At what value of x does graph of the following function f(x) have a vertical asymptote? F(x)= 1/x+3
1 answer:
Answer:
A
Step-by-step explanation:
Given
f(x) =
The denominator of f(x) cannot be zero as this would make f(x) undefined.
Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.
x + 3 = 0 → x = - 3 is the vertical asymptote
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Answer:
b=4
Step-by-step explanation:
So, we have the function . We need to find b such that the average rate of change or the slope is -1/8 between the intervel [2, b]. First, let's find f(2).
f(2) = 1/(2) = 1/2
So, we have the point (2, 1/2)
At point b, f(b) = 1/b.
Let's plug this into the slope formula:
Now, we just need to solve for b. First, let's multiply both the numerator and denominator by b (to get rid of the annoying fraction in the numerator).
Now, cross multiply.
Solve for b. Factor using the numbers -4 and -2.
Thus, b=4 or b=2.
However, b=2 is not a possible solution since the interval [2,2] means nothing. Thus, b=4.