# Analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of...

## Question:

Analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results.

{eq}\displaystyle y = \frac{x^2 - 6x + 12}{x - 4} {/eq}

## Point of inflection and concavity:

If {eq}x_0 {/eq} is a point of inflection of a function f twice derivable then {eq}f ''(x_0) = 0 {/eq}

If the function f has a point of inflection at {eq}x_0 {/eq} then the point P {eq}(x_0, f (x_0) {/eq} is called the inflection point of the graph of {eq}f(x) {/eq} In terms of the second derivative, a function {eq}f(x) {/eq} has a point of inflection at {eq}x_0 {/eq} if at this point there is a sign change of {eq}f ''(x) {/eq} At the point of inflection, continuity of the function is required.

Let {eq}f(x) {/eq} be a differentiable function on an interval I.

(a) We will say that the graph of {eq}f(x) {/eq} is concave up on I if {eq}f''(x) \geq 0 {/eq}

(b) We will say that the graph of {eq}f(x) {/eq} is concave down on I if {eq}f''(x) \leq 0 {/eq}

Domain {eq}\displaystyle y= \frac{x^2 - 6x + 12}{x - 4}\\ \boxed{\displaystyle D = \{ x \in R: \, x \, \neq \, 4 \}}\\ {/eq}

Intercepts {eq}\displaystyle y=\frac{x^2-6x+12}{x-4}\\ \displaystyle x=0 \,\,\, \rightarrow \,\,\, y=-3\\ \boxed{\displaystyle \left (0;-3 \right ) \,\,\, \rightarrow \,\,\, \textrm {interception point with the y-axis}}\\ \displaystyle y=0 \,\,\, \rightarrow \,\,\, \frac{x^2 - 6x + 12}{x - 4}=0\\ \displaystyle x^2-6x+12=0\\ \displaystyle D = b^2 - 4ac \,\,\, \rightarrow \,\,\, \textrm { discriminant formula} \\ \displaystyle D =(6)^2 -4(1)(12)\\ \displaystyle D =-12\\ \displaystyle D<0 \, \rightarrow \, \textrm {The function has no x-axis intercepts } {/eq}

Asymptotes

Vertical Asymptotes {eq}\textrm {For} \,\,\,\, x=4\\ \displaystyle \lim_{x \rightarrow 4^{-}} \frac{x^2 - 6x + 12}{x - 4}= -\infty \,\, \textrm {when the function approaches from the left to} \,\, x=4 \,\, \textrm {the function tends to} \,\, -\infty \displaystyle \lim_{x \rightarrow 4^{+}} \frac{x^2 - 6x + 12}{x - 4}= \infty \,\, \textrm {when the function approaches from the right to} \,\, x=4 \,\, \textrm {the function tends to} \,\, \infty\\ \displaystyle \boxed{ x=4} \,\,\, \rightarrow \,\,\, \textrm {Vertical asymptote of the function} {/eq}

Horizontal and Slant asymptotes {eq}\displaystyle m= \lim_{x \rightarrow \infty } \frac{y(x)}{x} \\ \displaystyle m= \lim_{x \rightarrow \infty } \frac{x^2 - 6x + 12}{x(x - 4)} \, \rightarrow \, \textrm {Applying l'Hospital's rule} \\ \displaystyle m= 1\\ \displaystyle b= \lim_{x \rightarrow \infty } (y(x)- mx)\\ \displaystyle b= \lim_{x \rightarrow \infty } \frac{x^2 - 6x + 12}{x - 4}-x\\ \displaystyle b= \lim_{x \rightarrow \infty } \frac{-2x + 12}{x - 4}\\ \displaystyle b=-2\\ \displaystyle \boxed{y=x-2} \,\, \rightarrow \,\, \textrm { Horizontal asymptote}\\ {/eq}

Conclusion (Horizontal asymptotes)

{eq}\displaystyle \boxed{y=\frac{3}{8}} \,\, \Longrightarrow \,\, \textrm { Horizontal asymptote}\\ {/eq}

Find all critical point

{eq}\displaystyle y(x) = \frac{x^2 - 6x + 12}{x - 4}\\ \displaystyle y'(x)= \frac{\left(2x-6\right)\left(x-4\right)-1\cdot \left(x^2-6x+12\right)}{\left(x-4\right)^2}\\ \displaystyle y'(x)= \frac{x^2-8x+12}{\left(x-4\right)^2}\\ \displaystyle y'(x)=0 \,\, \Rightarrow \,\, x^2-8x+12=0\\ \displaystyle \boxed{x=2 \,\,\, x=6} \,\, \Rightarrow \,\, \textrm {critical points } {/eq}

Second derivative test (local minimums and local maximum)

{eq}\displaystyle y'(x)= \frac{x^2-8x+12}{\left(x-4\right)^2}\\ \displaystyle y''(x)= \frac{\left(2x-8\right)\left(x-4\right)^2-2\left(x-4\right)\left(x^2-8x+12\right)}{\left(\left(x-4\right)^2\right)^2}\\ \displaystyle y''(x)= \frac{8}{\left(x-4\right)^3}\\ \displaystyle y''(2)=-1<0\\ \displaystyle y''(6)=1>0\\ \displaystyle \boxed{ \left(2,-2\right)} \,\, \Rightarrow \,\, \textrm {local maximum point of the function}\\ \displaystyle \boxed{ \left(6,6\right)} \,\, \Rightarrow \,\, \textrm {local minimum point of the function} {/eq}

To determine the points of inflection of the function, we find and analyze the second derivative:

{eq}\displaystyle y''(x)= \frac{8}{\left(x-4\right)^3}\\ \displaystyle y''(x) \neq 0 \,\,\,\, \forall \,\, x \, \in \, D \\ \displaystyle \boxed{\textrm {The function has no inflection point}} {/eq}

Sketch the graph