This question tests your understanding of how to graph absolute value functions and identify their key features like the vertex. Absolute value functions like f(x) = |x - h| + k create a V-shape with the vertex (point where direction changes) at (h, k). The graph is made of two linear pieces: one with positive slope for x ≥ h, one with negative slope for x < h. You can think of absolute value as 'distance from zero,' which is why |3| = 3 and |-3| = 3—both are 3 units away from zero! For g(x) = |x - 3| + 2, the vertex is at (3, 2)—this is where the inside equals zero. The graph makes a V: for x < 3, the slope is -1 (going down left toward vertex), and for x ≥ 3, the slope is +1 (going up right from vertex). Plot the vertex (3, 2), then draw two straight lines forming a V-shape: left side going down with slope -1, right side going up with slope +1. If there's a coefficient like 2|x - h|, the V is steeper! Choice C correctly locates the vertex at (3,2) because for |x - 3| + 2, the vertex is where x - 3 = 0, so x = 3, and then +2 gives y = 2. Choice A places the vertex at the wrong location: for |x - 3| + 2, the vertex is where the inside equals zero (x - 3 = 0), giving x = 3. So vertex is at (3, 2), not (3, -2)—watch the sign of the constant term! Absolute value vertex trick: the vertex is at (h, k) from f(x) = |x - h| + k, but watch the sign! |x - 3| has vertex at x = 3 (positive), while |x + 3| = |x - (-3)| has vertex at x = -3 (negative). The value that makes the inside equal zero is where the V points! Quick shape recognition: Absolute value = sharp V-shape with vertex at the point. Memorizing these characteristic shapes helps you sketch quickly and recognize function types from graphs!

This question tests your understanding of how to graph square root functions and identify their key features like starting point. Square root functions like f(x) = √(x - h) + k have a characteristic curved shape starting at the point (h, k)—that's where the expression under the radical equals zero. The domain is restricted to x ≥ h because we can't take the square root of negative numbers (in the real number system). The graph curves upward from the starting point but flattens out as it goes—it's increasing but at a decreasing rate. To graph f(x) = √(x + 3): (1) Find the starting point by setting (x + 3) = 0, giving x = -3, so we start at (-3, 0). (2) Find a few more points: when x = -2, f = √1 = 1; when x = 1, f = √4 = 2. (3) Plot these points and connect with a smooth curve that starts at (-3, 0) and curves upward to the right, flattening as it goes. (4) Remember: nothing to the left of x = -3 because the domain is x ≥ -3! Choice B correctly locates the starting point at (-3, 0) because solving x + 3 = 0 gives x = -3, and f(-3) = √0 = 0. Choice C places the starting point at the wrong location: for √(x + 3), the starting point is where x + 3 = 0, giving x = -3. So starting point is at (-3, 0), not (3, 0). The sign in (x + 3) can be tricky—it shifts left! For square root domain, remember: what's under the radical must be ≥ 0. For √(x - 3), set (x - 3) ≥ 0 and solve: x ≥ 3. For √(2x + 4), set (2x + 4) ≥ 0 and solve: x ≥ -2. This 'set inside ≥ 0' rule works every time! The graph starts where the domain starts.

This question tests your understanding of how to graph square root functions and identify their key features like domain. Square root functions like f(x) = √(x - h) + k have a characteristic curved shape starting at the point (h, k)—that's where the expression under the radical equals zero. The domain is restricted to x ≥ h because we can't take the square root of negative numbers (in the real number system). The graph curves upward from the starting point but flattens out as it goes—it's increasing but at a decreasing rate. To graph f(x) = √(5 - x): (1) Find the starting point by setting (5 - x) = 0, giving x = 5, so we start at (5, 0). (2) Find a few more points: when x = 4, f = √1 = 1; when x = 1, f = √4 = 2. (3) Plot these points and connect with a smooth curve that starts at (5, 0) and curves upward to the left, flattening as it goes. (4) Remember: nothing to the right of x = 5 because the domain is x ≤ 5! Choice A correctly identifies the domain as (-∞, 5] because solving 5 - x ≥ 0 gives x ≤ 5, including x = 5 where f(x) = 0. Choice B has the domain wrong: for f(x) = √(5 - x), we need what's under the radical to be non-negative: (5 - x) ≥ 0, which means x ≤ 5. This choice says [5, ∞). Always solve the inequality 'inside ≥ 0' to find the square root domain! For square root domain, remember: what's under the radical must be ≥ 0. For √(x - 3), set (x - 3) ≥ 0 and solve: x ≥ 3. For √(2x + 4), set (2x + 4) ≥ 0 and solve: x ≥ -2. This 'set inside ≥ 0' rule works every time! The graph starts where the domain starts.

This question tests your understanding of how to graph square root functions and identify their key features like the starting point. Square root functions like f(x) = √(x - h) + k have a characteristic curved shape starting at the point (h, k)—that's where the expression under the radical equals zero. The domain is restricted to x ≥ h because we can't take the square root of negative numbers (in the real number system). The graph curves upward from the starting point but flattens out as it goes—it's increasing but at a decreasing rate. To graph f(x) = √(x + 1): (1) Find the starting point by setting (x + 1) = 0, giving x = -1, so we start at (-1, 0). (2) Find a few more points: when x = 0, f = √1 = 1; when x = 3, f = √4 = 2. (3) Plot these points and connect with a smooth curve that starts at (-1, 0) and curves upward to the right, flattening as it goes. (4) Remember: nothing to the left of x = -1 because the domain is x ≥ -1! Choice B correctly identifies the starting point as (-1,0) because that's where x + 1 = 0 and f(x) = 0. Choice A places the starting point at the wrong location: for √(x + 1), the starting point is where the inside equals zero (x + 1 = 0), giving x = -1. So starting point is at (-1, 0), not (0,1)—solve for where inside=0! For square root domain, remember: what's under the radical must be ≥ 0. For √(x + 1), set (x + 1) ≥ 0 and solve: x ≥ -1. This 'set inside ≥ 0' rule works every time! The graph starts where the domain starts. Quick shape recognition: Square root = curved start at a point then gradually flatten upward. Memorizing these characteristic shapes helps you sketch quickly and recognize function types from graphs!

This question tests your understanding of how to graph square root functions and identify their key features like domain. Square root functions like f(x) = √(x - h) + k have a characteristic curved shape starting at the point (h, k)—that's where the expression under the radical equals zero. The domain is restricted to x ≥ h because we can't take the square root of negative numbers (in the real number system). The graph curves upward from the starting point but flattens out as it goes—it's increasing but at a decreasing rate. To graph f(x) = √(5 - x): (1) Find the starting point by setting (5 - x) = 0, giving x = 5, so we start at (5, 0). (2) Find a few more points: when x = 4, f = √1 = 1; when x = 1, f = √4 = 2. (3) Plot these points and connect with a smooth curve that starts at (5, 0) and curves upward to the left, flattening as it goes. (4) Remember: nothing to the right of x = 5 because the domain is x ≤ 5! Choice A correctly identifies the domain as (-∞, 5] because solving 5 - x ≥ 0 gives x ≤ 5, including x = 5 where f(x) = 0. Choice B has the domain wrong: for f(x) = √(5 - x), we need what's under the radical to be non-negative: (5 - x) ≥ 0, which means x ≤ 5. This choice says [5, ∞). Always solve the inequality 'inside ≥ 0' to find the square root domain! For square root domain, remember: what's under the radical must be ≥ 0. For √(x - 3), set (x - 3) ≥ 0 and solve: x ≥ 3. For √(2x + 4), set (2x + 4) ≥ 0 and solve: x ≥ -2. This 'set inside ≥ 0' rule works every time! The graph starts where the domain starts.

This question tests your understanding of how to graph square root functions and identify their key features like domain. Square root functions like f(x) = √(x - h) + k have a characteristic curved shape starting at the point (h, k)—that's where the expression under the radical equals zero. The domain is restricted to x ≥ h because we can't take the square root of negative numbers (in the real number system). The graph curves upward from the starting point but flattens out as it goes—it's increasing but at a decreasing rate. To graph f(x) = √(x + 5): (1) Find the starting point by setting (x + 5) = 0, giving x = -5, so we start at (-5, 0). (2) Find a few more points: when x = -4, f = √1 = 1; when x = -1, f = √4 = 2. (3) Plot these points and connect with a smooth curve that starts at (-5, 0) and curves upward to the right, flattening as it goes. (4) Remember: nothing to the left of x = -5 because the domain is x ≥ -5! Choice B correctly identifies the domain as [-5, ∞) because we need what's under the radical to be non-negative: (x + 5) ≥ 0, which means x ≥ -5, and it includes x = -5 where f(x) = 0. Choice C has the domain wrong: for f(x) = √(x + 5), we need (x + 5) ≥ 0, which means x ≥ -5, but this choice says (-5, ∞), excluding x = -5 where it's defined. Always solve the inequality 'inside ≥ 0' to find the square root domain, and check if the endpoint is included! For square root domain, remember: what's under the radical must be ≥ 0. For √(x + 5), set (x + 5) ≥ 0 and solve: x ≥ -5. This 'set inside ≥ 0' rule works every time! The graph starts where the domain starts.

This question tests your understanding of how to graph step functions and identify their key features like boundaries. Step functions are constant on intervals but jump to different values at certain points: the floor function f(x) = ⌊x⌋ gives the greatest integer less than or equal to x, creating horizontal segments that jump up by 1 at each integer. So f(2.7) = 2, f(3.0) = 3, f(3.8) = 3—it 'steps up' at whole numbers. These model situations like postage rates or parking fees where cost jumps at thresholds. The greatest integer function f(x) = ⌊x⌋ creates horizontal steps: for any x in the interval [n, n+1), the function value is n (the greatest integer ≤ x). So 0 ≤ x < 1 gives f(x) = 0, 1 ≤ x < 2 gives f(x) = 1, etc. On each interval, draw a horizontal segment at height n with a closed circle on the left endpoint and open circle on the right. The graph looks like stairs going up! Choice B correctly gives f(2.7) = 2 because 2 is the greatest integer less than or equal to 2.7. Choice A has the step function jumping at the wrong places or with wrong values. The floor function ⌊x⌋ equals 2 for all x in [2, 3), jumping to 3 exactly at x = 3. This choice has 3, but for 2.7 it's 2. Step functions need precise boundaries! Step function evaluation is straightforward: ⌊2.7⌋ = 2, ⌊5.1⌋ = 5, ⌊-1.3⌋ = -2. Find the greatest integer that's still less than or equal to your number. For positive decimals, just drop the decimal (2.7 → 2). For negative decimals, go down to next integer (-1.3 → -2, not -1). Graphing: horizontal segment from each integer to the next, jumping at integer values!

This question tests your understanding of how to graph absolute value functions and identify their key features like vertex. Absolute value functions like f(x) = |x - h| + k create a V-shape with the vertex (point where direction changes) at (h, k). The graph is made of two linear pieces: one with positive slope for x ≥ h, one with negative slope for x < h. You can think of absolute value as 'distance from zero,' which is why |3| = 3 and |-3| = 3—both are 3 units away from zero! For f(x) = |x - 4| + 2, the vertex is at (4, 2)—this is where the inside equals zero. The graph makes a V: for x < 4, the slope is -1 (going down left toward vertex), and for x ≥ 4, the slope is +1 (going up right from vertex). Plot the vertex (4, 2), then draw two straight lines forming a V-shape: left side going down with slope -1, right side going up with slope +1. If there's a coefficient like 2|x - h|, the V is steeper! Choice C correctly locates the vertex at (4, 2) because for |x - 4| + 2, the vertex is where x - 4 = 0, so x = 4, and f(4) = 2. Choice D places the vertex at the wrong location: for |x - 4| + 2, the vertex is where the inside equals zero (x - 4 = 0), giving x = 4. So vertex is at (4, 2), not (-4, 2). The sign in |x - h| can be tricky—positive h means vertex at positive h! Absolute value vertex trick: the vertex is at (h, k) from f(x) = |x - h| + k, but watch the sign! |x - 3| has vertex at x = 3 (positive), while |x + 3| = |x - (-3)| has vertex at x = -3 (negative). The value that makes the inside equal zero is where the V points!

This question tests your understanding of how to graph square root functions and identify their key features like starting point. Square root functions like $f(x) = \sqrt{x - h} + k$ have a characteristic curved shape starting at the point $(h, k)$—that's where the expression under the radical equals zero. The domain is restricted to $x \geq h$ because we can't take the square root of negative numbers (in the real number system). The graph curves upward from the starting point but flattens out as it goes—it's increasing but at a decreasing rate. To graph $f(x) = \sqrt{x + 3}$: (1) Find the starting point by setting $(x + 3) = 0$, giving $x = -3$, so we start at $(-3, 0)$. (2) Find a few more points: when $x = -2$, $f = \sqrt{1} = 1$; when $x = 1$, $f = \sqrt{4} = 2$. (3) Plot these points and connect with a smooth curve that starts at $(-3, 0)$ and curves upward to the right, flattening as it goes. (4) Remember: nothing to the left of $x = -3$ because the domain is $x \geq -3$! Choice B correctly locates the starting point at $(-3, 0)$ because solving $x + 3 = 0$ gives $x = -3$, and $f(-3) = \sqrt{0} = 0$. Choice C places the starting point at the wrong location: for $\sqrt{x + 3}$, the starting point is where $x + 3 = 0$, giving $x = -3$. So starting point is at $(-3, 0)$, not $(3, 0)$. The sign in $(x + 3)$ can be tricky—it shifts left! For square root domain, remember: what's under the radical must be $\geq 0$. For $\sqrt{x - 3}$, set $(x - 3) \geq 0$ and solve: $x \geq 3$. For $\sqrt{2x + 4}$, set $(2x + 4) \geq 0$ and solve: $x \geq -2$. This 'set inside $\geq 0$' rule works every time! The graph starts where the domain starts.

This question tests your understanding of how to graph absolute value functions and identify their key features like vertex. Absolute value functions like f(x) = |x - h| + k create a V-shape with the vertex (point where direction changes) at (h, k). The graph is made of two linear pieces: one with positive slope for x ≥ h, one with negative slope for x < h. You can think of absolute value as 'distance from zero,' which is why |3| = 3 and |-3| = 3—both are 3 units away from zero! For f(x) = |x - 4| + 2, the vertex is at (4, 2)—this is where the inside equals zero. The graph makes a V: for x < 4, the slope is -1 (going down left toward vertex), and for x ≥ 4, the slope is +1 (going up right from vertex). Plot the vertex (4, 2), then draw two straight lines forming a V-shape: left side going down with slope -1, right side going up with slope +1. If there's a coefficient like 2|x - h|, the V is steeper! Choice C correctly locates the vertex at (4, 2) because for |x - 4| + 2, the vertex is where x - 4 = 0, so x = 4, and f(4) = 2. Choice D places the vertex at the wrong location: for |x - 4| + 2, the vertex is where the inside equals zero (x - 4 = 0), giving x = 4. So vertex is at (4, 2), not (-4, 2). The sign in |x - h| can be tricky—positive h means vertex at positive h! Absolute value vertex trick: the vertex is at (h, k) from f(x) = |x - h| + k, but watch the sign! |x - 3| has vertex at x = 3 (positive), while |x + 3| = |x - (-3)| has vertex at x = -3 (negative). The value that makes the inside equal zero is where the V points!