Transformations of Functions and Graphs - Algebra 2

Which statement best compares the effects of $f(2x)$ and $2f(x)$ on the graph of a general function $f$ ?
Answer: $f(2x)$ compresses horizontally by 2, while $2f(x)$ stretches vertically by 2 This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. Specifically, f(2x) is inside, compressing horizontally by 2 (x scaled by 1/2), while 2f(x) is outside, stretching vertically by 2 (y scaled by 2). Choice B correctly compares the effects as f(2x) compressing horizontally by 2, while 2f(x) stretches vertically by 2. A distractor like A might swap compression and stretch, but remember |k|>1 inside compresses horizontally, and outside stretches vertically—you're getting the hang of it! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?
Determine whether the function $p(x)=x^3-x$ is even, odd, or neither.
Answer: Odd This question tests your understanding of how algebraic transformations of functions—like $f(x) + k$ , $k \cdot f(x)$ , $f(kx)$ , and $f(x + k)$ —affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: $f(-x) = f(x)$ for all x, meaning the left half of the graph is a mirror image of the right half. Examples include $f(x) = x^2$ , $x^4$ , $|x|$ , and $x^2 + 3$ . Odd functions have origin symmetry: $f(-x) = -f(x)$ , meaning rotating the graph 180° about the origin gives the same graph. Examples include $f(x) = x$ , $x^3$ , $1/x$ , and $x^3 - x$ . Most functions are neither even nor odd! For $p(x) = x^3 - x$ , compute $p(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x)$ , which equals $-p(x)$ , confirming origin symmetry without matching $p(x)$ . Choice B correctly determines the function is odd. An incorrect choice like A could stem from overlooking the signs after substitution—note the odd powers (3 and 1) flip signs appropriately for odd functions, but verification is key. For even/odd testing: (1) Take the given function $f(x)$ , (2) Find $f(-x)$ by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with $f(x)$ and $-f(x)$ : if $f(-x) = f(x)$ , it's even; if $f(-x) = -f(x)$ , it's odd; if neither match, it's neither. Example: $f(x) = x^2 - 3$ , so $f(-x) = (-x)^2 - 3 = x^2 - 3 = f(x)$ → even! Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation.
If $f(x)$ is a function where $f(2) = 5$ , and $g(x) = 3f(\frac{x}{2}) - 1$ , what is the value of $g(4)$ ?
Answer: 14 To find $g(4)$ , substitute into $g(x) = 3f(\frac{x}{2}) - 1$ : $g(4) = 3f(\frac{4}{2}) - 1 = 3f(2) - 1$ . Since $f(2) = 5$ , we have $g(4) = 3(5) - 1 = 15 - 1 = 14$ . Choice A results from computing $2f(2) + 1$ . Choice C results from computing $3f(2) + 2$ . Choice D results from computing $6f(2) - 1$ .
Let $f(x)=x^3-x$ . Show using symmetry rules whether $f$ is even, odd, or neither by considering $f(-x)$ relative to $f(x)$ .
Answer: Odd, because $f(-x)=-f(x)$ This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. Examples include f(x) = x², x⁴, |x|, and x² + 3. Odd functions have origin symmetry: f(-x) = -f(x), meaning rotating the graph 180° about the origin gives the same graph. Examples include f(x) = x, x³, 1/x, and x³ - x. Most functions are neither even nor odd! Computing f(-x) = $(-x)^3$ - (-x) = - $x^3$ + x = -( $x^3$ - x) = -f(x), showing origin symmetry—amazing job with the algebra! Choice B correctly determines it's odd because f(-x) = -f(x). Choice A could mislead if you only check even powers, but the odd powers make f(-x) = -f(x), not f(x)—keep verifying both conditions! For even/odd testing: (1) Take the given function f(x), (2) Find f(-x) by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with f(x) and -f(x): if f(-x) = f(x), it's even; if f(-x) = -f(x), it's odd; if neither match, it's neither. Example: f(x) = x² - 3, so f(-x) = (-x)² - 3 = x² - 3 = f(x) → even! Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation.
Show the symmetry type of $f(x)=x^3-x$ by identifying it as even, odd, or neither.
Answer: Odd, because $f(-x)=-f(x)$ This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. Examples include f(x) = x², x⁴, |x|, and x² + 3. Odd functions have origin symmetry: f(-x) = -f(x), meaning rotating the graph 180° about the origin gives the same graph. Examples include f(x) = x, x³, 1/x, and x³ - x. Most functions are neither even nor odd! To test f(x) = x³ - x, we find f(-x) = (-x)³ - (-x) = -x³ + x = -(x³ - x) = -f(x), confirming the function is odd since f(-x) equals -f(x). Choice B correctly identifies this as odd because f(-x) = -f(x). A common error is thinking that having both even and odd powers makes a function neither even nor odd, but the actual test is whether f(-x) equals f(x) or -f(x). For even/odd testing: (1) Take the given function f(x), (2) Find f(-x) by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with f(x) and -f(x): if f(-x) = f(x), it's even; if f(-x) = -f(x), it's odd; if neither match, it's neither.
The graph of $f(x)=x^2$ is shown. The graph of $g(x)$ has the same shape and vertex at $(0,-2)$ . Which equation matches $g(x)$ ?
Answer: $g(x)=f(x)-2$ This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. Moving the vertex to (0,-2) with the same shape means subtracting 2 from the quadratic's y-values, a downward vertical shift—outstanding reasoning! Choice C correctly identifies g(x) = f(x) - 2. Choice A would shift up instead, to (0,2)—just flip the sign for vertical shifts, and you've nailed it! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?
On the coordinate plane, the graph of $f(x)=|x|$ is shown along with the graph of $g(x)$ (a V-shape) whose vertex is at $(2,0)$ and which has the same shape as $f(x)$ . Which function could be $g(x)$ ?
Answer: $g(x)=f(x-2)$ This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. The vertex moving to (2,0) without shape change indicates a horizontal shift right by 2, which f(x-2) achieves by adjusting inputs—brilliant connection! Choice C correctly identifies g(x) = f(x-2). Choice A would shift left instead, landing the vertex at (-2,0)—remember the opposite sign for horizontal shifts! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?
Let the parent function be $f(x)=x^3$ . What is the effect on the graph of replacing $f(x)$ with $f(2x)$ ?
Answer: Horizontal compression by a factor of 2 This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). When we replace f(x) with f(2x), we're multiplying the input by 2, which means to get the same y-value, we need half the original x-value—this compresses the graph horizontally by a factor of 2. Choice A correctly identifies this as a horizontal compression by a factor of 2. Choice B incorrectly calls it a stretch (which would be f(x/2)), while C and D confuse horizontal transformations with vertical ones. Transformation memory aid: think 'outside affects y, inside affects x.' Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects).
Let $f(x)=x^3$ (the standard cubic). The function $g(x)=f(2x)$ is obtained from $f$ by which transformation?
Answer: Horizontal compression by factor $2$ (x-values are divided by $2$ ) This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. For g(x) = f(2x) with f(x) = x³, this replaces x with 2x, compressing the graph horizontally by a factor of 2, meaning x-values that produce the same y are half as large (e.g., g(1) = f(2) = 8, while f(1) = 1, so it squeezes toward the y-axis). Choice A correctly identifies this as a horizontal compression by factor 2, where x-values are divided by 2 to match the original function's output. An incorrect choice like B might mix up compression with stretch—since |2| > 1 in f(2x), it's compression, not stretch; stretches happen when |k| < 1. Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?
Let the parent function be $f(x)=\frac{1}{x}$ . Describe the transformations that change $f(x)$ to $g(x)=f(x)-2.$
Answer: Shift the graph down 2 units This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Since g(x) = f(x) - 2, we have the form f(x) + k where k = -2, which means we're adding -2 to all y-values, shifting every point on the graph down by 2 units. Choice A correctly identifies this as shifting the graph down 2 units. Choice B incorrectly suggests shifting up (which would be f(x) + 2), while C confuses vertical with horizontal shifts, and D mistakes it for compression. Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal.
Let the parent function be $f(x)=x^2$ . Which function represents a shift of the graph of $f$ right 3 units and down 2 units?
Answer: $g(x)=f(x-3)-2$ This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. To shift right 3 units, we need f(x - 3) (minus inside means right), and to shift down 2 units, we subtract 2 outside: g(x) = f(x - 3) - 2. Choice B correctly shows g(x) = f(x - 3) - 2 for a shift right 3 and down 2. Students often mix up the signs for horizontal shifts, thinking f(x + 3) shifts right, but remember: inside transformations work opposite to intuition. Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects).
Let $f(x)=x^3$ be the parent function. What transformation changes $f(x)$ to $g(x)=-f(x)$ ?
Answer: Reflection across the $x$ -axis This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. Here, g(x) = -f(x) multiplies the cubic by -1 outside, flipping it over the x-axis—terrific insight into reflections! Choice B correctly identifies the transformation as a reflection across the x-axis. Choice A confuses it with y-axis reflection, which would be f(-x) instead—keep those straight, and you're set! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?
Let the parent function be $f(x)=x^2$ . Describe the transformations from $f(x)$ to $g(x)=2f(x-1)+3.$
Answer: Shift right 1; vertical stretch by 2; shift up 3 This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). For g(x) = 2f(x-1) + 3, we apply transformations in order: (x-1) shifts right 1 unit, multiplying by 2 stretches vertically by factor 2, and adding 3 shifts up 3 units. Choice B correctly identifies all three transformations: shift right 1, vertical stretch by 2, shift up 3. Choice A incorrectly states left shift and down shift, C confuses vertical stretch with horizontal, and D has the wrong horizontal shift direction. When multiple transformations combine, work from inside out: first handle what's inside the parentheses (horizontal effects), then multiplication outside (vertical stretch/compression), then addition/subtraction outside (vertical shifts).
Let the parent function be $f(x)=x^3$ . What is the effect on the graph of replacing $f(x)$ with $g(x)=f(x+2)$ ?
Answer: Shift left 2 units This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). In g(x) = f(x + 2), we're adding 2 to the input before applying f, which shifts the graph LEFT by 2 units—to get the same output, we need an x-value that's 2 less than before. Choice B correctly identifies this as shifting left 2 units. Choice A incorrectly suggests right (remember: the sign is opposite for horizontal shifts!), while C and D confuse horizontal shifts with vertical ones. The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?
The parent function $f(x)=\sqrt{x}$ has its graph starting at $(0,0)$ and increasing to the right. Which description matches the transformation to $g(x)=f(x)+3$ ?
Answer: Shift the graph up 3 units This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. For g(x) = f(x) + 3, this is a vertical shift outside the function, moving the entire graph up by 3 units, so the starting point goes from (0,0) to (0,3). Choice A correctly identifies the transformation as shifting the graph up 3 units. If you picked D, you might have mixed up addition with multiplication, but +3 is a shift, while 3f(x) would be a vertical stretch—keep practicing to distinguish them! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?
If $g(x) = f(kx)$ where $k > 0$ , and the period of $f(x)$ is 8, what value of $k$ would make the period of $g(x)$ equal to 2?
Answer: $k = 4$ When $f(x)$ has period $P$ , then $f(kx)$ has period $\frac{P}{k}$ . Given that $f(x)$ has period 8, $g(x) = f(kx)$ has period $\frac{8}{k}$ . We want this period to equal 2, so $\frac{8}{k} = 2$ , which gives $k = 4$ . Choice A gives period $\frac{8}{1/4} = 32$ . Choice B gives period $\frac{8}{1/2} = 16$ . Choice C gives period $\frac{8}{2} = 4$ .
Determine whether the function $f(x)=x^4-2x^2$ is even, odd, or neither.
Answer: Even, because $f(-x)=f(x)$ This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. Examples include f(x) = x², x⁴, |x|, and x² + 3. Odd functions have origin symmetry: f(-x) = -f(x), meaning rotating the graph 180° about the origin gives the same graph. Examples include f(x) = x, x³, 1/x, and x³ - x. Most functions are neither even nor odd! To check, compute f(-x) = $(-x)^4$ - 2 $(-x)^2$ = $x^4$ - $2x^2$ , which equals f(x), confirming y-axis symmetry. Choice A correctly determines it's even because f(-x) = f(x). If you chose C thinking mixed powers mean neither, that's understandable, but actually all powers here are even, making it even—odd powers would suggest possible oddness! For even/odd testing: (1) Take the given function f(x), (2) Find f(-x) by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with f(x) and -f(x): if f(-x) = f(x), it's even; if f(-x) = -f(x), it's odd; if neither match, it's neither. Example: f(x) = x² - 3, so f(-x) = (-x)² - 3 = x² - 3 = f(x) → even! Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation.
Let $f(x)=\sqrt{x}$ be the parent function. Describe the effect of $g(x)=f(2x)$ on the graph of $f$ .
Answer: Horizontal compression by a factor of 2 This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. With g(x) = f(2x) for the square root function, multiplying x by 2 (inside) squeezes the graph horizontally toward the y-axis, making it steeper and narrower—excellent work recognizing this! Choice B correctly identifies the transformation as a horizontal compression by a factor of 2. Choice A might confuse compression with stretch, but since |2| > 1, it's compression, not stretch—you've got this! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?
Let the parent function be $f(x)=x^2$ (standard parabola with vertex at $(0,0)$ ). Which description correctly gives the transformations from $f$ to $g(x)=2f(x-3)+1$ ?
Answer: Shift right 3, shift up 1, and stretch vertically by factor $2$ This question tests your understanding of how algebraic transformations of functions—like $f(x) + k$ , $k \cdot f(x)$ , $f(kx)$ , and $f(x + k)$ —affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) $f(x) + k$ shifts the graph vertically (up if $k > 0$ , down if $k < 0$ ), (2) $k \cdot f(x)$ stretches vertically if $|k| > 1$ or compresses if $0 < |k| < 1$ (and reflects across x-axis if $k < 0$ ), (3) $f(x + k)$ shifts horizontally—LEFT if $k > 0$ , RIGHT if $k < 0$ (opposite of what you might expect!), (4) $f(kx)$ compresses horizontally if $|k| > 1$ or stretches if $0 < |k| < 1$ (and reflects across y-axis if $k < 0$ ). Outside the function ( $f(x) + k$ and $k \cdot f(x)$ ) affects y-values/vertical; inside the function ( $f(x + k)$ and $f(kx)$ ) affects x-values/horizontal. For $g(x) = 2f(x - 3) + 1$ with $f(x) = x^2$ , first $f(x - 3)$ shifts the graph right by 3 units (vertex from $(0,0)$ to $(3,0)$ ), then multiplying by 2 stretches vertically by a factor of 2 (making it taller and narrower in appearance), and adding 1 shifts up by 1 unit (vertex to $(3,1)$ ). Choice B correctly identifies the transformations as a right shift of 3, up shift of 1, and vertical stretch by factor 2. A common mistake, like in choice A, is confusing the horizontal shift direction— $f(x - 3)$ shifts right, not left, and the vertical factor is a stretch (not compression) since $|2| > 1$ . Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE $f$ (like $f(x) + 3$ or $2f(x)$ ) changes y-values (vertical effects). Anything done INSIDE the parentheses (like $f(x + 3)$ or $f(2x)$ ) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign— $f(x + 3)$ shifts LEFT 3 because you're subtracting 3 from x-coordinates, and $f(x - 2)$ shifts RIGHT 2. Think: what x-value gives the original function's behavior?
Which statement correctly compares the effects of $f(2x)$ and $2f(x)$ on the graph of a parent function $f(x)$ ?
Answer: $f(2x)$ compresses horizontally by 2, while $2f(x)$ stretches vertically by 2 This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. Comparing f(2x) and 2f(x): f(2x) has 2 multiplying x INSIDE, causing horizontal compression by factor 2 (x-values divided by 2), while 2f(x) has 2 multiplying OUTSIDE, causing vertical stretch by factor 2 (y-values multiplied by 2). Choice B correctly states that f(2x) compresses horizontally by 2, while 2f(x) stretches vertically by 2. This is a classic comparison showing how location (inside vs outside) determines direction (horizontal vs vertical) of the transformation. Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects).

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