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Algebra

Algebra Help: Graph Square Root And Piecewise Functions

Review real example questions for Graph Square Root And Piecewise Functions in Algebra.

Question 1

Where is the vertex of g(x)=∣x−3∣+2g(x)=|x-3|+2g(x)=∣x−3∣+2?​​

  1. (2,3)(2,3)(2,3)
  2. (3,−2)(3,-2)(3,−2)
  3. (3,2)(3,2)(3,2)
  4. (−3,2)(-3,2)(−3,2)
Explanation: 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!

Question 2

What is the starting point of the square root function f(x)=x+3f(x)=\sqrt{x+3}f(x)=x+3​?​

  1. (0,3)(0,3)(0,3)
  2. (−3,0)(-3,0)(−3,0)
  3. (3,0)(3,0)(3,0)
  4. (0,3)(0,\sqrt{3})(0,3​)
Explanation: 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.

Question 3

Graph the square root function f(x)=x−2+1f(x)=\sqrt{x-2}+1f(x)=x−2​+1. In your graph, show the starting point and indicate the domain.​

  1. Domain: (−∞,∞)(-\infty,\infty)(−∞,∞); starting point: (2,1)(2,1)(2,1)
  2. Domain: [0,∞)[0,\infty)[0,∞); starting point: (0,1)(0,1)(0,1)
  3. Domain: [−2,∞)[-2,\infty)[−2,∞); starting point: (−2,1)(-2,1)(−2,1)
  4. Domain: [2,∞)[2,\infty)[2,∞); starting point: (2,1)(2,1)(2,1)
Explanation: 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.

Question 4

What is the domain of the function f(x)=5−xf(x)=\sqrt{5-x}f(x)=5−x​?

  1. (−∞,5](-\infty,5](−∞,5]
  2. [5,∞)[5,\infty)[5,∞)
  3. (5,∞)(5,\infty)(5,∞)
  4. (−∞,∞)(-\infty,\infty)(−∞,∞)
Explanation: 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.

Question 5

What is the starting point of f(x)=x+1f(x)=\sqrt{x+1}f(x)=x+1​?​

  1. (0,1)(0,1)(0,1)
  2. (1,0)(1,0)(1,0)
  3. (−1,0)(-1,0)(−1,0)
  4. (−1,1)(-1,1)(−1,1)
Explanation: 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!

Question 6

Graph the step function f(x)=⌊x⌋f(x)=\lfloor x\rfloorf(x)=⌊x⌋ on the interval −2≤x≤3-2\le x\le 3−2≤x≤3. Which statement is true about the graph?

  1. On (−1,0)(-1,0)(−1,0) the graph is the horizontal segment y=0y=0y=0.
  2. On [1,2)[1,2)[1,2) the graph is the horizontal segment y=1y=1y=1.
  3. On [1,2)[1,2)[1,2) the graph is the horizontal segment y=2y=2y=2.
  4. At x=2x=2x=2 the function value is 111.
Explanation: 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!

Question 7

Graph the cube root function f(x)=x+83f(x)=\sqrt[3]{x+8}f(x)=3x+8​. What point is guaranteed to be on the graph?

  1. (−8,0)(-8,0)(−8,0)
  2. (0,0)(0,0)(0,0)
  3. (8,2)(8,2)(8,2)
  4. (−8,−2)(-8,-2)(−8,−2)
Explanation: 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!

Question 8

Which graph represents f(x)=∣x+2∣f(x)=|x+2|f(x)=∣x+2∣?

  1. A parabola opening upward with vertex at (−2,0)(-2,0)(−2,0).
  2. A V-shaped graph opening upward with vertex at (2,0)(2,0)(2,0).
  3. A V-shaped graph opening upward with vertex at (−2,0)(-2,0)(−2,0).
  4. A square-root curve starting at (−2,0)(-2,0)(−2,0) and increasing to the right.
Explanation: This question tests your understanding of how to graph absolute value functions and identify their key features like the vertex and shape. 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 + 2|, which is |x - (-2)|, the vertex is at (-2, 0)—this is where the inside equals zero. The graph makes a V: for x < -2, the slope is -1 (going down left toward vertex), and for x ≥ -2, the slope is +1 (going up right from vertex). Plot the vertex (-2, 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 A correctly shows a V-shaped graph opening upward with vertex at (-2,0) because for |x + 2|, the vertex is where x + 2 = 0, so x = -2, y=0. Choice B places the vertex at the wrong location: for |x + 2| = |x - (-2)|, the vertex is at x = -2, not x=2—watch the sign inside! The value that makes the inside equal zero is where the V points! 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!

Question 9

Graph the piecewise function

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?​
  1. There is a closed circle at (0,0)(0,0)(0,0) from the first piece and an open circle at (0,0)(0,0)(0,0) from the second piece.
  2. There are closed circles at (0,1)(0,1)(0,1) from both pieces.
  3. There are open circles at (0,1)(0,1)(0,1) from both pieces.
  4. There is an open circle at (0,1)(0,1)(0,1) from the first piece and a closed circle at (0,1)(0,1)(0,1) from the second piece.
Explanation: 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!

Question 10

Graph the cube root function f(x)=x+83f(x)=\sqrt[3]{x+8}f(x)=3x+8​. What point is guaranteed to be on the graph?​

  1. (0,0)(0,0)(0,0)
  2. (−8,−2)(-8,-2)(−8,−2)
  3. (8,2)(8,2)(8,2)
  4. (−8,0)(-8,0)(−8,0)
Explanation: 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!