We have the function:
[tex]f(x)=x+\frac{2}{x^2}-16.[/tex]We must find:
0. the intercepts,
,1. the vertical and horizontal asymptotes.
1) x-intercepts
The x-intercepts are given by the x values such that f(x) = 0. So we must find the values of x that satisfies the equation:
[tex]f(x)=x+\frac{2}{x^2}-16=0.[/tex]Solving for x, we get:
[tex]\begin{gathered} x+\frac{2}{x^2}-16=0 \\ x\cdot x^2+2-16\cdot x^2,\text{ }x\ne0, \\ x^3-16x^2+2=0. \end{gathered}[/tex]The real roots of this equation are:
[tex]\begin{gathered} x_1\approx15.9922, \\ x_2\approx0.35757, \\ x_3\approx-0.34975. \end{gathered}[/tex]So the x-intercepts are the points:
[tex]\begin{gathered} P_1=(15.9922,0), \\ P_2=(0.35757,0), \\ P_3=(-0.34975,0)\text{.} \end{gathered}[/tex]2) y-intercepts
The y-intercepts are given by the y values such that x = 0. Replacing x = 0 in the definition f(x), we see that the denominator of the second term diverges. So we conclude that there are no y-intercepts.
3) Vertical asymptotes
Vertical asymptotes are vertical lines near which the function grows without bound. From point 2, we know that the function grows without limit when x goes to zero. So one vertical asymptote is:
[tex]x=0.[/tex]4) Horizontal asymptotes
Horizontal asymptotes are horizontal lines that the graph of the function approaches when x → ±∞. We consider the limit of the function f(x) when x → ±∞:
[tex]\lim _{x\rightarrow\pm\infty}f(x)=\lim _{x\rightarrow\pm\infty}(x+\frac{2}{x^2}-16)\rightarrow\pm\infty.[/tex]We see that the function does not tend to any constant value when x → ±∞. So we conclude that there are no horizontal asymptotes.
5) Oblique asymptotes
When a linear asymptote is not parallel to the x- or y-axis, it is called an oblique asymptote or slant asymptote.
A function ƒ(x) is asymptotic to the straight line y = mx + n (m ≠ 0) if
[tex]{\displaystyle\lim _{x\to+\infty}\mleft[f(x)-(mx+n)\mright]=0\, {\mbox{ or }}\lim _{x\to-\infty}\mleft[f(x)-(mx+n)\mright]=0.}[/tex]We consider the line given by:
[tex]y=mx+n=x-16.[/tex]We compute the limit:
[tex]\begin{gathered} \lim _{x\rightarrow\pm\infty}(f(x)-(x-16)) \\ =\lim _{x\rightarrow\pm\infty}((x+\frac{2}{x^2}-16)-(x-16)) \\ =\lim _{x\rightarrow\pm\infty}(\frac{2}{x^2}) \\ =0. \end{gathered}[/tex]So we have proven that f(x) has the oblique asymptote:
[tex]y=x-16.[/tex]6) Graph
Plotting the intercepts and the asymptotes, we get the following graph:
Answer
1) x-intercepts: (-0.34975, 0), (0.35757, 0), (15.9922, 0)
2) y-intercepts: none
3) Vertical asymptotes: x = 0
4) Horizontal asymptotes: none
5) Oblique asympsotes: y = x -16
6) Graph
