Answer:
The correct option is D. Discontinuity at (1, 7), zero at (negative four thirds, 0)
Step-by-step explanation:
![\text{The function is given to be : }f(x)=\frac{3x^2+x-4}{x-1}\\\\\implies f(x)=\frac{(3x+4)(x-1)}{(x-1)}](https://tex.z-dn.net/?f=%5Ctext%7BThe%20function%20is%20given%20to%20be%20%3A%20%7Df%28x%29%3D%5Cfrac%7B3x%5E2%2Bx-4%7D%7Bx-1%7D%5C%5C%5C%5C%5Cimplies%20f%28x%29%3D%5Cfrac%7B%283x%2B4%29%28x-1%29%7D%7B%28x-1%29%7D)
To find the point of discontinuity :
Put the denominator equal to 0
⇒ x - 1 = 0
⇒ x = 1
Also, if the factor (x - 1) gets cancel, then it becomes a hole rather than a asymptote , ⇒ y = 3x + 4 at x = 1
⇒ y = 7
So, Point of discontinuity : (1, 7)
And the zero is : after cancelling the factor (x - 1) put the remaining factor = 0
⇒ 3x + 4 = 0
⇒ 3x = -4
⇒ x = negative four thirds ( zero of the function)
Therefore, The correct option is D. Discontinuity at (1, 7), zero at (negative four thirds, 0)