College Algebra Quiz: Combining Transformations

What this quiz covers

This quiz focuses on Combining Transformations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $a = 2, b = 3, h = 3, k = 4$ (correct answer)
2. $a = 6, b = 1, h = 3, k = 4$
3. $a = 2, b = 3, h = -3, k = 4$
4. $a = 2, b = 1, h = 3, k = 4$

Explanation: 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 $b$.

Question 2

1. $y = 10$ (correct answer)
2. $y = 1$
3. $y = 12$
4. $y = 14$

Explanation: 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: $4 \times 3 + 2 = 14$.

Question 3

1. $x = -1$
2. $x = -\frac{3}{2}$ (correct answer)
3. $x = \frac{1}{2}$
4. $x = -\frac{5}{2}$

Explanation: 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 $x = \frac{1-4}{2} = -\frac{5}{2}$.

Question 4

1. $(-3, 8)$ (correct answer)
2. $(-6, 8)$
3. $(-3, -8)$
4. $(3, 8)$

Explanation: 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.