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27 Oct 2025, 09:55
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Difficulty:
35%
(medium)
Question Stats:
57% (01:08) correct
42% (01:00) wrong based on 14 sessions
History
Date
Time
Result
Not Attempted Yet
The graphs \(y= x^2 + 4\) and \(y = |x| + 4\) have how many points in common?
A. none
B. one
C. two
D. three
E. four
A. none
B. one
C. two
D. three
E. four
02 Nov 2025, 00:32
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Official Explanation
Graph both equations. Each represents a standard graph, \(y = x^2 or y = |x|,\) shifted vertically upward by 4 units. The three points of intersection are (0, 4), (1, 5), and (-1, 5), so (D) is correct. Note that \(x^2 > |x|,\) except on the interval from –1 through 1. To understand why the graphs are equal when x = 0, 1, or –1, observe that \(0^2 = 0 = |0|,\) while \(1^2 = 1 = |1|,\) and \((–1)^2 = 1 = |–1|.\)
Answer: D
GMAT-Club-Forum-34bfo2s9.png [ 70.83 KiB | Viewed 317 times ]
Graph both equations. Each represents a standard graph, \(y = x^2 or y = |x|,\) shifted vertically upward by 4 units. The three points of intersection are (0, 4), (1, 5), and (-1, 5), so (D) is correct. Note that \(x^2 > |x|,\) except on the interval from –1 through 1. To understand why the graphs are equal when x = 0, 1, or –1, observe that \(0^2 = 0 = |0|,\) while \(1^2 = 1 = |1|,\) and \((–1)^2 = 1 = |–1|.\)
Answer: D
Attachment:
GMAT-Club-Forum-34bfo2s9.png [ 70.83 KiB | Viewed 317 times ]
