f(x) =
(3x-7)^2 (x^2-9x+18) / (divided by)
x(x^2-12x+40)(x^2-16)
all asymptotes and intercepts are asked for
To find the asymptotes and intercepts of the given function, f(x), we need to analyze its numerator and denominator separately.
1. Asymptotes:
An asymptote is a line that the function approaches but never touches. For rational functions like the one given, vertical asymptotes occur when the denominator is equal to zero. Let's find those values:
Denominator:
x(x^2-12x+40)(x^2-16) = 0
First, determine the values that make the denominator zero:
x = 0 (from the first term)
x^2 - 12x + 40 = 0 (from the second term)
x^2 - 16 = 0 (from the third term)
For the second term, use the quadratic formula:
x = (-b ± √(b^2-4ac))/2a
x = (12 ± √((-12)^2 - 4(1)(40)))/(2(1))
x = 6 ± √(36-160)/2
x = 6 ± √(-124)/2
x = 6 ± 2i√(31)
So, the function has vertical asymptotes at x = 0, x = 6 + 2i√(31), and x = 6 - 2i√(31) (due to the imaginary part).
2. Intercepts:
Intercepts occur when the value of the function is equal to zero.
Numerator:
(3x-7)^2 (x^2-9x+18) = 0
The numerator can be zero when:
3x - 7 = 0 (from the first term)
x^2 - 9x + 18 = 0 (from the second term)
For the first term:
3x - 7 = 0
3x = 7
x = 7/3
For the second term, use the quadratic formula:
x = (-b ± √(b^2-4ac))/(2a)
x = (9 ± √((-9)^2 - 4(1)(18)))/(2(1))
x = 9 ± √(81 - 72)/2
x = 9 ± √9/2
x = 9 ± 3/2
x = (9 + 3)/2 = 6
x = (9 - 3)/2 = 3
So, the function has x-intercepts at x = 7/3, x = 6, and x = 3.
To summarize:
Vertical asymptotes: x = 0, x = 6 + 2i√(31), and x = 6 - 2i√(31) (due to imaginary parts)
x-intercepts: x = 7/3, x = 6, and x = 3