Solution When sketching a rational function, we need to identify the critical features of the graph:

1. Vertical Asymptote(s)/Holes (Domain Restrictions)
2. End Behavior
3. Horizontal Intercepts
4. Vertical Intercepts

The order is not mandatory, rather we need all of the above before we can be confident in sketching the graph. In general, I recommend factoring the numerator/denominator of the rational function as our first step (although we will use the un-factored form too!)

\[ \solve{ f(x)&=&\dfrac{x^2+x-42}{x^2-x-30}\\ f(x)&=&\dfrac{(x+7)(x-6)}{(x+5)(x-6)} } \]

Now that the function is fully factored, it is easier to identify the zeros and domain restrictions, while the non-factored version of the equation is easier to identify the end behavior and vertical intercepts.

1. Domain Restrictions: Based on the fact there are no common factors between the numerator and denominator, there is one hole in the graph at \(x=6\). To calculate the \(y\) value of the hole, we simplify the rational function and then evaluate the result at \(x=6\): \(\dfrac{(x+7)(x-6)}{(x+5)(x-6)}=\dfrac{x+7}{x+5}\rightarrow y=\dfrac{{13}}{{11}}\). Thus, we have a hole located at \(\left(6,\frac{{13}}{{11}}\right)\). Based on the factored denominator, the vertical asymptote is \(x=-5\)
2. End Behavior: Using the non-factored form, we divide the leading terms: \(\dfrac{x^2}{x^2} = 1\). The constant result means we have a horizontal asymptote at \(y=1\).
3. Horizontal Intercepts: Using the factored form, we can see the zeros are at \(x= -7\). We must omit the other potential zero because \(x=6\) coincides with a domain restriction and is the location of a hole in the graph!.
4. Vertical Intercept: Using either form (although non-factored is easier) by setting \(x=0\) we find the vertical intercept. \(y=\dfrac{0^2+0-42}{0^2-0-30}=\dfrac{{42}}{{30}}=\dfrac{{7}}{{5}}\)

Now that we have all the features, we put it together for to sketch the graph.

Features:

Features + Graph: