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14 votes
14 votes
X2 - 1 y= OA. x" +2x-3 OB. ソラ 1 + 2 x+1 312 Ocソ C. 2x - 1 2x - 1 ODY x-1

X2 - 1 y= OA. x" +2x-3 OB. ソラ 1 + 2 x+1 312 Ocソ C. 2x - 1 2x - 1 ODY x-1-example-1
X2 - 1 y= OA. x" +2x-3 OB. ソラ 1 + 2 x+1 312 Ocソ C. 2x - 1 2x - 1 ODY x-1-example-1
X2 - 1 y= OA. x" +2x-3 OB. ソラ 1 + 2 x+1 312 Ocソ C. 2x - 1 2x - 1 ODY x-1-example-2
User Mudokonman
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1 Answer

19 votes
19 votes

From the graph, we can notice that the function is not defined at x = 1 and x = -3. A rational function is not defined when the denominator has a value of 0. From the given options, only option A is not defined at -3 and 1, as proven here

At x = -3

(-3)² + 2(-3) - 3 = 9 - 6 - 3 = 0

The denominator equals 0 at x = -3, then this function is not defined at x = -3

At x = 1

(1)² + 2(1) - 3 = 1 + 2 - 3 = 3 - 3 = 0

The denominator equals 0 at x = 1, then this function is not defined at x = 1

Then, the function y = x² - 1 / x² + 2x - 3 is the one of the graph. Then option A is the correct answer

User HEngi
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