What is the radius of the inscribed circle of HJK?
1 answer:
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Answer:
The correct option is D. Discontinuity at (1, 7), zero at (negative four thirds, 0)
Step-by-step explanation:
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)
Answer:
Step-by-step explanation:
then the equation reads:
The standard form of an quadratic function is:
So the answer is: