In Exercises 27–32, obtain a slope field and add to it graphs of the solution curves passing through the given points. 27. y' = y with c. (0,–1) a. (0, 1) b. (0, 2) 28. y' = 2(y – 4) with a. (0, 1) b. (0, 4) c. (0, 5) 29. y' = y(x + y) with b. (0, –2) c. (0, 1/4) d. (-1,–1) a. (0, 1) 30. y' = y? with a. (0, 1) b. (0, 2) c. (0,–1) d. (0, 0) 31. y' = (y - 1)(x + 2) with b. (0, 1) c. (0, 3) d. (1, – 1) a. (0, –1) ху 32. y' with x? + 4 c. (-2V3, -4) b. (0, –6) a. (0, 2)

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In Exercises 27–32, obtain a slope field and add to it graphs of the
solution curves passing through the given points.
27. y' = y with
c. (0,–1)
a. (0, 1)
b. (0, 2)
28. y' = 2(y – 4) with
a. (0, 1)
b. (0, 4)
c. (0, 5)
29. y' = y(x + y) with
b. (0, –2)
c. (0, 1/4)
d. (-1,–1)
a. (0, 1)
30. y' = y? with
a. (0, 1)
b. (0, 2)
c. (0,–1)
d. (0, 0)
31. y' = (y - 1)(x + 2) with
b. (0, 1)
c. (0, 3)
d. (1, – 1)
a. (0, –1)
ху
32. y'
with
x? + 4
c. (-2V3, -4)
b. (0, –6)
a. (0, 2)
Transcribed Image Text:In Exercises 27–32, obtain a slope field and add to it graphs of the solution curves passing through the given points. 27. y' = y with c. (0,–1) a. (0, 1) b. (0, 2) 28. y' = 2(y – 4) with a. (0, 1) b. (0, 4) c. (0, 5) 29. y' = y(x + y) with b. (0, –2) c. (0, 1/4) d. (-1,–1) a. (0, 1) 30. y' = y? with a. (0, 1) b. (0, 2) c. (0,–1) d. (0, 0) 31. y' = (y - 1)(x + 2) with b. (0, 1) c. (0, 3) d. (1, – 1) a. (0, –1) ху 32. y' with x? + 4 c. (-2V3, -4) b. (0, –6) a. (0, 2)
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