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Question 1
Let $f (x) = ∣ x^{2} - 4∣$ . Which of the following correctly describes all the corner points (sharp, non-differentiable points) of the graph of $y = f (x)$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
B $x = - 2$ and $x = 2$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the base function and its x-intercepts
The base function is $f (x) = x^{2} - 4$ , which factors as $(x - 2) (x + 2)$ . Its x-intercepts are at $x = - 2$ and $x = 2$ .
Step 2: Apply the rule for corner points of $y = |f(x)|$
For $y = ∣ f (x) ∣$ , corner points occur exactly at the x-intercepts of $f (x)$ , because that is where the sign of $f (x)$ changes. At these points, the graph transitions between $f (x)$ and $- f (x)$ , creating a sharp point.
Step 3: Check $x = 0$
At $x = 0$ , $f (0) = 0 - 4 = - 4 \neq = 0$ , so $x = 0$ is not an x-intercept and is not a corner point. The vertex of the base parabola does not produce a corner point.
Step 4: State the conclusion
Corner points occur at $x = - 2$ and $x = 2$ only.
Method #2Approach 2
Step 1: Identify what creates corner points
Corner points on $y = ∣ f (x) ∣$ occur at the zeros of $f (x)$ , where the negative portion is reflected upward.
Step 2: Eliminate '$x = 0$ only'
$f (0) = - 4 \neq = 0$ , so $x = 0$ is not an x-intercept of the base function and cannot be a corner point. This option is wrong.
Step 3: Eliminate '$x = -2$, $x = 0$, and $x = 2$'
This incorrectly includes $x = 0$ . Since $f (0) \neq = 0$ , $x = 0$ does not produce a corner point.
Step 4: Eliminate '$x = 2$ only'
Since the parabola $x^{2} - 4$ has two x-intercepts at $x = \pm 2$ , both produce corner points. Omitting $x = - 2$ is incorrect.
Step 5: Select the correct answer
The correct answer is ' $x = - 2$ and $x = 2$ ', as these are the only zeros of $f (x) = x^{2} - 4$ .
Question 2
The graph of $y = f (x)$ is sketched below, with x-intercepts at $x = - 3$ and $x = 1$ , and a minimum point at $(- 1, - 5)$ . Which of the following best describes the graph of $y = ∣ f (x) ∣$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
BThe portion of $y = f (x)$ between $x = - 3$ and $x = 1$ is reflected upward, giving a local maximum at $(- 1, 5)$ and corner points at $x = - 3$ and $x = 1$ .
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify where $f(x) < 0$
Since the x-intercepts are at $x = - 3$ and $x = 1$ and the minimum is at $(- 1, - 5)$ , the function $f (x) < 0$ on the interval $(- 3, 1)$ .
Step 2: Apply the $y = |f(x)|$ transformation
For $y = ∣ f (x) ∣$ , only the portion where $f (x) < 0$ is reflected upward. The portion on $(- 3, 1)$ is reflected, turning the minimum at $(- 1, - 5)$ into a local maximum at $(- 1, 5)$ .
Step 3: Identify corner points
Corner points appear at the x-intercepts $x = - 3$ and $x = 1$ , where the reflected and unreflected portions meet. The rest of the graph (outside $[- 3, 1]$ ) is unchanged.
Step 4: Confirm the correct description
The correct description is that the portion between $x = - 3$ and $x = 1$ is reflected upward, creating a local maximum at $(- 1, 5)$ and corner points at $x = - 3$ and $x = 1$ .
Method #2Approach 2
Step 1: Identify the key properties of $y = |f(x)|$
$y = ∣ f (x) ∣$ reflects negative portions upward; it does not shift the graph or necessarily create y-axis symmetry.
Step 2: Eliminate 'shifted upward by 5 units'
The modulus transformation is a reflection of negative parts, not a vertical translation. This option confuses modulus with a shift.
Step 3: Eliminate 'symmetric about the y-axis'
Symmetry about the y-axis is a property of $y = f (∣ x ∣)$ , not $y = ∣ f (x) ∣$ . This option confuses the two transformations.
Step 4: Eliminate 'entire graph reflected in the x-axis'
Only the portions where $f (x) < 0$ are reflected upward, not the entire graph. Reflecting the whole graph would give $y = - f (x)$ .
Step 5: Select the correct answer
The correct description is that the portion between $x = - 3$ and $x = 1$ is reflected upward, creating a local maximum at $(- 1, 5)$ and corner points at $x = - 3$ and $x = 1$ .

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13 more questions in this topic

Free preview 2/15

Question 1
Let $f (x) = ∣ x^{2} - 4∣$ . Which of the following correctly describes all the corner points (sharp, non-differentiable points) of the graph of $y = f (x)$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
B $x = - 2$ and $x = 2$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the base function and its x-intercepts
The base function is $f (x) = x^{2} - 4$ , which factors as $(x - 2) (x + 2)$ . Its x-intercepts are at $x = - 2$ and $x = 2$ .
Step 2: Apply the rule for corner points of $y = |f(x)|$
For $y = ∣ f (x) ∣$ , corner points occur exactly at the x-intercepts of $f (x)$ , because that is where the sign of $f (x)$ changes. At these points, the graph transitions between $f (x)$ and $- f (x)$ , creating a sharp point.
Step 3: Check $x = 0$
At $x = 0$ , $f (0) = 0 - 4 = - 4 \neq = 0$ , so $x = 0$ is not an x-intercept and is not a corner point. The vertex of the base parabola does not produce a corner point.
Step 4: State the conclusion
Corner points occur at $x = - 2$ and $x = 2$ only.
Method #2Approach 2
Step 1: Identify what creates corner points
Corner points on $y = ∣ f (x) ∣$ occur at the zeros of $f (x)$ , where the negative portion is reflected upward.
Step 2: Eliminate '$x = 0$ only'
$f (0) = - 4 \neq = 0$ , so $x = 0$ is not an x-intercept of the base function and cannot be a corner point. This option is wrong.
Step 3: Eliminate '$x = -2$, $x = 0$, and $x = 2$'
This incorrectly includes $x = 0$ . Since $f (0) \neq = 0$ , $x = 0$ does not produce a corner point.
Step 4: Eliminate '$x = 2$ only'
Since the parabola $x^{2} - 4$ has two x-intercepts at $x = \pm 2$ , both produce corner points. Omitting $x = - 2$ is incorrect.
Step 5: Select the correct answer
The correct answer is ' $x = - 2$ and $x = 2$ ', as these are the only zeros of $f (x) = x^{2} - 4$ .
Question 2
The graph of $y = f (x)$ is sketched below, with x-intercepts at $x = - 3$ and $x = 1$ , and a minimum point at $(- 1, - 5)$ . Which of the following best describes the graph of $y = ∣ f (x) ∣$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
BThe portion of $y = f (x)$ between $x = - 3$ and $x = 1$ is reflected upward, giving a local maximum at $(- 1, 5)$ and corner points at $x = - 3$ and $x = 1$ .
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify where $f(x) < 0$
Since the x-intercepts are at $x = - 3$ and $x = 1$ and the minimum is at $(- 1, - 5)$ , the function $f (x) < 0$ on the interval $(- 3, 1)$ .
Step 2: Apply the $y = |f(x)|$ transformation
For $y = ∣ f (x) ∣$ , only the portion where $f (x) < 0$ is reflected upward. The portion on $(- 3, 1)$ is reflected, turning the minimum at $(- 1, - 5)$ into a local maximum at $(- 1, 5)$ .
Step 3: Identify corner points
Corner points appear at the x-intercepts $x = - 3$ and $x = 1$ , where the reflected and unreflected portions meet. The rest of the graph (outside $[- 3, 1]$ ) is unchanged.
Step 4: Confirm the correct description
The correct description is that the portion between $x = - 3$ and $x = 1$ is reflected upward, creating a local maximum at $(- 1, 5)$ and corner points at $x = - 3$ and $x = 1$ .
Method #2Approach 2
Step 1: Identify the key properties of $y = |f(x)|$
$y = ∣ f (x) ∣$ reflects negative portions upward; it does not shift the graph or necessarily create y-axis symmetry.
Step 2: Eliminate 'shifted upward by 5 units'
The modulus transformation is a reflection of negative parts, not a vertical translation. This option confuses modulus with a shift.
Step 3: Eliminate 'symmetric about the y-axis'
Symmetry about the y-axis is a property of $y = f (∣ x ∣)$ , not $y = ∣ f (x) ∣$ . This option confuses the two transformations.
Step 4: Eliminate 'entire graph reflected in the x-axis'
Only the portions where $f (x) < 0$ are reflected upward, not the entire graph. Reflecting the whole graph would give $y = - f (x)$ .
Step 5: Select the correct answer
The correct description is that the portion between $x = - 3$ and $x = 1$ is reflected upward, creating a local maximum at $(- 1, 5)$ and corner points at $x = - 3$ and $x = 1$ .

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Step-by-step walkthrough

Step 1: Identify the base function and its x-intercepts

Step 2: Apply the rule for corner points of $y = |f(x)|$

Step 3: Check $x = 0$

Step 4: State the conclusion

Step 1: Identify what creates corner points

Step 2: Eliminate '$x = 0$ only'

Step 3: Eliminate '$x = -2$, $x = 0$, and $x = 2$'

Step 4: Eliminate '$x = 2$ only'

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Identify where $f(x) < 0$

Step 2: Apply the $y = |f(x)|$ transformation

Step 3: Identify corner points

Step 4: Confirm the correct description

Step 1: Identify the key properties of $y = |f(x)|$

Step 2: Eliminate 'shifted upward by 5 units'

Step 3: Eliminate 'symmetric about the y-axis'

Step 4: Eliminate 'entire graph reflected in the x-axis'

Step 5: Select the correct answer

13 more questions in this topic

Step-by-step walkthrough

Step 1: Identify the base function and its x-intercepts

Step 2: Apply the rule for corner points of $y = |f(x)|$

Step 3: Check $x = 0$

Step 4: State the conclusion

Step 1: Identify what creates corner points

Step 2: Eliminate '$x = 0$ only'

Step 3: Eliminate '$x = -2$, $x = 0$, and $x = 2$'

Step 4: Eliminate '$x = 2$ only'

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Identify where $f(x) < 0$

Step 2: Apply the $y = |f(x)|$ transformation

Step 3: Identify corner points

Step 4: Confirm the correct description

Step 1: Identify the key properties of $y = |f(x)|$

Step 2: Eliminate 'shifted upward by 5 units'

Step 3: Eliminate 'symmetric about the y-axis'

Step 4: Eliminate 'entire graph reflected in the x-axis'

Step 5: Select the correct answer

13 more questions in this topic