How do i find the asymptotes?
y=(1/x-4)+3
vertical asymptotes where the denominator is zero.
To find horizontal asymptotes, find y when x gets really huge. Remember that
1/∞ = 0
∞-4 = ∞
So an asymptote would be X=2.75
heavens! How did you get that?
y = 1/(x-4) + 3
x-4=0 when x=4
So, there is a vertical asymptote at x=4
Now, what happens when x->∞?
1/(x-4) -> 1/∞ -> 0
So, y=0+3 = 3
is a horizontal asymptote
See the graph at
http://www.wolframalpha.com/input/?i=1%2F%28x-4%29+%2B+3
To find the asymptotes of the given function, y = (1/x - 4) + 3, follow these steps:
1. Vertical Asymptote: Determine the values of x that would make the denominator of the fraction (1/x) equal to zero. In this case, the denominator is x - 4. Set it equal to zero and solve for x:
x - 4 = 0
x = 4
So, there is a vertical asymptote at x = 4.
2. Horizontal Asymptote: Investigate the behavior of the function as x approaches positive infinity and negative infinity. To do this, observe that as x gets larger, the fraction (1/x) gets smaller and approaches zero. Adding 3 to this result gives the horizontal asymptote:
y = 0 + 3
y = 3
Therefore, there is a horizontal asymptote at y = 3.
To summarize, the function y = (1/x - 4) + 3 has a vertical asymptote at x = 4 and a horizontal asymptote at y = 3.