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In the 𝑥𝑦-plane, the graph of a rational function 𝑓 has a hole at 𝑥=2. Input values of 𝑓 sufficiently close to 2 correspond to output values arbitrarily close to 6. Which of the following could define 𝑓𝑥 ?Responses 𝑓𝑥=6𝑥-2𝑥+3𝑥-3𝑥-2f open parentheses x close parentheses equals fraction numerator 6 open parentheses x minus 2 close parentheses open parentheses x plus 3 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses x minus 2 close parentheses end fraction 𝑓𝑥=𝑥-2𝑥+4𝑥-2𝑥-1f open parentheses x close parentheses equals fraction numerator open parentheses x minus 2 close parentheses open parentheses x plus 4 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses x minus 1 close parentheses end fraction 𝑓𝑥=𝑥-6𝑥+4𝑥-6𝑥-1f open parentheses x close parentheses equals fraction numerator open parentheses x minus 6 close parentheses open parentheses x plus 4 close parentheses over denominator open parentheses x minus 6 close parentheses open parentheses x minus 1 close parentheses end fraction

Question

In the 𝑥𝑦-plane, the graph of a rational function 𝑓 has a hole at 𝑥=2. Input values of 𝑓 sufficiently close to 2 correspond to output values arbitrarily close to 6. Which of the following could define 𝑓𝑥 ?Responses 𝑓𝑥=6𝑥-2𝑥+3𝑥-3𝑥-2f open parentheses x close parentheses equals fraction numerator 6 open parentheses x minus 2 close parentheses open parentheses x plus 3 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses x minus 2 close parentheses end fraction 𝑓𝑥=𝑥-2𝑥+4𝑥-2𝑥-1f open parentheses x close parentheses equals fraction numerator open parentheses x minus 2 close parentheses open parentheses x plus 4 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses x minus 1 close parentheses end fraction 𝑓𝑥=𝑥-6𝑥+4𝑥-6𝑥-1f open parentheses x close parentheses equals fraction numerator open parentheses x minus 6 close parentheses open parentheses x plus 4 close parentheses over denominator open parentheses x minus 6 close parentheses open parentheses x minus 1 close parentheses end fraction

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Solution

The function that could define 𝑓𝑥 is 𝑓𝑥=𝑥-2𝑥+4𝑥-2𝑥-1.

Here's why:

A hole in the graph of a rational function occurs when the same factor is in the numerator and the denominator. In this case, the factor is (x-2), which is present in the numerator and denominator of the function 𝑓𝑥=𝑥-2𝑥+4𝑥-2𝑥-1.

Furthermore, when x is close to 2, the function should be close to 6. If we substitute x=2 in the function 𝑓𝑥=𝑥-2𝑥+4𝑥-2𝑥-1, we get 6, which satisfies the condition.

Therefore, the function 𝑓𝑥=𝑥-2𝑥+4𝑥-2𝑥-1 could define 𝑓𝑥.

This problem has been solved

Similar Questions

The function 𝑓 is given by 𝑓⁡(𝑥)=2(3⁢𝑥). Which of the following statements describes characteristics of the graph of 𝑓 in the 𝑥⁢𝑦-plane?ResponsesThe graph of 𝑓 is a vertical dilation of 𝑦=2𝑥, and 𝑓⁡(𝑥) is equivalent to 8𝑥.The graph of  f  is a vertical dilation of  y equals 2 to the power of x , and  f open parentheses x close parentheses  is equivalent to  8 to the power of x .The graph of 𝑓 is a vertical dilation of 𝑦=2𝑥, and 𝑓⁡(𝑥) is equivalent to 8·2𝑥.The graph of  f  is a vertical dilation of  y equals 2 to the power of x , and  f open parentheses x close parentheses  is equivalent to  8 times 2 to the power of x .The graph of 𝑓 is a horizontal dilation of 𝑦=2𝑥, and 𝑓⁡(𝑥) is equivalent to 8𝑥.The graph of  f  is a horizontal dilation of  y equals 2 to the power of x , and  f open parentheses x close parentheses  is equivalent to  8 to the power of x .The graph of 𝑓 is a horizontal dilation of 𝑦=2𝑥, and 𝑓⁡(𝑥) is equivalent to 8·2𝑥.

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QUESTION 4Given: and𝑓(𝑥) = 3𝑥−2 + 1 ℎ(𝑥) =− 1𝑥+524.1 Write down the equation of the asymptotes of 𝑓. (2)4.2 Determine the intercepts of𝑥 − 𝑓. (3)4.3 Determine the intercepts of𝑦 − 𝑓. (2)4.4 Sketch the graph of and on the same set of axes, clearly indicating𝑓 ℎthe distinctive features of the functions. (5)4.5 Determine the values for which𝑥 𝑓(𝑥) < ℎ(𝑥). (2)4.6 The line cuts at and . Write down the𝑦 =− 34 𝑥 + 254 𝑓 𝐸(3; 4) 𝐹coordinates of 𝐹. (7)

If (6,2) is a point on the graph of 𝑓(𝑥), give the coordinates of one point that must be on the graph of 𝑓−1(𝑥) if 𝑓−1(𝑥) exists.

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