Can you please find the hole(s) in these three functions?
f(x) = x^2-x-6 / 2x^2 -x- 6
f(x) = -x^3-7x^2-12x / 3x^2-11x-20
f(x) = x^2+4x-5 / x^4-4x^3+3x^2
Thanks so much! If you can't do one don't worry about it! All the work helps!
Can you please find the hole(s) in these three functions?
f(x) = x^2-x-6 / 2x^2 -x- 6
f(x) = -x^3-7x^2-12x / 3x^2-11x-20
f(x) = x^2+4x-5 / x^4-4x^3+3x^2
Thanks so much! If you can't do one don't worry about it! All the work helps!
Hello, Sbach2010!
Find the hole(s) in these three functions.
. . $\displaystyle f(x) \:= \:\frac{x^2-x-6}{2x^2 -x- 6}$
Factor: .$\displaystyle f(x) \;=\;\frac{(x+2)(x-3)}{(x-2)(2x+3)}$
It has vertical asymptotes at: .$\displaystyle x \,=\,2\text{ and }x \,=\,-\tfrac{3}{2}$
. . but no holes
$\displaystyle f(x) \:= \:\frac{-x^3-7x^2-12x}{3x^2-11x-20}$
Factor: .$\displaystyle f(x)\;=\;\frac{-x(x+3)(x+4)}{(x-5)(3x+4)} $
It has vertical asymptotes at: .$\displaystyle x\,=\,5\text{ and }x\,=\,-\tfrac{4}{3}$
. . but no holes.
$\displaystyle f(x) \:= \:\frac{x^2+4x-5}{x^4-4x^3+3x^2}$
Factor: .$\displaystyle f(x)\;=\;\frac{(x-1)(x+5)}{x^2(x-1)(x-3)} $
It has vertical asymptotes at: .$\displaystyle x \,=\,0\text{ and }x\,=\,3$
. . and a hole at (1, -3).
To find the y-intecept, you need to set x = 0 for the equation
To find the x-intecept, you need to set y = 0 for the equation
To find the domain of the function, you can refer to this useful link for guidance 2a. Domain and Range of a Function
Hope it helps.