Question

Determine the horizontal vertical and slant asymptote y=x^2+2x-3/x-7

A.Vertical:x=7
Slant:y=x+9
B.horizontal:x=7
C.vertical:x=-1,x=3
Horizontal:y=7
D.vertical:x=7
Horizontal:y=-1,y=3
Slant:y=x-5

Answer 1

A.Vertical:x=7

Slant:y=x+9

Explanation:

`f(x)=(x^2+2x-3)/(x-7)\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_(x\to\pm\infty)(x^2+2x-3)/(x-7)=\lim\limits_(x\to\pm\infty)(x^2\left(1+(2)/(x)-(3)/(x^2)\right))/(x\left(1-(7)/(x)\right))=\lim\limits_(x\to\pm\infty)(x\left(1+(2)/(x)-(3)/(x^2)\right))/(1-(7)/(x))=\pm\infty\\\\\boxed{not\ exist}`

`slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_(x\to\pm\infty)(f(x))/(x)\\\\b=\lim\limits_(x\to\pm\infty)(f(x)-ax)\\\\a=\lim\limits_(x\to\pm\infty)((x^2+2x-3)/(x-7))/(x)=\lim\limits_(x\to\pm\infty)(x^2+2x-3)/(x(x-7))=\lim\limits_(x\to\pm\infty)(x^2+2x-3)/(x^2-7x)\\\\=\lim\limits_(x\to\pm\infty)(x^2\left(1+(2)/(x)-(3)/(x^2)\right))/(x^2\left(1-(7)/(x)\right))=\lim\limits_(x\to\pm\infty)(1+(2)/(x)-(3)/(x^2))/(1-(7)/(x))=(1)/(1)=1`

`b=\lim\limits_(x\to\pm\infty)\left((x^2+2x-3)/(x-7)-1x\right)=\lim\limits_(x\to\pm\infty)\left((x^2+2x-3)/(x-7)-(x(x-7))/(x-7)\right)\\\\=\lim\limits_(x\to\pm\infty)\left((x^2+2x-3)/(x-7)-(x^2-7x)/(x-7)\right)=\lim\limits_(x\to\pm\infty)(x^2+2x-3-(x^2-7x))/(x-7)\\\\=\lim\limits_(x\to\pm\infty)(x^2+2x-3-x^2+7x)/(x-7)=\lim\limits_(x\to\pm\infty)(9x-3)/(x-7)=\lim\limits_(x\to\pm\infty)(x\left(9-(3)/(x)\right))/(x\left(1-(7)/(x)\right))`

`=\lim\limits_(x\to\pm\infty)(9-(3)/(x))/(1-(7)/(x))=(9)/(1)=9\\\\\boxed{y=1x+9}`