Question

A rational function, R(x) has the following characteristics:

a vertical asymptote at x = 3,
a horizontal asymptote at y = 2,
and a hole at (2, −2).

Sketch the function and determine what it could be using the following steps:

Put in the factor that would account for the vertical asymptote at x = 3.
Add in the factors that would account for a hole at x = 2.
Determine what must be true about the numerator and denominator for there to be a horizontal asymptote at y = 2.
Add the factors that would account for the horizontal asymptote at y = 2.
Describe what you must do in order for the hole to appear at (2, −2).
Write the completed function.

Answer 1

Step-by-step explanation:

Step-by-step explanation:

$\begin{gathered}\sf rationalizing\:the\:denominator \:means\:making\:the\:denominator\:rational \\ \\ \sf (i) \frac{1}{\sqrt{7}+\sqrt{2}} \\ \\ \sf = \frac{(\sqrt{7}-\sqrt{2})}{(\sqrt{7} + \sqrt{2})(\sqrt{7}-\sqrt{2})} \\ \\ \sf = \frac{(\sqrt{7}-\sqrt{2})}{{(\sqrt{7})}^{2}-{(\sqrt{2})}^{2}} \\ \\ \sf = \frac{(\sqrt{7}-\sqrt{2})}{7-2} \\ \\ \sf = \frac {(\sqrt{7}-\sqrt{2})}{5}\\ \\ \sf (ii) \frac{3}{2\sqrt{5}-3\sqrt{2}} \\ \\ \sf = \frac{3(2\sqrt{5}+3\sqrt{2})}{(2\sqrt{5}+3\sqrt{2})(2\sqrt{5}-3\sqrt{2})}\\ \\ \sf = \frac{3(2\sqrt{5}+3\sqrt{2})}{{(2\sqrt{5})}^{2} - {(3\sqrt{2})}^{2}}\\ \\ \sf = \frac{3(2\sqrt{5}+3\sqrt{2})}{20 - 18 } \\ \\ \sf = \frac{3(2\sqrt{5}+3\sqrt{2})}{6}\\ \\ \sf = \frac{(2\sqrt{5}+3\sqrt{2})}{2} \\ \\ \sf (iii) \frac{4}{7+4\sqrt{3}} \\ \\ \sf = \frac{4(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})} \\ \\ \sf = \frac{4(7-4\sqrt{3})}{{7}^{2}-{(4\sqrt{3})}^{2}} \\ \\ \sf = \frac{4(7-4\sqrt{3})}{49-48} \\ \\ \sf = \frac{4(7-4\sqrt{3})}{1} \\ \\ \sf = 4(7 -4\sqrt{3})\end{gathered}$

Answer 2

Answer:

here(-2,2) are coolinear. mark as brainlist