Question

What is the discontinuity and zero of the function f(x) = 3x^2+x-4/x-1 ?

a. Discontinuity at (-1, 1), zero at ( 4/3 , 0)
b. Discontinuity at (-1, 1), zero at ( -4/3 , 0)
c. Discontinuity at (1, 7), zero at ( 4/3 , 0)
d. Discontinuity at (1, 7), zero at ( -4/3 , 0)

Answer 1

Set the denominator equal to zero and solve in order to find the x location of the discontinuity. Plug that value back in to the function in order to find the y location.

Answer 2

f(x) = 3x^2+x-4/x-1
= (3x+4)(x-1)/(x-1)
= 3x + 4
3x + 4 =0
3x = -4
x = -4/3
that is your zero.

Answer 3

The discontinuity. Like euler has said is simply
x - 1 =0
x = 1
so.
(1,-1), and (-4/3,0)

Answer 4

The function to plug the discontinuity into is the one that shamil98 found after factoring and simplifying... 3x+4. This returns 7 for the y value. If you had plugged that into the original, you would get 0/0 (whoops, minor oversight ;) )