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a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) =

The asymptote is given as x → -∞, f(x) = → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = = 9

The y-intercept is f(x) = (0, 9)

b) f(x) =

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = - 1

The asymptote is  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =   - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, - 1 = 0

= 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f)

The asymptote is → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) =

The y-intercept is (0, 6)

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)