How do you find the asymptotes for #f(x)=( -10x+3)/(8x+2)#?

vertical asymptotes occur as the denominator of a rational

function tends to zero.

solving 8x + 2 = 0 will give the asymptote

If the degree of numerator and denominator are equal

Which they are in this case , both of degree 1. then the

equation can be found by taking the ratio of leading

coefficients.

Here is the graph 0f f(x) graph{(-10x+3)/(8x+2) [-20, 20, -10, 10]}

To find the asymptotes of ( f(x) = \frac{-10x + 3}{8x + 2} ), we examine the behavior of the function as ( x ) approaches positive or negative infinity.

Vertical asymptotes occur where the denominator of the rational function equals zero and the numerator is nonzero. Thus, set the denominator equal to zero and solve for ( x ):

[ 8x + 2 = 0 ] [ x = -\frac{1}{4} ]

So, there is a vertical asymptote at ( x = -\frac{1}{4} ).

Horizontal asymptotes are found by comparing the degrees of the numerator and denominator of the rational function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at ( y = 0 ). If the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

In this case, the degree of the numerator is 1 and the degree of the denominator is also 1. So, to find the horizontal asymptote, divide the leading coefficient of the numerator by the leading coefficient of the denominator:

[ \frac{-10}{8} = -\frac{5}{4} ]

Therefore, the horizontal asymptote is ( y = -\frac{5}{4} ).