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.

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! For h(x) = x⁴ - 2x², compute h(-x) = (-x)⁴ - 2(-x)² = x⁴ - 2x², which equals h(x), confirming y-axis symmetry without matching -h(x) = -x⁴ + 2x². Choice A correctly determines the function is even. A distractor like choice B might result from not fully simplifying h(-x) or confusing even with odd—note that the powers are all even (4 and 2), which often indicates even functions, but always verify algebraically. 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.

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 f(x) = x², g₁(x) = 2f(x) = 2x² multiplies y-values by 2 (vertical stretch by 2, making the parabola taller), while g₂(x) = f(2x) = (2x)² = 4x² compresses x-values by factor 2 (horizontal compression, squeezing toward y-axis so it appears steeper). Choice B correctly states that g₁ is a vertical stretch by factor 2, while g₂ is a horizontal compression by factor 2. A distractor like choice A swaps the effects—remember, multipliers outside affect vertical (y), inside affect horizontal (x), so g₁ is vertical and g₂ is horizontal. 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?

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?

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) = (1/2) f(x), multiplying the absolute value function by 1/2 (which is between 0 and 1) scales every y-value down, compressing the V-shape vertically toward the x-axis—you're doing awesome spotting these details! Choice B correctly determines this as a vertical compression by a factor of 1/2. Choice C might tempt you if you misread the coefficient as greater than 1, but since it's less than 1, it's compression, not stretch—great job double-checking! 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?

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.

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?

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. Breaking down g(x) = 2f(x-1) + 3: f(x-1) shifts the parabola right by 1, multiplying by 2 stretches it vertically, and adding 3 shifts it up—super job piecing this together! Choice B correctly identifies the transformation as shift right 1 unit, vertical stretch by 2, then shift up 3 units. Choice A mixes up directions and types, like shifting left instead of right and compressing instead of stretching—remember the signs and orders, and you're golden! 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?

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?

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) + 3 adds 3 to every y-value of the parent quadratic, resulting in a vertical shift upward by 3 units—keep up the great work noticing that! Choice B correctly identifies the transformation as a shift up 3 units. A common distractor like choice A might confuse the sign, suggesting a downward shift, but since we're adding a positive 3 outside, it's definitely upward—don't worry, practicing more will make this intuitive! 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?