Question

How do you find the vertical, horizontal or slant asymptotes for `f(x)= (4x+8)/(x-3)`?

Answer 1

`VA: x-3=0->x=3,HA:f(x)=lim x->prop(4x)/x=4`

Explanation:

Take the denominator set it equal to 0 and solve for x to find the vertical asymptote. To find the horizontal asymptote find the limit as x goes to infinity and use the end behavior of a rational function. That is, pick the highest degree term from the top and bottom then simplify

Answer 2

`x=3` is a Vertical Asymptote
`y=4` is a Horizontal Asymptote
Slant Asymptote: None

Explanation:

At `x=3` the value of the function is not defined. Therefore the line approached by the graph but will never touch it no matter how far is `x=3`

Similarly, no matter how far the graph of the function increases or decreases to the right or to the left respectively, the line approached by the graph is `y=4`

Therefore `y=4` is a horizontal asymptote

See the graph of `y= (4x+8)/(x-3)`

graph{(y- (4x+8)/(x-3))=0[-40,40,-20,20]}

See also the graph of the asymptotes `x=3` and `y=4`

graph{(y-4)(y+(1000)x-3*(1000))=0[-40,40,-20,20]}

God bless ... I hope the explanation is useful