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. 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. To test f(x) = x³ - x, we find f(-x) = (-x)³ - (-x) = -x³ + x = -(x³ - x) = -f(x), confirming it's odd. Choice B correctly identifies this function as odd (symmetric about the origin). The function cannot be even since f(-x) = -f(x) ≠ f(x), and it clearly satisfies the odd function definition. Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation. Most functions are neither even nor odd!

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. 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.

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·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 this order: (1) f(x-1) shifts right 1 unit (since we have x minus 1), (2) multiplying by 2 gives 2f(x-1), which stretches vertically by factor 2, (3) adding 3 gives 2f(x-1) + 3, which shifts up 3 units. Choice B correctly identifies this sequence: shift right 1, stretch vertically by factor 2, then shift up 3. Choice A incorrectly states shift left instead of right, while C and D misidentify the transformation types entirely. 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.

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), we're multiplying all output values by -1, which reflects the graph across the x-axis—every point (x, y) becomes (x, -y). Choice A correctly identifies this as a reflection across the x-axis. Choice B incorrectly suggests y-axis reflection (that would be f(-x)), while C and D confuse reflections with shifts or compressions. Transformation memory aid: negative sign outside the function (-f(x)) flips y-values, creating x-axis reflection; negative sign inside the function (f(-x)) flips x-values, creating y-axis reflection.

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) = 2f(x-1) + 3, we apply transformations in this order: (1) f(x-1) shifts right 1 unit (minus inside means right), (2) 2f(x-1) stretches vertically by factor 2, (3) 2f(x-1) + 3 shifts up 3 units. Choice B correctly identifies these as shift right 1, shift up 3, then vertical stretch by a factor of 2. Note that the order listed in choice B differs slightly from the mathematical order of operations, but all three transformations are correctly identified. 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).

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. In g(x) = f(x) + 3, we're adding 3 OUTSIDE the function f(x), which means we're adding 3 to all y-values, shifting every point on the graph up by 3 units. Choice B correctly identifies the transformation as shifting the graph up 3 units. A common error is thinking +3 shifts down (that would be f(x) - 3), or confusing this with 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).

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. In g(x) = f(2x), we're multiplying x by 2 INSIDE the function, which compresses the graph horizontally by a factor of 2—to get the same y-value, we now need only half the x-value, so all x-coordinates are divided by 2. Choice B correctly describes this as horizontal compression by a factor of 2 (x-values are divided by 2). A common mistake is thinking f(2x) stretches horizontally (that would be f(x/2)) or affects the graph vertically. 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?

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 $$\frac{8}{2} = 4$$.