Answer:
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) = [tex]3^{x + 2}[/tex]
The asymptote is given as x → -∞, f(x) = [tex]3^{x + 2}[/tex] → 0
∴ The horizontal asymptote is f(x) = y = 0
The y-intercept is given when x = 0, we get;
f(x) = [tex]3^{0 + 2}[/tex] = 9
The y-intercept is f(x) = (0, 9)
b) f(x) = [tex]5^{1 - x}[/tex]
The asymptote is fx) = 0 as x → ∞
The asymptote is y = 0
Similar to question (1) above, the y-intercept is f(x) = [tex]5^{1 - 0}[/tex] = 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) = [tex]2^{x + 3}[/tex] - 1
The asymptote is [tex]2^{x + 3}[/tex] → 0 and f(x) → -1 as x → -∞
The horizontal asymptote is y = -1
The y-intercept is f(x) = [tex]2^{0 + 3}[/tex] - 1 = 7
The y-intercept is (0, 7)
When f(x) = 0, [tex]2^{x + 3}[/tex] - 1 = 0
[tex]2^{x + 3}[/tex] = 1
x + 3 = 0, x = -3
The x-intercept is (-3, 0)
f) [tex]f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2[/tex]
The asymptote is [tex]\left (\dfrac{1}{2} \right)^{x - 2}[/tex] → 0 and f(x) → 2 as x → ∞
The asymptote is y = 2
The y-intercept is f(x) = [tex]f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6[/tex]
The y-intercept is (0, 6)
