Solve the following equation for a. Be sure to take into account whether a letter is
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)
Answer:
Step-by-step explanation:
The range is the set of output values for the domain (the set of inputs).
- When x=-5, y=
.
- When x=0,
.
- When x=2,
So, the range is
Answer:
No because it contains points (3,4) and (3,-4). You cannot have a x assigned to more than one y-value if you want it to be a function.
Step-by-step explanation:
A function has one output per input.
If we are trying to determine if the given is a function of x, then x is the input.
However I can get two outputs from plugging in x=3.
Subtract 9 on both sides:
Take the square root of both sides:
.
So input x=3 yields y=4 and y=-4.
Since this input has more than one output then the given is not a function of x.
----Also!
If you graph the equation, it is a circle with radius 5 and center (0,0). So I could I plug in any number for x between -5 and 5 excluding -5 and 5 which would yield only one output each. Since plugging in either one gives:
Subtract 25 on both sides:
Simplify:
There is only one value y such that when you square it gives you 0. That is 0.
x=5 only gives y=0 and x=-5 only gives y=0.
There is no circle, unless it is a circle with radius 0 which means it really wouldn't be a circle, that is a function.