Multiple Choice Identify the choice that best completes the statement or answers the question. Find any points of discontinuity for the rational function. 1. \( y=\frac{(x-7)(x+2)(x-9)}{(x-5)(x-2)} \) a. \( x=-5, x=-2 \) b. \( x=5, x=2 \) c. \( x=-7, x=2, x=-9 \) d. \( x=7, x=-2, x=9 \) 2. \( y=\frac{(x+7)(x+4)(x+2)}{(x+5)(x-3)} \) a. \( x=-5, x=3 \) b. \( x=7, x=4, x=2 \) c. \( x=-7, x=-4, x=-2 \) d. \( x=5, x=-3 \) 3. \( y=\frac{x+4}{x^{2}+8 x+15} \) a. \( x=-5, x=-3 \) b. \( x=-4 \) c. \( x=-5, x=3 \) d. \( x=5, x=3 \) 4. \( y=\frac{x-3}{x^{2}+3 x-10} \) a. \( x=-5, x=2 \) b. \( x=5, x=-2 \) c. \( x=3 \) d. \( x \) \( =-5, x=-2 \) 6. What are the points of discontinuity? Are they all removable? \[ y=\frac{(x-4)}{x^{2}-13 x+36} \] a. \( x=-9, x=-4, x=8 \); yes b. \( x=1, x=8, x= \) -8; no c. \( x=9, x=4 \); no d. \( x=-9, x=-4 \); no 7. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \). a. asymptote: \( x=2 \) and hole: \( x=-5 \) b. asymptotes: \( x=-2 \) and hole: \( x=-5 \) c. asymptote: \( x=-2 \) and hole: \( x=5 \) d. asymptote: \( x=-2 \) and hole: \( x=-2 \) a. \( x=-3, x=-8 \); no b. \( x=5, x=-7, x=1 \); no c. \( x=-5, x=7, x=-1 \); yes d. \( x=3, x=8 \); yes 8. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{x+1}{x^{2}+6 x+5} \). a. asymptote: \( x=-1 \) and hole: \( x=-1 \) b. asymptote: \( x=-1 \) and hole: \( x=-1 \) c. asymptotes: \( x=-1,-1 \) and hole: \( x=-1 \) d. asymptote: -5 and hole: \( x=-1 \), 9. Find the horizontal asymptote of the graph of \( y=\frac{7 x^{6}+7 x+3}{9 x^{5}+7 x+3} \). a. \( y=0 \) b. \( y=\frac{7}{9} \) c. no horizontal asymptote d. \( y=\frac{6}{5} \)
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Beyond the Answer
To find points of discontinuity in rational functions, first identify values of \( x \) that make the denominator equal to zero, since these are where the function cannot be defined. For option 1, when you set \( (x-5)(x-2)=0 \), you get \( x=5 \) and \( x=2 \) as points of discontinuity. Similarly, apply this method to other functions to arrive at the correct answers. Vertical asymptotes occur at those values of \( x \) that make the denominator zero but are not cancelled out by the numerator. If both the numerator and denominator have a common factor that cancels, it creates a removable discontinuity (or a hole) at that x-value. For instance, in the function \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \), \( x=5 \) is a hole while \( x=-2 \) is a vertical asymptote. Stay playful with these concepts, and soon you'll be spotting discontinuities like a pro!