Question

student: class: date rational expressions and functions student activity sheet 5; exploring \asymptotes and removable discontinuities\ page 7 of 11 10. what is a removable discontinuity? what is its cause? 11. describe the behavior of ( h(x) ) at ( x = 2 ). explain your answer. 12. complete the tables and answer the related questions. ( h(x)=\frac{5 x - 10}{x^{2}+x - 6} ) as ( x ) gets very large, what do the function values approach? ( h(x)=\frac{5 x - 10}{x^{2}+x - 6} ) as ( x ) gets very small, what do the function values approach?

Explanation:

Step1: Calculate function values for positive \(x\)

• Substitute \(x = 400,600,800,1000\) into \(h(x)=\frac{5x - 10}{x^{2}+x - 6}\)
• Use the formula \(h(x)=\frac{5x - 10}{x^{2}+x - 6}\), for \(x = 400\): \(h(400)=\frac{5\times400 - 10}{400^{2}+400 - 6}\), simplify numerator \(5\times400-10 = 2000 - 10=1990\), denominator \(400^{2}+400 - 6=160000 + 400-6 = 160394\), then \(h(400)=\frac{1990}{160394}\approx0.0124\)
• Similarly for \(x = 600\): numerator \(5\times600-10=2990\), denominator \(600^{2}+600 - 6 = 360000+600 - 6=360594\), \(h(600)=\frac{2990}{360594}\approx0.0083\)
• For \(x = 800\): numerator \(5\times800-10 = 3990\), denominator \(800^{2}+800 - 6=640000+800 - 6 = 640794\), \(h(800)=\frac{3990}{640794}\approx0.0062\)
• For \(x = 1000\): numerator \(5\times1000-10=4990\), denominator \(1000^{2}+1000 - 6=1000000+1000 - 6=1000994\), \(h(1000)=\frac{4990}{1000994}\approx0.0050\)

Step2: Analyze the limit as \(x\to+\infty\)

• For a rational function \(y=\frac{f(x)}{g(x)}\) where \(f(x)=5x - 10\) (degree \(n = 1\)) and \(g(x)=x^{2}+x - 6\) (degree \(m=2\), \(n
• By the rule of limits for rational functions \(\lim_{x\to+\infty}\frac{ax^{n}+bx^{n - 1}+\cdots}{cx^{m}+dx^{m - 1}+\cdots}\), when \(n
• Another way: \(h(x)=\frac{5x - 10}{x^{2}+x - 6}=\frac{x(5-\frac{10}{x})}{x^{2}(1+\frac{1}{x}-\frac{6}{x^{2}})}=\frac{5-\frac{10}{x}}{x(1+\frac{1}{x}-\frac{6}{x^{2}})}\), as \(x\to+\infty\), \(\frac{10}{x}\to0\), \(\frac{1}{x}\to0\), \(\frac{6}{x^{2}}\to0\), so \(h(x)\to0\)

Step3: Calculate function values for negative \(x\)

• Substitute \(x=-400,-600,-800,-1000\) into \(h(x)=\frac{5x - 10}{x^{2}+x - 6}\)
• For \(x=-400\): numerator \(5\times(-400)-10=-2010\), denominator \((-400)^{2}+(-400)-6 = 160000-400 - 6=159594\), \(h(-400)=\frac{-2010}{159594}\approx - 0.0126\)
• For \(x=-600\): numerator \(5\times(-600)-10=-3010\), denominator \((-600)^{2}+(-600)-6=360000-600 - 6=359394\), \(h(-600)=\frac{-3010}{359394}\approx - 0.0084\)
• For \(x=-800\): numerator \(5\times(-800)-10=-4010\), denominator \((-800)^{2}+(-800)-6=640000-800 - 6=639194\), \(h(-800)=\frac{-4010}{639194}\approx - 0.0063\)
• For \(x=-1000\): numerator \(5\times(-1000)-10=-5010\), denominator \((-1000)^{2}+(-1000)-6=1000000-1000 - 6=998994\), \(h(-1000)=\frac{-5010}{998994}\approx - 0.0050\)

Step4: Analyze the limit as \(x\to-\infty\)

• Using the same rule for rational - function limits (\(n = 1\), \(m = 2\), \(n
• Also, \(h(x)=\frac{5x - 10}{x^{2}+x - 6}=\frac{x(5-\frac{10}{x})}{x^{2}(1+\frac{1}{x}-\frac{6}{x^{2}})}=\frac{5-\frac{10}{x}}{x(1+\frac{1}{x}-\frac{6}{x^{2}})}\), as \(x\to-\infty\), \(\frac{10}{x}\to0\), \(\frac{1}{x}\to0\), \(\frac{6}{x^{2}}\to0\), and \(x\to-\infty\), so \(h(x)\to0\)

Answer:

For \(x = 400\), \(h(400)=\frac{5\times400 - 10}{400^{2}+400 - 6}=\frac{2000 - 10}{160000+400 - 6}=\frac{1990}{160394}\approx0.0124\)

For \(x = 600\), \(h(600)=\frac{5\times600 - 10}{600^{2}+600 - 6}=\frac{3000 - 10}{360000+600 - 6}=\frac{2990}{360594}\approx0.0083\)

For \(x = 800\), \(h(800)=\frac{5\times800 - 10}{800^{2}+800 - 6}=\frac{4000 - 10}{640000+800 - 6}=\frac{3990}{640794}\approx0.0062\)

For \(x = 1000\), \(h(1000)=\frac{5\times1000 - 10}{1000^{2}+1000 - 6}=\frac{5000 - 10}{1000000+1000 - 6}=\frac{4990}{1000994}\approx0.0050\)

As \(x\) gets very large (\(x\to+\infty\)), we use the fact that for a rational function \(y = \frac{ax + b}{cx^{2}+dx+e}\), when \(x\to\pm\infty\), \(y\approx\frac{ax}{cx^{2}}=\frac{a}{cx}\). For \(h(x)=\frac{5x - 10}{x^{2}+x - 6}\), as \(x\to+\infty\), \(h(x)\approx\frac{5x}{x^{2}}=\frac{5}{x}\to0\)

For \(x=- 400\), \(h(-400)=\frac{5\times(-400)-10}{(-400)^{2}+(-400)-6}=\frac{-2000 - 10}{160000-400 - 6}=\frac{-2010}{159594}\approx - 0.0126\)

For \(x=-600\), \(h(-600)=\frac{5\times(-600)-10}{(-600)^{2}+(-600)-6}=\frac{-3000 - 10}{360000-600 - 6}=\frac{-3010}{359394}\approx - 0.0084\)

For \(x=-800\), \(h(-800)=\frac{5\times(-800)-10}{(-800)^{2}+(-800)-6}=\frac{-4000 - 10}{640000-800 - 6}=\frac{-4010}{639194}\approx - 0.0063\)

For \(x=-1000\), \(h(-1000)=\frac{5\times(-1000)-10}{(-1000)^{2}+(-1000)-6}=\frac{-5000 - 10}{1000000-1000 - 6}=\frac{-5010}{998994}\approx - 0.0050\)

As \(x\) gets very small (\(x\to-\infty\)), \(h(x)\approx\frac{5x}{x^{2}}=\frac{5}{x}\to0\)