Question

Math question

Answer 1

I'm sorry, option B. is a wrong option

Answer 2

we have no vertical asymptotes, since the equation \(x^2+9=0\) has no real solutions

Answer 3

You need to try and calculate the limit of the function as it approaches infinity and x approaches negative infinity.

Answer 4

Sorry, meant as x approaches infinity.

Answer 5

hint:
if we divide both numerator and denominator by \(x^2\), we get:
\[f\left( x \right) = \frac{{{x^2} - 4}}{{{x^2} + 9}} = \frac{{1 - \left( {4/{x^2}} \right)}}{{1 - \left( {9/{x^2}} \right)}}\]
Now it is simple to compute the limit value of \(f(x)\) as \(x\) goes to \(\pm \infty\)

Answer 6

After you that, then you evaluate the function. I think. I'll just let @Michele_Laino take over. :)

Answer 7

to find vertical asy
we should set the denominator equal to zero and then solve for the variable \[x^2+9=0\]
to solve for x you should subtract 9 both sides
so u get x^2 = -9 ,(since it's negative ) as mentioned above there is no vertical asy

for\(\color{green}{\rm Horizontal ~asy.}\)
focus on highest degrees
~if the highest degree of the numerator is greater than the denominator then `No horizontal asy.` \[\color{reD}{\rm N}>\color{blue}{\rm D}\]
~if the highest degree of the denominator is greater than the highest degree of the numerator then `y=0` would be horizontal asy.
\[\rm \color{reD}{N}<\color{blue}{\rm D}\]~if both degrees are the same then divide the leading coefficient of the numerator by the leading coefficient of the denominator
\[\rm \color{red}{N}=\color{blue}{D}\]