f(x) = (4x² - 3x + 5)/(2x² + x - 1). Find and clearly show the horizontal asymptote, identify all vertical asymptotes,

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Let's find the asymptotes of the rational function f of x equals 4x squared minus 3x plus 5, divided by 2x squared plus x minus 1. First, we'll find the horizontal asymptote. Since both the numerator and denominator have the same degree, which is 2, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is 4, and the leading coefficient of the denominator is 2. Therefore, the horizontal asymptote is y equals 2. We can see this on the graph as the green line that the function approaches as x approaches positive or negative infinity. Now, let's find the vertical asymptotes of our rational function. Vertical asymptotes occur where the denominator equals zero, provided the numerator is not also zero at those points. We need to solve the equation: 2x squared plus x minus 1 equals 0. Factoring this expression, we get (2x minus 1) times (x plus 1) equals 0. Solving for x, we get x equals one-half or x equals negative 1. Let's check if the numerator is zero at these values. For both x equals one-half and x equals negative 1, the numerator is not zero. Therefore, we have vertical asymptotes at x equals one-half and x equals negative 1, shown as the red dashed lines on our graph. Let's verify our findings. For the horizontal asymptote, we take the limit of the function as x approaches positive or negative infinity. Since both numerator and denominator have the same degree, we take the ratio of the leading coefficients, which is 4 divided by 2, giving us 2. This confirms our horizontal asymptote is y equals 2. For the vertical asymptotes, we check what happens as x approaches negative 1 or one-half. In both cases, the denominator approaches zero while the numerator remains non-zero, causing the function to approach positive or negative infinity. This confirms our vertical asymptotes at x equals negative 1 and x equals one-half. In summary, our rational function has one horizontal asymptote at y equals 2 and two vertical asymptotes at x equals negative 1 and x equals one-half. Let's examine the behavior of our function near the asymptotes. For the horizontal asymptote, as x approaches positive or negative infinity, the function value approaches 2. We can see this on both the left and right sides of our graph. For the vertical asymptote at x equals negative 1, the function behaves differently depending on which side we approach from. As x approaches negative 1 from the left, the function decreases without bound, approaching negative infinity. But as x approaches negative 1 from the right, the function increases without bound, approaching positive infinity. Similarly, for the vertical asymptote at x equals one-half, as x approaches one-half from the left, the function increases without bound, approaching positive infinity. And as x approaches one-half from the right, the function decreases without bound, approaching negative infinity. These behaviors create the characteristic shape of our rational function graph. Let's summarize what we've learned about the asymptotes of our rational function f of x equals 4x squared minus 3x plus 5, divided by 2x squared plus x minus 1. We found one horizontal asymptote at y equals 2. This was determined by comparing the degrees of the numerator and denominator, which are both 2, and then taking the ratio of their leading coefficients, 4 divided by 2, which equals 2. We also identified two vertical asymptotes: x equals negative 1 and x equals one-half. These occur at the values where the denominator equals zero. To find these values, we factored the denominator 2x squared plus x minus 1 as (2x minus 1) times (x plus 1) equals 0, giving us x equals one-half and x equals negative 1. The key methods we used were comparing polynomial degrees for horizontal asymptotes and finding where the denominator equals zero for vertical asymptotes. These techniques can be applied to analyze any rational function.