Problem 96 - A Graphical Approach to Precalculus with Limits

Question

Use a calculator to graph rational function in the window indicated. Then (a) give the \(x\) - and y-intercepts, (b) explain why there are no vertical asymptotes, (c) give the equation of the oblique asymptote, and (d) give the domain and range. $$f(x)=\frac{-x^{3}-7 x^{2}+16 x+112}{x^{2}+x+28} ;[-15,10] \text { by }[-5,15]$$

Step-by-Step Solution

Answer

x-intercepts: -8, 2, -7; y-intercept: (0,4); No vertical asymptotes; Oblique asymptote: y = -x - 8; Domain: all reals; Range: all reals.

Step 1: Determine the x-intercepts

To find the x-intercepts of the function \(f(x) = \frac{-x^3 - 7x^2 + 16x + 112}{x^2 + x + 28}\), we set the numerator equal to zero and solve the equation \(-x^3 - 7x^2 + 16x + 112 = 0\). Using a calculator or synthetic division, we may find that \(x = -8\), \(x = 2\), and \(x = -7\) are the x-intercepts.

Step 2: Determine the y-intercept

To find the y-intercept, set \(x = 0\) in the function. \[ f(0) = \frac{0^3 - 7(0)^2 + 16(0) + 112}{0^2 + 0 + 28} = \frac{112}{28} = 4. \] So, the y-intercept is \((0,4)\).

Step 3: Identify if vertical asymptotes exist

Vertical asymptotes occur where the denominator is zero but not the numerator. Solve \(x^2 + x + 28 = 0\). The discriminant \(b^2 - 4ac = 1 - 112 = -111\) is less than zero, indicating no real roots. Hence, there are no vertical asymptotes as the denominator does not equal zero for any real x.

Step 4: Determine the oblique asymptote

Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is an oblique asymptote. Perform polynomial long division of \(-x^3 - 7x^2 + 16x + 112\) by \(x^2 + x + 28\) to obtain the quotient, which is the equation of the oblique asymptote: \(y = -x - 8\).

Step 5: Identify the domain and range

The domain of \(f(x)\) is all real numbers except where the denominator equals zero. From step 3, there are no real x-values that make the denominator zero, so the domain is all real numbers. For the range, the lack of real vertical asymptotes and the presence of an oblique asymptote suggest that, as \(x\) approaches positive or negative infinity, \(f(x)\) will approach \(-x-8\) without any restrictions, thus covering all real numbers. Therefore, the range is all real numbers.

Key Concepts

Understanding X-Intercepts in Rational Functions

X-intercepts are the points where a graph crosses the x-axis. For the rational function \(f(x) = \frac{-x^3 - 7x^2 + 16x + 112}{x^2 + x + 28}\), we find the x-intercepts by setting the numerator equal to zero. We are essentially looking for the values of \(x\) that make the whole expression zero. This can often involve solving a polynomial equation, and sometimes requires the use of tools like synthetic division or a calculator for complex polynomials.In this function, calculating \(-x^3 - 7x^2 + 16x + 112 = 0\) yields three solutions: \(x = -8, x = 2,\) and \(x = -7\). These solutions mean that the x-intercepts of this function are \((-8, 0)\), \((2, 0)\), and \((-7, 0)\). Every zero of the numerator contributes to an x-intercept, dictating where the graph will touch or cross the x-axis.

Discovering the Y-Intercept

The y-intercept of a graph is the point where it crosses the y-axis. This intersects when \(x = 0\). To find the y-intercept of a rational function, simply substitute \(x = 0\) into the equation and solve for \(f(x)\).For our function \(f(x) = \frac{-x^3 - 7x^2 + 16x + 112}{x^2 + x + 28}\), plugging in \(x = 0\) yields:\[f(0) = \frac{0^3 - 7(0)^2 + 16(0) + 112}{0^2 + 0 + 28} = \frac{112}{28} = 4\]Thus, the y-intercept is at the point \((0, 4)\). This is where the graph will touch or cross the y-axis, indicating the value of \(f(x)\) when \(x\) is zero, giving us a straightforward point on the graph.

Identifying the Oblique Asymptote

An oblique asymptote occurs in a rational function when the degree of the numerator is one higher than the degree of the denominator. It represents the behavior of the function as \(x\) approaches infinity or negative infinity.For the function \(f(x) = \frac{-x^3 - 7x^2 + 16x + 112}{x^2 + x + 28}\), the numerator is of degree 3, and the denominator is of degree 2. This indicates the presence of an oblique asymptote.To find the equation of the oblique asymptote, perform polynomial long division of the numerator by the denominator. The quotient will provide the equation. In this case, dividing \(-x^3 - 7x^2 + 16x + 112\) by \(x^2 + x + 28\) gives us the oblique asymptote equation: \[y = -x - 8\]This oblique asymptote describes the slanted line that the function graph approaches, offering insight into the end behavior of the graph.

Exploring Domain and Range

The domain of a rational function consists of all the x-values for which the function is defined. This is generally all real numbers except where the denominator equals zero. For the function \(f(x) = \frac{-x^3 - 7x^2 + 16x + 112}{x^2 + x + 28}\), solving the equation \(x^2 + x + 28 = 0\) shows no real solutions since the discriminant is negative. Thus, the denominator never equals zero, and the domain includes all real numbers.The range of a rational function is the set of possible y-values. Since this function has no vertical asymptotes and features an oblique asymptote, the function approaches \(-x - 8\) as \(x\) tends toward infinity in either direction. Consequently, the function spans all real numbers vertically, indicating that the range is also all real numbers.Understanding domain and range helps visualize how the graph behaves horizontally and vertically, providing critical insight into the overall aspect of the function on a coordinate plane.