Question

what is the horizontal asymptote for the graph of y=3- ( (x-b)/(x+c) )


the answer is 2 however i do not know how to get it

Answer 1

lim_(x->±infinity) (3-(-b+x)/(c+x)) = 2

Answer 2

$$\lim_{x\to\pm\infty}\left(3-\frac{x-b}{x+c}\right)=3-\lim_{x\to\infty}\frac{x-b}{x+c}$$Recognize we can rewrite that rational function as:$$\frac{x-b}{x+c}=\frac{x+c-b-c}{x+c}=\frac{x+c}{x+c}-\frac{b+c}{x+c}=1-\frac{b+c}{x+c}$$Taking the limit of that as \(x\to\pm\infty\),$$\lim_{x\to\pm\infty}\left(1-\frac{b+c}{x+c}\right)=1-\lim_{x\to\pm\infty}\frac{b+c}{x+c}=1$$putting it all together:$$\lim_{x\to\pm\infty}\left(3-\frac{x-b}{x+c}\right)=3-1=2$$

Answer 3

here is a really simple method... which doesn't need limits

rewrite the equation with a common denominator

\[y = \frac{3(x + c)}{x + c} - \frac{-b+ x}{x + c}\]

which simplifies to

\[y = \frac{3x + 3x +b - x}{x + c}...or.... y=\frac{2x +3c +b}{x + c}\]

the numerator and the denominator are the same degree... so look at the coefficients of the leading terms to get the hoizontal asymptote...

asymptote is at \[y = \frac{2}{1} ... or.... y = 2\]

quite simple really

Answer 4

finally understood! thank you