I asked this question and got an answer but am more confusted. Can someone please explain with the work and explanation of each one.

Question

Find the specified asymptotes of the following functions. Recall that asymptotes are lines therefore the answer must be given as an equation of a line. All should be done with answers, show work, explain in words. And answer them with (ex. a, b, c etc)

a) Find the vertical asymptote of the function f(x) = 4 over x + 5

b) Find the horizontal asymptote of the function g(x) = 5x^2 -4 over x + 1

c) Find the vertical and horizontal asymptote of the function f(x) = 3x – 1 over x +4

#c needs both answers and show work and explain in words.

d) Find the vertical and horizontal asymptote of the function g(x) = _x + 7_ over x2 - 4

#d needs both answers and show work and explain in words.

Answer 1

Char,

An asymptote of a curve is a line such that the distance between the curve and the line approaches zero as they tend to infinity. Horizontal asymptotes are parallel to the x-axis (fixed y number) that the graph of the function approaches as x tends to +∞ or −∞. Vertical asymptotes are vertical lines (perpendicular to the x-axis, fixed x number) near which f(x) tends to +∞ or −∞.

a) Find the vertical asymptote of the function f(x) = 4 over x + 5 = 4/(x+5)

Vertical, we are looking for an x number where f(x) -> ∞. In this case it happens when denominator=0, f(x)->∞, which is when x=-5.

b) Find the horizontal asymptote of the function g(x) = 5x^2 -4 over x + 1=(5x²-4)/(x+1)

Horizontal, we are looking for the g(x) when x tends to ∞. When x increases, g(x) -> 5x ( the -4 and +1 matter much less with x->∞, we ignore them). In this case, g(x) increases with x and does not -> to a fixed number (horizontal line). Therefore, there is no horizontal asymptote for this function.

c) Find the vertical and horizontal asymptote of the function f(x) = 3x – 1 over x +4=(3x-1)/(x+4)

Vertical, f(x) -> ∞, that happens again when denominator=x+4=0, therefore the line is x=-4.

Horizontal, x -> ∞, f(x) -> 3x/x=3 (again, with x -> ∞, -1 and +4 matter very little). The line is f(x)=3

d) Find the vertical and horizontal asymptote of the function g(x) = _x + 7_ over x2 - 4=(x+7)/(x²-4)

Vertical, g(x) -> ∞, that happens again when denominator=x²-4=0, and x=2 or x=-2, these are two vertical lines.

Horizontal, x -> ∞, g(x) ->x/x²=1/x, which ->0 with x -> ∞, therefore the horizontal line is g(x)=0.

Hope this helps.

Heng

Answer 2

a.  To find the vertical asymptote, set the denominator equal to zero and solve:

x + 5 =0

x = -5 is the vertical asymptote

 

b.  If the degree of the expression in the numerator is greater than the degree of the expression in the denominator, there is no horizontal asymptote.  If the degree of the numerator is one higher than the degree of the  denominator, there is an oblique asymptote.

Since the numerator has x to the second power, but the denominator only has x to the first power, the numerator has a degree that is one higher than the denominator.

There is no horizontal asymptote.  There is an oblique asymptote.

c.  Find the vertical asymptote as explained in part a:

x + 4 = 0

x = -4 vertical asymptote

To find the horizontal asymptote, divide the leading coefficient (number that multiplies by the highest power of variable) of the numerator by the leading coefficient of the denominator.

3x/x

y = 3 horizontal asymptote

 

d.  To find the vertical asymptote, we will use the method given in part a.

x² - 4 = 0

x² = 4

x = +/-2

The vertical asymptotes are at x = 2 and x = -2

If the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote will be y =0.

Since the numerator has an exponent of 1 and the denominator has an exponent of 2, the denominator has a higher degree than the numerator.

y = 0 is the horizontal asymptote