Where are the asymptotes for the following function located? f(x)=14/(x-5)(x+1)
2 answers:
Given:
<span>f(x)=14/(x-5)(x+1)
</span><span>The asymptotes for the following function is located in </span><span>a. x=-1 and x=5.
x = -1 and x = 5 is the vetical asymptotes.
the horizontal aymptotes is located at y = 0.
there is no oblique asymptotes.
Pls. see attachment for addtl explanation. </span>
<span>f(x)=14/(x-5)(x+1)
</span><span>The asymptotes for the following function is located in </span><span>a. x=-1 and x=5.
x = -1 and x = 5 is the vetical asymptotes.
the horizontal aymptotes is located at y = 0.
there is no oblique asymptotes.
Pls. see attachment for addtl explanation. </span>
Answer:
Option a is correct.
x=-1 and x=5
Step-by-step explanation:
To find the asymptotes for the rational function:
The vertical asymptotes for this function is to set the denominator equals to 0.
The horizontal asymptotes for this given fucntion is 0.
Denominator of the given function is:
By definition of asymptotes;
By zero product property:
and
⇒x = 5 and x= -1
Therefore, the asymptotes for the following function located are: x = -1 and x =5
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Answer:
1)
2)
Step-by-step explanation:
Assuming that our function is for the first case and
for the second case.
Part 1
We can rewrite the expression like this:
And we can reorder the terms like this:
Now if we apply integral in both sides we got:
And after do the integrals we got:
Now we can use the initial condition
And the final solution would be:
Part 2
We can rewrite the expression like this:
And we can reorder the terms like this:
Now if we apply integral in both sides we got:
And after do the integrals we got:
Now we can use the initial condition
And the final solution would be: