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

1.

$$y=\dfrac{x^2+3x+5}{2x-3}$$

$2x-3\not=0=>x\not=\dfrac{3}{2}$

$$x=0: y=\dfrac{(0)^2+3(0)+5}{2(0)-3}=-\dfrac{5}{3}$$

$$y=0:\dfrac{x^2+3x+5}{2x-3}=0$$

$$x^2+3x+5=0, x\not=\dfrac{3}{2}$$

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

No solution.

$$\lim\limits_{x\to(3/2)^-}\dfrac{x^2+3x+5}{2x-3}=-\infin$$

$$\lim\limits_{x\to(3/2)^+}\dfrac{x^2+3x+5}{2x-3}=\infin$$

Vertical asymptote: $x=3/2$

$$\lim\limits_{x\to-\infin}\dfrac{x^2+3x+5}{2x-3}=-\infin$$

$$\lim\limits_{x\to\infin}\dfrac{x^2+3x+5}{2x-3}=\infin$$

There is no horizontal asymptote.

$$\dfrac{x^2+3x+5}{2x-3}=\dfrac{x^2-(3/2)x+(9/2)x-27/4+47/4}{2x-3}$$

$$=\dfrac{1}{2}x+\dfrac{9}{4}+\dfrac{47/4}{2x-3}$$

Slant (oblique) asymptote: $y=\dfrac{1}{2}x+\dfrac{9}{4}$

$$y'=(\dfrac{x^2+3x+5}{2x-3})'=$$

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

$$=\dfrac{4x^2-9-2x^2-6x-10}{(2x-3)^2}$$

$$=\dfrac{2x^2-6x-19}{(2x-3)^2}$$

Find the critical number(s)

$$y'=0=>\dfrac{2x^2-6x-19}{(2x-3)^2}=0$$ $$2x^2-6x-19=0, x\not=\dfrac{3}{2}$$

$$D=(-6)^2-4(2)(-19)=188>0$$

$$x=\dfrac{6\pm\sqrt{188}}{2(2)}=\dfrac{3\pm\sqrt{47}}{2}$$

Critical numbers:

$$\dfrac{3-\sqrt{47}}{2}, \dfrac{3}{2}, \dfrac{3+\sqrt{47}}{2}$$

If $x<\dfrac{3-\sqrt{47}}{2}, y'>0, y$ increases.

If $\dfrac{3-\sqrt{47}}{2}<x<\dfrac{3}{2}, y'<0, y$ decreases.

If $\dfrac{3}{2}<x<\dfrac{3+\sqrt{47}}{2}, y'<0, y$ decreases.

If $x>\dfrac{3+\sqrt{47}}{2}, y>0, y$ increases.

$$y(\dfrac{3-\sqrt{47}}{2})=\dfrac{(\dfrac{3-\sqrt{47}}{2})^2+3(\dfrac{3-\sqrt{47}}{2})+5}{2(\dfrac{3-\sqrt{47}}{2})-3}$$

$$=3-\dfrac{\sqrt{47}}{2}$$

$$y(\dfrac{3+\sqrt{47}}{2})=\dfrac{(\dfrac{3+\sqrt{47}}{2})^2+3(\dfrac{3+\sqrt{47}}{2})+5}{2(\dfrac{3+\sqrt{47}}{2})-3}$$

$$=3+\dfrac{\sqrt{47}}{2}$$

The function $y$ has a local maximum with value of $3-\dfrac{\sqrt{47}}{2}$ at $x=\dfrac{3-\sqrt{47}}{2}.$

The function $y$ has a local minimum with value of $3+\dfrac{\sqrt{47}}{2}$ at $x=\dfrac{3+\sqrt{47}}{2}.$

If $x<\dfrac{3}{2},$ the function $y$ increases from $-\infin$ to $3-\dfrac{\sqrt{47}}{2},$ and then decreases from $3-\dfrac{\sqrt{47}}{2}$ to $-\infin.$

Hence $y\in(-\infin, 3-\dfrac{\sqrt{47}}{2}],$ when $x\in(-\infin, \dfrac{3}{2}).$

If $x>\dfrac{3}{2},$ the function $y$ decreases from $\infin$ to $3+\dfrac{\sqrt{47}}{2},$ and then increases from $3+\dfrac{\sqrt{47}}{2}$ to $\infin.$

Hence $y\in[3+\dfrac{\sqrt{47}}{2},\infin)$ when $x\in(\dfrac{3}{2},\infin, ).$

(a) Domain:

$$(-\infin, \dfrac{3}{2})\cup (\dfrac{3}{2}, \infin).$$

(b) Range:

$$(-\infin, 3-\dfrac{\sqrt{47}}{2})\cup (3+\dfrac{\sqrt{47}}{2}, \infin).$$

(c) Determine the intercepts of the function.

$y$ - intercept: $(0, -\dfrac{5}{3})$

There are no $x$ - intercepts.

(d)

Vertical asymptote: $x=3/2$

There is no horizontal asymptote.

Slant (oblique) asymptote: $y=\dfrac{1}{2}x+\dfrac{9}{4}$

(e)

Turning point $(\dfrac{3-\sqrt{47}}{2}, 3-\dfrac{\sqrt{47}}{2})$ is a local maximum.

Turning point $(\dfrac{3+\sqrt{47}}{2}, 3+\dfrac{\sqrt{47}}{2})$ is a local minimum.

(f) Manually sketch the graph of the function.