Question

Create a rational function with the following characteristics: three real zeros, one of multiplicity \(2 ; y\) -intercept 1 ; vertical asymptotes, \(x=-2\) and \(x=3 ;\) oblique asymptote, \(y=2 x+1\). Is this rational function unique? Compare your function with those of other students. What will be the same as everyone else's? Add some more characteristics, such as symmetry or naming the real zeros. How does this modify the rational function?

Short answer

The rational function is: \(f(x) = \frac{3(x+1)(x-1)(x-2)^2}{2(x+2)(x-3)}\).

Step-by-step solution

Identify Key Characteristics

First, identify the key characteristics of the rational function. The given properties include: three real zeros (one of multiplicity 2), y-intercept at 1, vertical asymptotes at \(x=-2\) and \(x=3\), and an oblique asymptote at \(y=2x+1\).

Determine the Zeros and Factors

Since there are three real zeros, let's denote them as \(a\), \(b\), and \(c\) with \(c\) having a multiplicity of 2. An example choice could be \(a=-1\), \(b=1\), and \(c=2\). Hence, the factors related to these zeros will be \((x+1)\), \((x-1)\), and \((x-2)^2\).

Define the Vertical Asymptotes

Given the vertical asymptotes at \(x=-2\) and \(x=3\), include factors in the denominator that will be zero at these x-values: \((x+2)\) and \((x-3)\). Thus, the denominator can be expressed as \((x+2)(x-3)\).

Determine the Oblique Asymptote

The oblique asymptote \(y=2x+1\) implies that the long division of the rational function’s numerator by its denominator must result in \(2x+1\). This suggests the degree of the numerator should be one higher than the degree of the denominator. If the denominator's degree is 2 (from \((x+2)(x-3)\)), the numerator must be a cubic polynomial.

Construct the Rational Function

Combine the numerator and the denominator while considering the zeros and asymptotes. The rational function can be constructed as follows: \(f(x) = \frac{k(x+1)(x-1)(x-2)^2}{(x+2)(x-3)}\).

Determine the Constant k

Use the y-intercept to find the constant \(k\). When \(x=0\), \(f(0)=1\): \(f(0) = \frac{k(0+1)(0-1)(0-2)^2}{(0+2)(0-3)} = \frac{k(-1)(4)}{-6} = \frac{4k}{6} = \frac{2k}{3}\). Set \(f(0)=1\): \(1 = \frac{2k}{3} \rightarrow k = \frac{3}{2}\).

Verify and Write the Rational Function

After determining \(k\), the rational function becomes: \(f(x) = \frac{\frac{3}{2}(x+1)(x-1)(x-2)^2}{(x+2)(x-3)}\). Multiplying through by \(2\) to clear the fraction: \(f(x) = \frac{3(x+1)(x-1)(x-2)^2}{2(x+2)(x-3)}\).

Key concepts

Real Zeros

Real zeros in a rational function are the values of x that make the numerator equal to zero. These zeros are the x-intercepts of the function graph. For a rational function to have three real zeros, it might look like this: consider a cubic polynomial as the numerator.
For example, given the solution above, we identified three real zeros:

• -\(\tau= -1\) (Single Zero)
• \(\beta=1\) (Single Zero)
• \(\gamma=2\) (Multiplicity 2)

The multiplicity of 2 for one zero means the function will just touch the x-axis at \(\tau=2\), and not cross it. In general, real zeros are crucial because they guide the shape of the graph and indicate where the function equals zero.

Vertical Asymptotes

Vertical asymptotes occur where the denominator of a rational function equals zero, creating undefined points that the graph will approach but never reach. In our example, the vertical asymptotes are at \(x=-2\) and \(x=3\). These points make the factors denoted in the denominator turn to zero:

• \((x+2) \to 0\) at \(x=-2\)
• \((x-3) \to 0\) at \(x= 3\)A rational function will display these asymptotes as sharp increases or decreases toward infinity at these x-values. They indicate breaks in the graph where the function can no longer continue smoothly.
  

Oblique Asymptote

An oblique (or slant) asymptote occurs when the degree of the polynomial in the numerator is exactly one more than the degree of the polynomial in the denominator. In our rational function, the oblique asymptote is \(y=2x+1\). This suggests we did a long division of the numerator by the denominator that resulted in a linear polynomial.This oblique asymptote gives a general direction the curve follows as \(x\) approaches either positive or negative infinity. Contrary to vertical asymptotes, an oblique asymptote is a skewed line running diagonally across the graph. It helps visualize the end-behavior of the function, showing how, very far from the x-axis, the function behaves similarly to the line \(2x+1\).

• Oblique asymptote: guides the function's behavior for very large or very small \(x\) values

Y-Intercept

The y-intercept of a rational function is the point where the graph crosses the y-axis. It occurs when \(x=0\). In this example, the y-intercept is given as \((0,1)\). We use this property to determine the constant \(k\) in the rational function.To find the y-intercept, substitute \(x=0\) into the function and solve for \(f(0)\). With \(x=0\):$$f(0)= \frac{k(0+1)(0-1)(0-2)^2}{(0+2)(0-3)}=\frac{k(-1)(4)}{-6}=\frac{4k}{6}=\frac{2k}{3}.$$ We know $$f(0)=1$$ so;$$1=\frac{2k}{3} \rightarrow k= \frac{3}{2}.$$Thus, by determining \(k\), we can ensure the rational function properly intersects the y-axis at \((0,1)\). The y-intercept is an essential feature because it gives a tangible start point of the graph on the Cartesian plane.

• The y-intercept tells you where the function touches the y-axis.