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Section 1.1: Four Ways to Represent a Function Edit

Exercise 1 Edit

a) y = f(-1) = -2

b) y = f(2) ≈ 2.8

c) This can be solved by determining the values of x where y = 2, which yields x = {-3, 1}.

d) This can be solved using the above method, so x ≈ {-2.5, 0.3}.

e) The x-values exist from -3 to 3, so the domain is -3 ≤ x ≤ 3, or [-3, 3]. The y-values exist from -2 to 3, so the range is -2 ≤ y ≤ 3, or [-2, 3].

f) f has a positive slope from -1 to 3, so the interval is [-1, 3].

Exercise 2 Edit

a) f(-4) = -2, g(3) = 4

b) This is solved by determining the values of x where f and g intersect, which yields x = {-2, 2}.

c) This can be solved by determining the values of x where y = -1, which yields x = {-3, 4}.

d) f has a negative slope from 0 to 4, so the interval is [0, 4].

e) The x-values of f exist from -4 to 4, so the domain is -4 ≤ x ≤ 4, or [-4, 4]. The y-values of f exist from -2 to 3, so the range is -2 ≤ y ≤ 3, or [-2, 3].

f) The x-values exist from -3 to 3, so the domain is -4 ≤ x ≤ 3, or [-4, 3]. The y-values exist from -2 to 3, so the range is -0.5 ≤ y ≤ 4, or [-0.5, 4].

Exercise 3 Edit

Figure 1: The lowest point is at about (12, -85) while the peak is at about (17, 115). Thus, the range is -85 ≤ a ≤ 115, or [-85, 115].

Figure 10: Using the above method, the range is -325 ≤ a ≤ 485, or [-85, 115].

Figure 11: Per above, the range is -210 ≤ a ≤ 200, or [-210, 200].

Exercise 5 Edit

No, this is not a function because it is possible for a vertical line to intersect the curve more than once.

Exercise 6 Edit

Yes, this is a function because a vertical line cannot intersect the curve more than once. The domain is [-2, 2] and range is [-1, 2].

Exercise 7 Edit

Yes, this is a function per exercise 6. The domain is [-3, 2] and range is [-3, 2) ∪ [-1, 3].

Exercise 8 Edit

No, this is not a function because there are vertical line segments, which result in an infinite number points for the same value of x.

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