Question

f(x) = 3(x-1)(x+2)/ (x-4) (x+2)

a) state vertical asymptotes

b) state the roots

Answer 1

ANSWER. a)The line $x=4$ is the vertical asymptote

b) $x=1$ is the only root of the function.

EXPLANATION

The function $f$ is a rational function since it has the form $f(x)=\frac { P(x) }{ Q(x) }$ where $P$ and $Q$ are polynomials. $P(x)=3(x-1)(x+2), Q(x)=(x-4)(x+2)$ .The domain of definition of function $f$ is the set $D=\left( -\infty ,+\infty \right) \diagdown \left\{ x:Q(x)=0 \right\} =\left( -\infty ,-2 \right) \cup \left( -2,4 \right) \cup \left( 4,+\infty \right) \ .$ If $x\in D$ then $f(x)=\frac { 3(x-1)\ }{ (x-4)\ } .$ In the domain of the definition of the function $f$ is continuous. Since $\lim _{ x\rightarrow -2 }{ f(x)=\lim _{ x\rightarrow -2 }{ \frac { 3(x-1)\quad }{ (x-4)\quad } = } \frac { 3\cdot (-2-1) }{ (-2-4) } =\frac { 3 }{ 2 } }$ , then $x=-2$ point of removable of discontinuity . a)The line $x=4$ is the vertical asymptote, because $\lim _{ x\rightarrow 4 }{ f(x)=\lim _{ x\rightarrow 4 }{ \frac { 3(x-1)\quad }{ (x-4)\quad } =\infty } \ }.$

b) $f(x)=0$ if and only if $x=1.$ So, $x=1$ is the only root of the function.