f(x) = (2x³ - 5x² + 3x - 1)/(x² + 2x + 5). Identify the oblique (slant) asymptote, show there are no vertical asymptotes, and demonstrate how the function behaves as x approaches ±∞.

视频信息

答案文本 复制

视频字幕 复制

Let's analyze the function f(x) equals 2x cubed minus 5x squared plus 3x minus 1, divided by x squared plus 2x plus 5. First, we note that the degree of the numerator is 3, while the degree of the denominator is 2. Since the degree of the numerator is exactly one more than the degree of the denominator, this function has an oblique asymptote. To find it, we perform polynomial long division, which gives us 2x minus 9 plus a remainder term. Therefore, the oblique asymptote is y equals 2x minus 9, shown by the red line. As x approaches positive or negative infinity, the function approaches this slant asymptote. Now, let's examine why this function has no vertical asymptotes. Vertical asymptotes occur at values where the denominator equals zero. The denominator of our function is x squared plus 2x plus 5. To find its roots, we calculate the discriminant: b squared minus 4ac, which is 2 squared minus 4 times 1 times 5, giving us 4 minus 20, which equals negative 16. Since the discriminant is negative, this quadratic equation has no real roots. Looking at the graph of the denominator, we can see it's always positive and never touches the x-axis. The minimum value occurs at x equals negative 1, where the function value is 4. Since the denominator is never zero for any real value of x, our original rational function has no vertical asymptotes. To find the oblique asymptote, we need to perform polynomial long division. Since the degree of the numerator is one more than the denominator, we divide 2x cubed minus 5x squared plus 3x minus 1 by x squared plus 2x plus 5. Working through the division, we get a quotient of 2x minus 9 with a remainder of 11x plus 44. This means our function can be rewritten as 2x minus 9 plus the remainder term 11x plus 44 divided by x squared plus 2x plus 5. As x approaches infinity, this remainder term approaches zero, because the degree of the numerator is less than the degree of the denominator. Therefore, the oblique asymptote is y equals 2x minus 9. On the graph, we can see that as x gets larger in magnitude, the difference between the function and the asymptote gets smaller and smaller. Now, let's analyze the behavior of our function as x approaches positive or negative infinity. We've rewritten the function as 2x minus 9 plus a remainder term. To determine the limit of this remainder term as x approaches infinity, we divide both numerator and denominator by the highest power of x, which is x squared. This gives us 11 over x plus 44 over x squared in the numerator, and 1 plus 2 over x plus 5 over x squared in the denominator. As x approaches infinity, all terms with x in the denominator approach zero. Therefore, the limit of the remainder term is zero. This means that as x approaches positive or negative infinity, our function approaches the line y equals 2x minus 9. We can visualize this on the graph by looking at the difference between the function and the asymptote. As we move further out on the x-axis in either direction, this difference gets closer and closer to zero. Let's summarize our findings about the function f(x) equals 2x cubed minus 5x squared plus 3x minus 1, divided by x squared plus 2x plus 5. First, we determined that this function has an oblique asymptote with equation y equals 2x minus 9. This was found by performing polynomial long division. Second, we proved there are no vertical asymptotes because the denominator has no real roots, as confirmed by its negative discriminant of negative 16. Third, we showed that as x approaches positive or negative infinity, the function approaches the line y equals 2x minus 9. Finally, we demonstrated that the difference between the function and its asymptote approaches zero as the absolute value of x increases. These properties completely characterize the asymptotic behavior of our rational function.