Question

1. Consider the function y = x2 + 3x + 5. 2x−3

(a) Determine the domain of the function.

(b) Determine the range of the function.

(c) Determine the intercepts of the function.

(d) Find the asymptotes if they exist.

(e) Find the turning points (if they exist) and determine the type of turning points they are.

(f) Manually sketch the graph of the function.

Answer 1

$f(x)=\frac{x^2+3x+5}{2x-3}$

a)

denominator of f(x): $2x-3=0$ at $x=3/2$ , so

domain: $x\isin (-\infin,3/2)\cup (3/2,\infin)$

b)

$f(x)\to -\infin$ for $x\to -\infin$

$f(x)\to \infin$ for $x\to \infin$

$f(x)\to -\infin$ for $x\to (3/2)^-$

$f(x)\to \infin$ for $x\to (3/2)^+$

local maximum: $f(-1.93)=-0.43$

local minimum: $f(4.93)=6.43$

so:

range: $x\isin (-\infin,-0.43]\cup [6.43,\infin)$

c)

y-intercept: $(0,-5/3)$

for $x^2 + 3x + 5=0$ :

discriminant $D=9-20=-11<0$

so, $x^2 + 3x + 5>0$ and there are no x-intercepts

d)

$\displaystyle\lim_{x\to (3/2)^-}f(x)=-\infin$

$\displaystyle\lim_{x\to (3/2)^+}f(x)=\infin$

$x=3/2$ is vertical asymptote

for oblique asymptote:

$y=mx+n$

$m=\displaystyle\lim_{x\to (3/2)}f(x)/x=\displaystyle\lim_{x\to (3/2)}\frac{x^2+3x+5}{x(2x-3)}=\displaystyle\lim_{x\to (3/2)}\frac{2x+3}{4x-3}=2$

$n=\displaystyle\lim_{x\to (3/2)}(f(x)-mx)=\displaystyle\lim_{x\to (3/2)}(\frac{x^2+3x+5}{2x-3}-2x)=$

$=\displaystyle\lim_{x\to (3/2)}\frac{-6x+9}{x}=0$

$y=2x$ is oblique asymptote

e)

$f'(x)=\frac{(2x+3)(2x-3)-2(x^2+3x+5)}{(2x-3)^2}=\frac{2x^2-6x-19}{(2x-3)^2}=0$

$2x^2-6x-19=0$

$x=\frac{6\pm \sqrt{36+152}}{4}$

$x_1=-1.93,x_2=4.93$

$f(-1.93)=-0.43$

$f(4.93)=6.43$

at $x_1=-1.93$ f'(x) changes sign from "+" to "-", so this is local maximum

at $x_2=4.93$ f'(x) changes sign from "-" to "+", so this is local minimum

f)