Question

What is the discontinuity and zero of the function f(x) = the quantity of 3 x squared plus x minus 4, all over x minus 1? A. Discontinuity at (-1, 1), zero at (four thirds, 0) B. Discontinuity at (-1, 1), zero at (negative four thirds , 0) C. Discontinuity at (1, 7), zero at (four thirds , 0) D. Discontinuity at (1, 7), zero at (negtive four thirds, 0)

Answer 1

The discontinuity is at (1, 7) and the zero is at (negative four thirds, 0) for the function f(x) = the quantity of 3 x squared plus x minus 4, all over x minus 1.

 

The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

Answer 2

Answer:

The correct option is D. Discontinuity at (1, 7), zero at (negative four thirds, 0)  

Step-by-step explanation:

$\text{The function is given to be : }f(x)=\frac{3x^2+x-4}{x-1}\\\\\implies f(x)=\frac{(3x+4)(x-1)}{(x-1)}$

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)