Given:
The function is
![F(x)=\dfrac{1}{x+2}](https://tex.z-dn.net/?f=F%28x%29%3D%5Cdfrac%7B1%7D%7Bx%2B2%7D)
To find:
The vertical asymptote of the given function.
Solution:
Vertical asymptote are the vertical line passes thought the values for which the function is not defined.
To find the vertical asymptote, equate the denominator equal to 0.
We have,
![F(x)=\dfrac{1}{x+2}](https://tex.z-dn.net/?f=F%28x%29%3D%5Cdfrac%7B1%7D%7Bx%2B2%7D)
Denominator is (x+2).
![x+2=0](https://tex.z-dn.net/?f=x%2B2%3D0)
![x=-2](https://tex.z-dn.net/?f=x%3D-2)
The vertical asymptote is
.
Therefore, the correct option is B.
Answer:
the ansqer is g(x) = 4^x-4 + 2
Answer:8
Step-by-step explanation:
Answer:
Step 2
Step-by-step explanation:
Step 2 is not correct. It should be abs(25 - - 26) = abs(25 + 26) = 51
Given that Step 2 is accepted (as - 1) then three is OK and and so is 4. But technically they are incorrect as well. I think you are likely to say that step 2 is the problem.
6x^2-x-2 this is of the form ax^2+bx+c
jk=ac=-12 and j+k=b=-1 so j and k must be -4 and 3 so
6x^2+3x-4x-2
3x(2x+1)-2(2x+1)
(3x-2)(2x+1)