Graph Square Root and Piecewise Functions - Algebra 1

What is the starting point of $f(x)=\sqrt{x+1}$ ?
Answer: $(-1,0)$ 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!
A piecewise function $p(x)$ is defined such that $p(2) = 5$ , $p(4) = 3$ , and the function decreases linearly from $x = 2$ to $x = 4$ . For $x < 2$ , $p(x) = \sqrt{x+7}$ . What is the value of $p(2) + p(-2)$ ?
Answer: $5 + \sqrt{5}$ Given p(2) = 5. For p(-2), since -2 < 2, use p(x) = √(x+7). So p(-2) = √(-2+7) = √5. Therefore p(2) + p(-2) = 5 + √5. Choice B would require √5 = 3, which is incorrect. Choice C uses √7 instead of √5. Choice D assumes p(-2) = 5, which is incorrect.
A company's profit function is modeled by $P(x) = \begin{cases} -x^2 + 8x & \text{if } 0 \leq x \leq 4 \\ \sqrt{x + 12} + 12 & \text{if } x > 4 \end{cases}$ where $x$ represents months since January and $P(x)$ represents profit in thousands of dollars. What is the profit in May ( $x = 4$ ) minus the profit in July ( $x = 6$ )?
Answer: $4 - \sqrt{18}$ thousand dollars When you encounter a piecewise function, the key is identifying which piece to use for each input value by checking the given conditions carefully. To find the profit in May ( $x = 4$ ) minus the profit in July ( $x = 6$ ), you need to evaluate $P(4) - P(6)$ using the appropriate pieces of the function. For May ( $x = 4$ ): Since $0 \leq 4 \leq 4$ , use the first piece: $P(4) = -4^2 + 8(4) = -16 + 32 = 16$ thousand dollars. For July ( $x = 6$ ): Since $6 > 4$ , use the second piece: $P(6) = \sqrt{6 + 12} + 12 = \sqrt{18} + 12$ thousand dollars. Therefore: $P(4) - P(6) = 16 - (\sqrt{18} + 12) = 16 - \sqrt{18} - 12 = 4 - \sqrt{18}$ thousand dollars. Choice A ( $16 - \sqrt{18} - 12$ ) shows the intermediate step but wasn't simplified to $4 - \sqrt{18}$ . Choice B ( $-2 - \sqrt{18}$ ) likely results from incorrectly calculating $P(4)$ as 14 instead of 16, perhaps from an arithmetic error like $8 \times 4 = 24$ instead of 32. Choice D ( $\sqrt{18} - 4$ ) reverses the subtraction order, calculating $P(6) - P(4)$ instead of $P(4) - P(6)$ . Always double-check which piece of a piecewise function applies by carefully reading the conditions, and pay close attention to the order of operations when subtracting function values.
Where is the vertex of the absolute value function $g(x)=|x-4|+2$ ?
Answer: $(4,2)$ 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!
Graph the piecewise function $f(x)=\begin{cases}x+1 & \text{if } x<0\\2 & \text{if } 0\le x\le 2\\-x+4 & \text{if } x>2\end{cases}$ Which statement about endpoints is correct?
Answer: There is an open circle at $(0,1)$ and a closed circle at $(0,2)$ . This question tests your understanding of how to graph piecewise functions and identify their key features like boundaries and endpoints. Piecewise functions use different formulas on different parts of their domain: f(x) = {formula₁ if condition₁; formula₂ if condition₂} means 'use formula₁ when condition₁ is true, use formula₂ when condition₂ is true.' To graph them: graph each piece on its specified interval, paying attention to whether endpoints are included (closed circle •) or excluded (open circle ○). To graph this f(x): (1) Graph x+1 only for x-values where x < 0, checking the endpoint—use an open circle ○ at x=0 since it's strict inequality. (2) Graph 2 on 0 ≤ x ≤ 2 with closed circles • at both ends. (3) Graph -x+4 for x > 2 with open circle ○ at x=2. The result may have a jump discontinuity (if pieces don't connect) or be continuous (if they meet). This graph has jumps at x=0 and continuity at x=2. Choice A correctly shows an open circle at (0,1) for the left piece and a closed circle at (0,2) for the middle piece because at x=0, the left condition is x<0 (strict, so open) approaching y=1, and the middle includes x=0 with y=2 (closed). Choice B gets the piecewise boundaries wrong: it uses closed at (0,1) and open at (0,2), but since the left is x<0 (strict inequality), use open ○ at (0,1); the middle says x ≥ 0 (includes equality), so closed • at (0,2). These circles matter! For piecewise functions: make a plan before graphing: (1) Identify each piece and its interval, (2) Graph each piece ONLY on its interval, (3) Check endpoints—closed circle • means 'include this point,' open circle ○ means 'don't include,' (4) See if pieces connect (continuous) or jump (discontinuous). Being methodical with boundaries prevents errors! Quick shape recognition: Piecewise = combination of pieces (could be lines, curves, etc.). Memorizing these characteristic shapes helps you sketch quickly and recognize function types from graphs!
Graph the square root function $f(x)=\sqrt{x-2}+1$ . In your graph, show the starting point and indicate the domain.
Answer: Domain: $[2,\infty)$ ; starting point: $(2,1)$ This question tests your understanding of how to graph square root functions and identify their key features like domain and 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 - 2) + 1: (1) Find the starting point by setting (x - 2) = 0, giving x = 2, so we start at (2, 1). (2) Find a few more points: when x = 3, f = √1 + 1 = 2; when x = 6, f = √4 + 1 = 3. (3) Plot these points and connect with a smooth curve that starts at (2, 1) and curves upward to the right, flattening as it goes. (4) Remember: nothing to the left of x = 2 because the domain is x ≥ 2! Choice B correctly identifies the domain as [2, ∞) and the starting point at (2, 1) because solving x - 2 ≥ 0 gives x ≥ 2, and at x = 2, f(x) = 1. Choice C has the domain wrong: for f(x) = √(x - 2), we need what's under the radical to be non-negative: (x - 2) ≥ 0, which means x ≥ 2. This choice says [-2, ∞). 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.
What is the starting point of the square root function $f(x)=\sqrt{x+3}$ ?
Answer: $(-3,0)$ 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.
Graph the step function $f(x)=\lfloor x\rfloor$ on the interval $-2\le x\le 3$ . Which statement is true about the graph?
Answer: On $[1,2)$ the graph is the horizontal segment $y=1$ . 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 states that on [1,2) the graph is the horizontal segment y=1 because for x in [1,2), ⌊x⌋ = 1. Choice A has the step function jumping at the wrong places or with wrong values. The floor function ⌊x⌋ equals 1 for all x in [1, 2), jumping to 2 exactly at x = 2. This choice has y=2 there instead. 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!
Which statement about the graph of $f(x) = 2\sqrt{x-3} + 1$ is true?
Answer: The graph has domain $x \geq 3$ and passes through point $(7, 5)$ Domain: x - 3 ≥ 0, so x ≥ 3. Check point (7,5): f(7) = 2√(7-3) + 1 = 2√4 + 1 = 2(2) + 1 = 5. So (7,5) is on the graph. Check other points: f(4) = 2√1 + 1 = 3, so (4,3) is correct but choice A has right domain. Choice B has wrong domain. Choice C: f(7) = 5 ≠ 4, so (7,4) is not on the graph.
Graph the piecewise function $f(x)=\begin{cases} 2x+1 & \text{if } x<0\\ -x+1 & \text{if } x\ge 0 \end{cases}$ Which statement correctly describes what happens at $x=0$ on the graph?
Answer: There is an open circle at $(0,1)$ from the first piece and a closed circle at $(0,1)$ from the second piece. This question tests your understanding of how to graph piecewise functions and identify their key features like boundaries. Piecewise functions use different formulas on different parts of their domain: f(x) = {formula₁ if condition₁; formula₂ if condition₂} means 'use formula₁ when condition₁ is true, use formula₂ when condition₂ is true.' To graph them: graph each piece on its specified interval, paying attention to whether endpoints are included (closed circle •) or excluded (open circle ○). To graph f(x) = {2x + 1 if x < 0; -x + 1 if x ≥ 0}: (1) Graph 2x + 1 only for x-values where x < 0, checking the endpoint—use an open circle ○ at x = 0 since it's strict inequality < 0. (2) Graph -x + 1 on x ≥ 0 with closed circle • at x = 0. (3) The result may have a jump discontinuity (if pieces don't connect) or be continuous (if they meet). This graph is continuous since both pieces approach y = 1 at x = 0. Choice A correctly describes an open circle at (0,1) from the first piece and a closed circle at (0,1) from the second piece because the first condition is x < 0 (strict, so open) and the second is x ≥ 0 (includes equality, so closed). Choice B gets the piecewise boundaries wrong: it uses a closed circle at (0,0) from the first piece, but at x = 0, the first piece would be 2(0) + 1 = 1, not 0, and it's open anyway. When the condition says 'x < 0' (strict inequality), use an open circle ○; when it says 'x ≥ 0' (includes equality), use a closed circle •. These circles matter! For piecewise functions: make a plan before graphing: (1) Identify each piece and its interval, (2) Graph each piece ONLY on its interval, (3) Check endpoints—closed circle • means 'include this point,' open circle ○ means 'don't include,' (4) See if pieces connect (continuous) or jump (discontinuous). Being methodical with boundaries prevents errors!
For which value of $k$ will the function $g(x) = \sqrt{x + k}$ have a domain of $x \geq -5$ ?
Answer: $k = 5$ When working with square root functions, you need to remember that the expression under the radical must be non-negative (zero or positive) for the function to be defined in the real numbers. This constraint determines the domain of the function. For $g(x) = \sqrt{x + k}$ , the expression under the radical is $x + k$ . For the function to be defined, we need $x + k \geq 0$ . Solving this inequality for $x$ , we get $x \geq -k$ . This means the domain is all real numbers greater than or equal to $-k$ . Since we want the domain to be $x \geq -5$ , we need $-k = -5$ , which gives us $k = 5$ . Let's verify: if $k = 5$ , then $g(x) = \sqrt{x + 5}$ , and we need $x + 5 \geq 0$ , so $x \geq -5$ . Perfect! Looking at the wrong answers: Choice A gives $k = -5$ , which would create the domain $x \geq 5$ (much more restrictive than wanted). Choice B gives $k = 0$ , creating the domain $x \geq 0$ (also too restrictive). Choice D gives $k = 10$ , creating the domain $x \geq -10$ (too permissive, allowing values that should be excluded). Study tip: For square root domain problems, always set the expression under the radical ≥ 0, solve for x, and remember that adding a positive constant to x shifts the domain left (making it less restrictive), while adding a negative constant shifts it right (making it more restrictive).
Graph the piecewise function $f(x)=\begin{cases} 2x+1 & \text{if } x<0\\ -x+1 & \text{if } x\ge 0 \end{cases}$ Which statement correctly describes what happens at $x=0$ on the graph?
Answer: There is an open circle at $(0,1)$ from the first piece and a closed circle at $(0,1)$ from the second piece. This question tests your understanding of how to graph piecewise functions and identify their key features like boundaries. Piecewise functions use different formulas on different parts of their domain: f(x) = {formula₁ if condition₁; formula₂ if condition₂} means 'use formula₁ when condition₁ is true, use formula₂ when condition₂ is true.' To graph them: graph each piece on its specified interval, paying attention to whether endpoints are included (closed circle •) or excluded (open circle ○). To graph f(x) = {2x + 1 if x < 0; -x + 1 if x ≥ 0}: (1) Graph 2x + 1 only for x-values where x < 0, checking the endpoint—use an open circle ○ at x = 0 since it's strict inequality < 0. (2) Graph -x + 1 on x ≥ 0 with closed circle • at x = 0. (3) The result may have a jump discontinuity (if pieces don't connect) or be continuous (if they meet). This graph is continuous since both pieces approach y = 1 at x = 0. Choice A correctly describes an open circle at (0,1) from the first piece and a closed circle at (0,1) from the second piece because the first condition is x < 0 (strict, so open) and the second is x ≥ 0 (includes equality, so closed). Choice B gets the piecewise boundaries wrong: it uses a closed circle at (0,0) from the first piece, but at x = 0, the first piece would be 2(0) + 1 = 1, not 0, and it's open anyway. When the condition says 'x < 0' (strict inequality), use an open circle ○; when it says 'x ≥ 0' (includes equality), use a closed circle •. These circles matter! For piecewise functions: make a plan before graphing: (1) Identify each piece and its interval, (2) Graph each piece ONLY on its interval, (3) Check endpoints—closed circle • means 'include this point,' open circle ○ means 'don't include,' (4) See if pieces connect (continuous) or jump (discontinuous). Being methodical with boundaries prevents errors!
For the floor (greatest integer) function $f(x)=\lfloor x\rfloor$ , what is $f(2.7)$ ?
Answer: $2$ 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!
What is the starting point of the square root function $f(x)=\sqrt{x+3}$ ?
Answer: $(-3,0)$ 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.
Graph the cube root function $f(x)=\sqrt[3]{x+8}$ . What point is guaranteed to be on the graph?
Answer: $(-8,0)$ This question tests your understanding of how to graph cube root functions and identify their key features like starting point. Cube root functions like f(x) = ∛(x - h) + k have an S-like shape passing through the point (h, k), and unlike square roots, they're defined for all real numbers since cube roots work with negatives. The graph goes through the point where the inside is zero, and curves gently, steeper in the middle. To graph f(x) = ∛(x + 8): (1) Find the point by setting (x + 8) = 0, giving x = -8, so at (-8, 0). (2) Find more points: when x = -8 + 1 = -7, ∛1 = 1; when x = -8 -1 = -9, ∛(-1) = -1. (3) Plot these and connect with a smooth S-curve through (-8, 0), going up to the right and down to the left. (4) It extends infinitely in both directions! Choice A correctly identifies the point (-8, 0) on the graph because at x = -8, ∛(0) = 0. Choice B places the point at the wrong location: for ∛(x + 8), the key point is where x + 8 = 0, so x = -8, not (0, 0). The sign in (x + 8) means shift left by 8! Quick shape recognition: Square root = curved start at a point then gradually flatten upward. Absolute value = sharp V-shape with vertex at the point. Piecewise = combination of pieces (could be lines, curves, etc.). Step function = horizontal stairs with jumps. Cube root = S-curve through origin. Memorizing these characteristic shapes helps you sketch quickly and recognize function types from graphs!
Graph $f(x)=|x+1|$ . Which point is the vertex of the graph?
Answer: $(-1,0)$ 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 + 1|, which is |x - (-1)| + 0, the vertex is at (-1, 0)—this is where the inside equals zero. The graph makes a V: for x < -1, the slope is -1 (going down left toward vertex), and for x ≥ -1, the slope is +1 (going up right from vertex). Plot the vertex (-1, 0), 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 (-1, 0) because for |x + 1|, the vertex is where x + 1 = 0, so x = -1, and f(-1) = 0. Choice B places the vertex at the wrong location: for |x + 1|, the vertex is where the inside equals zero (x + 1 = 0), giving x = -1. So vertex is at (-1, 0), not (1, 0). The sign in |x + h| can be tricky—it shifts left! 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!
Consider the piecewise function $f(x) = \begin{cases} 2\sqrt{x-1} & \text{if } x \geq 1 \\ -x + 3 & \text{if } x < 1 \end{cases}$ . What is the value of $f(5) - f(-1)$ ?
Answer: $0$ For f(5): Since 5 ≥ 1, use f(x) = 2√(x-1). So f(5) = 2√(5-1) = 2√4 = 2(2) = 4. For f(-1): Since -1 < 1, use f(x) = -x + 3. So f(-1) = -(-1) + 3 = 1 + 3 = 4. Therefore f(5) - f(-1) = 4 - 4 = 0. Choice B results from calculating f(5) correctly but making an error with f(-1). Choice C comes from adding instead of subtracting. Choice D results from errors in both function evaluations.
What are the key features of the cube root parent function $f(x)=\sqrt[3]{x}$ ?
Answer: Domain all real numbers, range all real numbers, increasing everywhere, S-shaped through $(0,0)$ . This question tests your understanding of how to graph cube root functions and identify their key features like domain and shape. Cube root functions like f(x) = ∛(x - h) + k have an S-shaped curve passing through (h, k), defined for all real numbers since cube roots handle negatives. The graph increases everywhere but flattens on the ends—it's like a stretched S through the point. You can think of cube root as the inverse of cubing: ∛(-8) = -2 because $(-2)^3$ = -8. To graph f(x) = ∛x: (1) It passes through (0,0). (2) Find points: at x=1, ∛1=1; at x=8, ∛8=2; at x=-1, ∛-1=-1; at x=-8, ∛-8=-2. (3) Connect with a smooth S-curve: steeper in the middle, flattening out left and right. (4) Domain all reals, range all reals—goes forever in all directions! Choice B correctly identifies domain all real numbers, range all real numbers, increasing everywhere, S-shaped through (0,0) because cube roots are defined for negatives and create that characteristic S. Choice A has the domain wrong: for ∛x, it's all reals, not x ≥ 0 like square roots. Always remember cube roots allow negatives! Quick shape recognition: Cube root = S-curve through origin. Memorizing these characteristic shapes helps you sketch quickly and recognize function types from graphs! For domain, cube roots have no restrictions—unlike square roots which need inside ≥ 0.
Where is the vertex of $g(x)=|x-3|+2$ ?
Answer: $(3,2)$ 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!
Graph $h(x)=\sqrt{x}-2$ . Which point lies on the graph?
Answer: $(4,0)$ This question tests your understanding of how to graph square root functions and identify their key features like points on the graph. 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 h(x) = √x - 2: (1) Find the starting point by setting x = 0, giving (0, -2). (2) Find a few more points: when x = 1, h = √1 - 2 = -1; when x = 4, h = √4 - 2 = 0. (3) Plot these points and connect with a smooth curve that starts at (0, -2) and curves upward to the right, flattening as it goes. (4) Remember: nothing to the left of x = 0 because the domain is x ≥ 0! Choice A correctly shows the point (4,0) on the graph because when x=4, √4 - 2 = 2 - 2 = 0. Choice B shows a point that's not on the graph: for (1,0), but √1 - 2 = 1 - 2 = -1, not 0. Make sure to plug in x and compute y to verify points! For square root domain, remember: what's under the radical must be ≥ 0. For √x, set x ≥ 0. 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!

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