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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)

Verified Answer :

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
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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)

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