Graph $f(x) =\dfrac{x^{2}-2x+6}{x-3}$.

Solution

Step 1 $f$ is a rational function; the domain of $f$ is $\{ x|x\neq 3\}$. There are no $x$-intercepts, since $x^{2}-2x+6=0$ has no real solutions. (Its discriminant is negative.) The $y$ -intercept is $f( 0) =-2$. Plot the intercept $( 0,-2).$

Step 2 We identify any vertical asymptotes by checking for infinite limits ($3$ is the only number to check). Since $\lim\limits_{x\rightarrow 3^{-}}\left( \dfrac{x^{2}-2x+6}{x-3}\right) =-\infty$ and $\lim\limits_{x\rightarrow 3^{+}}\left( \dfrac{x^{2}-2x+6}{x-3 }\right) =\infty$, the line $x=3$ is a vertical asymptote.

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The degree of the numerator of $f$ is 1 more than the degree of the denominator, so $f$ will have an oblique asymptote. We divide $x^{2}-2x+6$ by $x-3$ to find the line $y=mx+b$. $$f(x) =\frac{x^{2}-2x+6}{x-3}=x+1+\frac{9}{x-3}$$

Since $\lim\limits_{x\rightarrow \infty }[ f(x) -( x+1) ] =\lim\limits_{x\rightarrow \infty }\dfrac{9}{x-3}=0,$ then $y=x+1$ is an oblique asymptote of the graph of $f$. Draw the asymptotes on the graph.

Step 3 $$\begin{eqnarray*} f^\prime (x) &=&\dfrac{d}{\textit{dx}}\left( \dfrac{x^{2}-2x+6}{x-3}\right) =\dfrac{( 2x-2) (x-3) -( x^{2}-2x+6) ( 1) }{(x-3) ^{2}}\\ &=&\dfrac{ x^{2}-6x}{(x-3)^{2}}=\dfrac{x( x-6) }{(x-3) ^{2}}\\ f^{\prime \prime} (x) &=&\dfrac{d}{\textit{dx}}\left[ \dfrac{x^{2}-6x}{ (x-3) ^{2}}\right] =\dfrac{( 2x-6) (x-3) ^{2}-( x^{2}-6x) [ 2(x-3) ] }{(x-3) ^{4}} \\ &=&(x-3) \dfrac{( 2x^{2}-12x+18) -( 2x^{2}-12x) }{(x-3) ^{4}}=\dfrac{18}{(x-3) ^{3} } \end{eqnarray*}$$

$f^\prime (x) =0$ at $x=0$ and at $x=6$. So, $0$ and $6$ are critical numbers. The tangent lines are horizontal at $0$ and at $6.$ Since $3$ is not in the domain of $f$, $3$ is not a critical number.

Step 4 To apply the Increasing/Decreasing Function Test, we use the numbers $0,$ $3,$ and $6$ to form four intervals.

Step 5 Using the First Derivative Test, we find that $f( 0) =-2$ is a local maximum value and $f( 6) =10$ is a local minimum value. Plot the points $\left( 0,-2\right)$ and $\left( 6,10\right)$.

Step 6 $f^{\prime \prime} (x) =\dfrac{18}{(x-3) ^{3}}$. If $x<3$, $f^{\prime \prime} (x) <0$, and if $x>3$ , $f^{\prime \prime} (x) >0$. From the Test for Concavity, we conclude $f$ is concave down on the interval $\left( -\infty ,3\right)$ and $f$ is concave up on the interval $(3,\infty)$. The concavity changes at $3$, but $3$ is not in the domain of $f$. So, the graph of $f$ has no inflection point.

Step 7 We use the information to complete the graph of the function. See Figure 48.

Figure 48 $f(x)=\dfrac{x^2-2x+6}{x-3}$