Which is the correct reason why (x+y)^2 is not equal to x^2+y^2
1 answer:
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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) =
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)
