I need hep to find the vertical and horizental asymptotes. f(x) = x − 3/x + 3, f(x)=x^2/x^2-9, f(x)=1-x^2/x^2-2x+1, f(x)=x^2-4/x^4-16, f(x)=x(x-1)(x-2)/x^4-x^2, f(x)=e^x, f(x) = ln x

Answer 1

Vertical asymptotes of rational functions occur where factors of a denominator is equal to zero (division by zero is undefined). If the numerator and the denominator have a common factor the graph will have a hole at the x-value where the factor would equal zero.

Ther are three cases to consider for horizontal asymptotes of rational functions.

(1) degrees of the numerator and denominator are equal will have a non-zero asymptote equal to the highest power of the numerator and denominator simplified to simplest form

(2) degree of the numerator is less than the degree of the denominator will have y = 0 as its horizontal asymptote.

(3) degree of the numerator is greater than the degree of the denominator will not have a horizontal asymptote.

(1) f(x) = (x-3) / (x + 3) VA: x = 3 HA: y = 1

(2) f(x) = (x²) / (x² - 9)

= (x²) / (x - 3)(x + 3) VA: x = -3, x = 3 HA: y = 1

(3) f(x) = (1 - x² ) / (x² - 2x + 1)

= (1 - x)(1 + x) / (x - 1)²

= -1 / (x - 1) VA: x = 1 HA: y = -1

(4) f(x) = (x² - 4) / (x⁴ - 16)

= (x - 2)(x + 2) / (x² -4)(x² + 4)

= (x - 2)(x + 2) / (x - 2)(x + 2)(x² + 4)

= 1 / (x² + 4) VA: none the graph of y = f(x) would have holes at x = 1 and x = -1

HA: y = 0

(5) f(x) = x(x - 1)(x - 2) / (x⁴ - x²)

= x(x - 1)(x - 2) / (x²)(x²- 1)

= x(x - 1)(x - 2) / (x²)(x - 1)(x + 1)

= (x - 2) / (x)(x + 1) VA: x = 0, x = -1 the graph of y = f(x) also has a hole at x = 1

HA: y = 0

(6) f(x) = e^x VA: none HA: y = 0

(7) f(x) = ln x VA: x = 0 HA: none