + 5$. What is the y-intercept of $h(x)$?
   Answer: $-43$ When you encounter a transformed exponential function, finding the y-intercept means evaluating the function at $x = 0$. This tests your ability to work with complex exponential transformations and perform accurate substitution. To find the y-intercept of $h(x) = -3 \cdot 2^{-2x+4} + 5$, substitute $x = 0$: $h(0) = -3 \cdot 2^{-2(0)+4} + 5$ $h(0) = -3 \cdot 2^{4} + 5$ $h(0) = -3 \cdot 16 + 5$ $h(0) = -48 + 5 = -43$ The y-intercept is $-43$, which is answer choice D. Looking at the wrong answers: Choice A ($53$) likely comes from incorrectly making the coefficient positive, getting $3 \cdot 16 + 5 = 53$. Choice B ($-11$) suggests an error in calculating $2^4$, perhaps using $2^4 = 8$ instead of $16$, giving $-3 \cdot 8 + 5 = -19$, or making other computational mistakes. Choice C ($17$) could result from sign errors combined with calculation mistakes, possibly computing $3 \cdot 4 + 5 = 17$ by confusing $2^4$ with $2 \times 4$. Remember this key strategy: for any function's y-intercept, always substitute $x = 0$ and carefully track your signs and order of operations. With transformed exponential functions, pay special attention to negative coefficients and ensure you're calculating the exponential term correctly before applying the outer transformations.
3. The function $f(x) = x^3 - 2x^2 + x - 3$ undergoes the transformation $g(x) = -2f(\frac{x-4}{3}) + 1$. Which statement best describes the combined effect of all transformations applied to $f(x)$?
   Answer: Horizontal stretch by factor 3, right shift 4 units, vertical stretch by factor 2, reflection over x-axis, up shift 1 unit To find the transformations, we analyze $g(x) = -2f(\frac{x-4}{3}) + 1$. The input transformation $\frac{x-4}{3}$ means we first shift right by 4 (from $x-4$), then stretch horizontally by factor 3 (from division by 1/3). The output transformation $-2f(...) + 1$ means we stretch vertically by factor 2, reflect over the x-axis (from the negative), and shift up 1 unit. Choice B incorrectly identifies the horizontal shift direction. Choice C incorrectly calculates the horizontal shift as 12. Choice D incorrectly identifies compression instead of stretch and reflection over y-axis instead of x-axis.
4. The function $f(x) = \frac{x^2-4}{x-1}$ is transformed to create $h(x) = 2f(\frac{x+3}{2}) - 1$. What is the equation of the vertical asymptote of $h(x)$?
   Answer: $x = -1$ The vertical asymptote of $f(x) = \frac{x^2-4}{x-1}$ occurs where the denominator equals zero, at $x = 1$. For $h(x) = 2f(\frac{x+3}{2}) - 1$, we need to find where the input to $f$ makes its denominator zero. This happens when $\frac{x+3}{2} - 1 = 0$, which gives us $\frac{x+3}{2} = 1$, so $x + 3 = 2$, therefore $x = -1$. Choice B gives the original asymptote location. Choices C and D result from incorrect transformation calculations.
5. A polynomial function $p(x)$ has roots at $x = -1, 2,$ and $4$. After applying the transformation $q(x) = p(2x-3)$, what are the roots of $q(x)$?
   Answer: $x = 1, 2.5, 3.5$ To find the roots of $q(x) = p(2x-3)$, we need to solve $q(x) = 0$, which means $p(2x-3) = 0$. This occurs when $2x-3$ equals any root of $p(x)$. So we solve: $2x-3 = -1 \Rightarrow x = 1$, $2x-3 = 2 \Rightarrow x = 2.5$, and $2x-3 = 4 \Rightarrow x = 3.5$. Therefore, the roots of $q(x)$ are $x = 1, 2.5, 3.5$. Choice A incorrectly calculates some roots. Choice B uses the wrong transformation formula. Choice D also incorrectly applies the transformation.
6. Consider the rational function $h(x) = \frac{2x - 6}{x + 1}$. If this function is transformed by shifting 3 units right and 4 units down, what is the equation of the resulting vertical asymptote?
   Answer: $x = 2$ The original function h(x) has a vertical asymptote where the denominator equals zero: x + 1 = 0, so x = -1. When we shift right 3 units, we replace x with (x-3), giving us h(x-3) = (2(x-3)-6)/((x-3)+1) = (2x-12)/(x-2). The vertical asymptote occurs where x-2 = 0, so x = 2. The vertical shift doesn't affect the location of vertical asymptotes. Choice A incorrectly subtracts 3 from the original asymptote. Choice C gives the original asymptote location. Choice D incorrectly adds 3 to the original asymptote.
7. If $g(x) = (x - 3)^3 + 2$ is obtained by transforming $f(x) = x^3$, and then $g(x)$ is transformed to produce $h(x)$ by reflecting across the line $y = 2$, what is $h(x)$?
   Answer: $h(x) = -(x - 3)^3 + 2$ Reflecting across the horizontal line y = 2 means that a point (a, b) becomes (a, 4-b). For the function g(x) = (x-3)³ + 2, we need to reflect each y-value across y = 2. If y = g(x), then after reflection y = 4 - g(x) = 4 - ((x-3)³ + 2) = 4 - (x-3)³ - 2 = 2 - (x-3)³ = -(x-3)³ + 2. Choice B reflects across the x-axis instead. Choice C incorrectly calculates the reflection. Choice D reflects across the y-axis instead of y = 2.
8. The polynomial $p(x) = x^3 - 6x^2 + 9x - 2$ has a local maximum at $x = 1$ and a local minimum at $x = 3$. After the transformation $q(x) = 2p(\frac{x-6}{3}) + 4$, at what x-value does the local maximum of $q(x)$ occur?
   Answer: $x = 9$ The transformation $q(x) = 2p(\frac{x-6}{3}) + 4$ involves several steps. The input transformation $\frac{x-6}{3}$ represents a horizontal stretch by factor 3 and a right shift by 6. To find where the local maximum occurs in $q(x)$, we need to find where $\frac{x-6}{3} = 1$ (since the original maximum is at $x = 1$). Solving: $\frac{x-6}{3} = 1 \Rightarrow x-6 = 3 \Rightarrow x = 9$. Choice A results from incorrect application of transformations. Choice C incorrectly calculates the combined effect. Choice D applies the transformations in the wrong order.
9. A function $f(x)$ has the property that $f(2) = 7$. After applying the transformation $y = 3f\left(\frac{x-1}{2}\right) - 4$, what is the y-coordinate of the point on the transformed graph that corresponds to the original point $(2, 7)$?
   Answer: $17$ When you encounter function transformations, you need to trace how a specific point moves through each transformation step by step. The key insight is that transformations change both the input values and output values in predictable ways. Starting with the original point $(2, 7)$ where $f(2) = 7$, let's see where this point goes under the transformation $y = 3f\left(\frac{x-1}{2}\right) - 4$. First, we need to find what x-value in the transformed function will use our known input of $x = 2$. Setting $\frac{x-1}{2} = 2$ and solving: $x - 1 = 4$, so $x = 5$. This means the point that originally was at $x = 2$ is now located at $x = 5$ on the transformed graph. Next, we apply the transformations to the y-value. Since $f(2) = 7$, we have: $y = 3f(2) - 4 = 3(7) - 4 = 21 - 4 = 17$ So the transformed point is $(5, 17)$, making the y-coordinate $17$, which is choice C. Looking at the wrong answers: Choice A ($25$) likely comes from calculating $3 \times 7 + 4 = 25$, incorrectly adding instead of subtracting the vertical shift. Choice B ($13$) results from $3 \times 7 - 8 = 13$, possibly doubling the vertical shift. Choice D ($21$) comes from forgetting the $-4$ vertical shift entirely, giving just $3 \times 7 = 21$. Remember: when tracking points through transformations, work systematically through each piece—horizontal shifts affect where you'll find the point, while vertical stretches and shifts transform the output value.
10. The function $f(x) = \sqrt{x}$ undergoes the following sequence of transformations: first a horizontal compression by a factor of $\frac{1}{4}$, then a reflection across the y-axis, then a vertical stretch by a factor of 3, and finally a shift up 2 units. What is the domain of the resulting function?
    Answer: $x \leq 0$ Starting with f(x) = √x: (1) Horizontal compression by 1/4 gives √(4x), (2) Reflection across y-axis gives √(4(-x)) = √(-4x), (3) Vertical stretch by 3 gives 3√(-4x), (4) Shift up 2 gives 3√(-4x) + 2. For the square root to be defined, we need -4x ≥ 0, which means x ≤ 0. Choice A gives the original domain. Choice C incorrectly relates the vertical shift to domain restrictions. Choice D ignores the square root's domain restrictions.
11. A rational function has the form $r(x) = \frac{ax + b}{cx + d}$ where $a, b, c, d$ are constants. After applying a horizontal shift of 2 units left and a vertical shift of 3 units up, the resulting function has a horizontal asymptote at $y = 5$ and a vertical asymptote at $x = 1$. What was the horizontal asymptote of the original function?
    Answer: $y = 2$ After transformation, the function is r(x+2) + 3. The vertical asymptote moves from cx + d = 0 to c(x+2) + d = 0, which gives x = -d/c - 2. Since this equals 1, we have -d/c - 2 = 1, so -d/c = 3. The horizontal asymptote of r(x) = (ax+b)/(cx+d) is y = a/c. After adding 3, the horizontal asymptote becomes y = a/c + 3 = 5, so a/c = 2. Therefore, the original horizontal asymptote was y = 2. Choice B incorrectly assumes horizontal asymptotes don't change with vertical shifts. Choice C adds instead of subtracts the shift. Choice D gives the amount of vertical shift.
12. Given the piecewise function $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ 2x + 1 & \text{if } x \geq 0 \end{cases}$, what is the equation of the function obtained by applying the transformation $y = f(2x - 4)$?
    Answer: $y = \begin{cases} (2x-4)^2 & \text{if } x < 2 \\ 2(2x-4) + 1 & \text{if } x \geq 2 \end{cases}$ When transforming piecewise functions, you need to handle two key elements: substituting the transformation into the function expressions AND adjusting the domain conditions to match your new input variable. For $y = f(2x - 4)$, you substitute $(2x - 4)$ everywhere you see the input variable in the original function. This gives you: - When the condition is met: $y = (2x-4)^2$ - When the condition isn't met: $y = 2(2x-4) + 1$ The crucial step is determining when to use each piece. The original function uses $x^2$ when the input is negative, so you need $2x - 4 < 0$. Solving this inequality: $2x < 4$, which means $x < 2$. Similarly, you use the linear piece when $2x - 4 \geq 0$, which gives $x \geq 2$. Choice A incorrectly keeps the condition as $2x - 4 < 0$ instead of solving for $x$. This makes the function difficult to evaluate since you'd need to substitute $x$ values and check the inequality $2x - 4$ each time. Choice B makes an algebra error, getting $x < 4$ instead of $x < 2$ when solving $2x - 4 < 0$. Choice D correctly identifies the boundary at $x = 2$ but incorrectly simplifies the function expressions to $(x-2)^2$ and $2(x-2) + 1$, which would correspond to $f(x-2)$, not $f(2x-4)$. Study tip: Always solve inequality conditions for the outer variable ($x$) and keep function expressions in terms of the full transformation $(2x-4)$.

Start this quiz