If the question is
g(x) = 1/(x-3) + 1 then
the vertical asymptote is x = 3
and the horizontal is y = 1
(I believe Damon asked you yesterday for a clarification about the confusing denominator, but today you typed it exactly the same way.)
give the domain of the function and identify any vertical or horizontal asymptotes.
g(x)= 1 / x-3 + 1
My answer was x = o and y=3 but this was wrong. Please explain!!
3 answers
There are no parentheses used in the equation. Its just 1 over x-3 and then plus 1 is on the right side.
when you say in "words"
"Its just 1 over x-3 and then plus 1" we know exactly what it means, namely
1/(x-3) + 1 as I guessed
but when you write
1/x - 3 + 1, the 1 is only divided by the x and then 3 is subtracted.
In your text it probably has a fraction bar to show the fraction. Since we cannot create a "fraction bar" in this format, we have to use brackets to clarify what we mean.
"Its just 1 over x-3 and then plus 1" we know exactly what it means, namely
1/(x-3) + 1 as I guessed
but when you write
1/x - 3 + 1, the 1 is only divided by the x and then 3 is subtracted.
In your text it probably has a fraction bar to show the fraction. Since we cannot create a "fraction bar" in this format, we have to use brackets to clarify what we mean.
1 answer