Need help in finding all vertical, horizontal, and slant asymptotes, x- and y-intercepts, and symmetries of f(x)=(x-3)/(x-4)
using quotient rule,
dy/dx - (x-4 - (x-3))/(x-4)^2
VA: x=4
HA: y=1
so slants
for y-intercept, let x = 0, then y =3/4
(0, 3/4)
for x-intercept , let y = 0
verification:
http://www.wolframalpha.com/input/?i=plot+y+%3D%28x-3%29%2F%28x-4%29
(x-3)/(x-4) = 0
x = 3 ---> x-intercept is (3, 0)
thank you!
To find the vertical asymptotes, set the denominator equal to zero and solve for x. In this case, x-4=0, so x=4. Hence, there is a vertical asymptote at x=4.
To find the horizontal asymptote, we need to compare the degrees of the numerator and denominator. The degree of the numerator is 1, and the degree of the denominator is also 1. Since the degrees are the same, we can find the horizontal asymptote by taking the ratio of the leading coefficients. In this case, the leading coefficients are both 1. Therefore, the horizontal asymptote is y=1.
To find the slant asymptote, we divide the numerator by the denominator using long division or synthetic division. Performing the division on (x-3) divided by (x-4), we get (1 + 1/(x-4)). This indicates a slant asymptote at y=x+1.
To find the x-intercept, we set y=0 and solve for x. In this case, 0=(x-3)/(x-4). Since the numerator can only be zero if x-3=0, we have x-3=0. Solving for x gives x=3. Therefore, the x-intercept is (3, 0).
To find the y-intercept, we set x=0 and solve for y. In this case, y=(0-3)/(0-4). Evaluating this expression gives y=3/4. Therefore, the y-intercept is (0, 3/4).
The function f(x)=(x-3)/(x-4) is not symmetric about the x-axis, the y-axis, or the origin.
Therefore, the properties of the function are as follows:
- Vertical asymptote: x=4
- Horizontal asymptote: y=1
- Slant asymptote: y=x+1
- x-intercept: (3, 0)
- y-intercept: (0, 3/4)
- No symmetry about the x-axis, y-axis, or origin.