Combining Transformations - College Algebra

1. The function $f(x) = x^2$ is transformed to create $g(x) = -(2x + 6)^2 + 8$. What is the vertex of the parabola represented by $g(x)$?
   Answer: $(-3, 8)$ To find the vertex of $g(x) = -(2x + 6)^2 + 8$, we need to identify the transformations applied to $f(x) = x^2$. First, rewrite the expression: $g(x) = -(2x + 6)^2 + 8 = -[2(x + 3)]^2 + 8 = -4(x + 3)^2 + 8$. This shows that the function is in vertex form $a(x - h)^2 + k$ where $a = -4$, $h = -3$, and $k = 8$. Therefore, the vertex is at $(-3, 8)$. The transformations applied are: horizontal shift left 3 units (due to $x + 3$), horizontal compression by factor $\frac{1}{2}$ (due to the factor of 2 inside), vertical stretch by factor 4, reflection across the x-axis (due to the negative sign), and vertical shift up 8 units. Option B results from incorrectly identifying the horizontal shift as 6 units instead of 3. Option C results from incorrectly applying the reflection to the y-coordinate of the vertex. Option D results from incorrectly determining the direction of the horizontal shift.
2. The graph of $y = f(x)$ passes through the point $(2, 5)$ and has a horizontal asymptote at $y = 3$. After applying the transformations to create $y = 4f(\frac{1}{2}x - 1) - 2$, what is the equation of the new horizontal asymptote?
   Answer: $y = 10$ To find the new horizontal asymptote, we need to understand how the transformations affect the original horizontal asymptote at $y = 3$. The transformation $y = 4f(\frac{1}{2}x - 1) - 2$ includes: a horizontal stretch by factor 2 and horizontal shift (which don't affect horizontal asymptotes), a vertical stretch by factor 4, and a vertical shift down by 2 units. The horizontal asymptote is affected only by the vertical transformations. Starting with the original asymptote $y = 3$: first apply the vertical stretch by factor 4, giving $y = 4 \times 3 = 12$, then apply the vertical shift down by 2 units, giving $y = 12 - 2 = 10$. Therefore, the new horizontal asymptote is $y = 10$. Option B results from applying the transformations incorrectly: $3 - 2 = 1$, forgetting the vertical stretch. Option C results from applying only the vertical stretch: $4 \times 3 = 12$, forgetting the vertical shift. Option D results from incorrectly adding instead of subtracting: $4 \times 3 + 2 = 14$.
3. Consider the function $j(x) = 2|3x - 9| + 4$. If this function is written in the form $j(x) = a|b(x - h)| + k$, what are the values of $a$, $b$, $h$, and $k$?
   Answer: $a = 2, b = 3, h = 3, k = 4$ To write $j(x) = 2|3x - 9| + 4$ in the form $j(x) = a|b(x - h)| + k$, we need to factor the expression inside the absolute value. Starting with $3x - 9$, we factor out the 3: $3x - 9 = 3(x - 3)$. Therefore, $j(x) = 2|3(x - 3)| + 4$. Comparing this to the form $a|b(x - h)| + k$, we have $a = 2$, $b = 3$, $h = 3$, and $k = 4$. Option B incorrectly combines the coefficients 2 and 3 to get $a = 6$ and sets $b = 1$. Option C incorrectly identifies $h = -3$, confusing the sign inside the parentheses. Option D incorrectly sets $b = 1$, failing to recognize that the factor of 3 should remain as the coefficient $b$.
4. The function $g(x) = f(2x + 4) - 3$ is obtained from $f(x)$ by applying two transformations. If the original function $f(x)$ has a vertical asymptote at $x = 1$, what is the equation of the corresponding vertical asymptote in $g(x)$?
   Answer: $x = -\frac{3}{2}$ To find where the vertical asymptote occurs in $g(x) = f(2x + 4) - 3$, we need to determine when the input to $f$ equals 1 (the location of the original asymptote). So we solve $2x + 4 = 1$. This gives $2x = 1 - 4 = -3$, so $x = -\frac{3}{2}$. Therefore, the vertical asymptote of $g(x)$ is at $x = -\frac{3}{2}$. The vertical shift of -3 doesn't affect the location of vertical asymptotes, only horizontal transformations do. Option A would result from incorrectly thinking $2x + 4 = 1$ gives $x = -1$. Option C would result from solving $2x = 1$ and ignoring the +4. Option D would result from incorrectly applying the transformation $x = \frac{1-4}{2} = -\frac{5}{2}$.

Start this quiz