Question

How do you find domain, asymptotes, holes, intercepts for the given function \[f\left( x \right)=\dfrac{x+3}{{{x}^{2}}+8x+15}\]?

Answer 1

Hint: We start solving the problem by making use of the fact that the function $\dfrac{g\left( x \right)}{h\left( x \right)}$ defines only if $h\left( x \right)\ne 0$ to find the domain of the given function. We then factorize the denominator and then equate the common factors present in the numerator and denominator (if any) to zero to get the hole. We then simplify the given equation and equate the remaining factors of the denominator to 0 to get the asymptote(s) of the given function. We then substitute $x=0$ to get the y-intercept and then substitute $f\left( x \right)=0$ to get the x-intercept of the given function.

Complete step by step answer:
According to the problem, we are asked to find the domain, asymptotes, holes, intercepts for the given function \[f\left( x \right)=\dfrac{x+3}{{{x}^{2}}+8x+15}\].
We know that the function $\dfrac{g\left( x \right)}{h\left( x \right)}$ defines only if $h\left( x \right)\ne 0$. Now, let us find the values of x such that ${{x}^{2}}+8x+15=0$.
$\Rightarrow {{x}^{2}}+5x+3x+15=0$.
$\Rightarrow x\left( x+5 \right)+3\left( x+5 \right)=0$.
$\Rightarrow \left( x+3 \right)\left( x+5 \right)=0$.
$\Rightarrow x+3=0$, $x+5=0$.
$\Rightarrow x=-3$, $x=-5$.
So, the given function defines for every real value of x except at $x=-3,-5$.
So, the domain of the given function is $R-\left\{ -3,-5 \right\}$.
Now, let us simplify the given function \[f\left( x \right)=\dfrac{x+3}{\left( x+3 \right)\left( x+5 \right)}\]. We can see that the numerator and the denominator have a common factor \[\left( x+3 \right)\]. Now, let us equate this factor to zero to get the hole. So, we get $x+3=0\Leftrightarrow x=-3$.
So, the hole of the given function \[f\left( x \right)=\dfrac{x+3}{{{x}^{2}}+8x+15}\] as $x=-3$.
Now, the simplified function is \[f\left( x \right)=\dfrac{1}{x+5}\]. This makes that the function has asymptote at $x=-5$ as the function has hole discontinuity at $x=-3$.
We know that in order to find y-intercept, we substitute $x=0$ in the simplified function.
So, the y-intercept is \[f\left( 0 \right)=\dfrac{1}{0+5}=\dfrac{1}{5}\].
We know that in order to find x-intercept, we substitute $f\left( x \right)=0$ in the simplified function.
So, the x-intercept is \[0=\dfrac{1}{x+5}\], which is undefined.

Note:
We should perform each step carefully in order to avoid confusion and calculation mistakes. We should not report $x=-3$ as an asymptote which is the common mistake done by students. We should keep in mind that the domain of the given function should be found without simplifying it first. We should not report $x=\infty $ or $x=-\infty $ as x-intercept which is another mistake done by students. Similarly, we can expect problems to find the domain, asymptotes, holes, intercepts for the given function \[f\left( x \right)=\dfrac{{{x}^{3}}+x}{{{x}^{3}}-16{{x}^{2}}+64x}\].