Math Problem Statement

List the domain and the​ x- and​ y-intercepts of the following function. Graph the function. Be sure to label all the asymptotes.

f left parenthesis x right parenthesis equals StartFraction x minus 1 Over x squared minus 5 x minus 6 EndFraction

Solution

We are given the function:

f(x)=x1x25x6f(x) = \frac{x - 1}{x^2 - 5x - 6}

Step 1: Domain of the function

The domain of a function refers to the set of all xx-values where the function is defined. For rational functions like this one, the function is undefined where the denominator is zero.

Let's find the values of xx that make the denominator zero:

x25x6=0x^2 - 5x - 6 = 0

Factor the quadratic:

(x6)(x+1)=0(x - 6)(x + 1) = 0

The solutions are x=6x = 6 and x=1x = -1.

Thus, the function is undefined at x=6x = 6 and x=1x = -1. Therefore, the domain is:

Domain: (,1)(1,6)(6,)\text{Domain: } (-\infty, -1) \cup (-1, 6) \cup (6, \infty)

Step 2: xx- and yy-intercepts

  • xx-intercepts: To find the xx-intercept(s), set f(x)=0f(x) = 0. A rational function equals zero when the numerator equals zero, as long as the denominator is non-zero.

Set the numerator equal to zero:

x1=0x - 1 = 0

Thus, x=1x = 1. So, the xx-intercept is at x=1x = 1, or the point (1,0)(1, 0).

  • yy-intercept: To find the yy-intercept, set x=0x = 0 and solve for f(0)f(0):

f(0)=01025(0)6=16=16f(0) = \frac{0 - 1}{0^2 - 5(0) - 6} = \frac{-1}{-6} = \frac{1}{6}

Thus, the yy-intercept is at the point (0,16)\left(0, \frac{1}{6}\right).

Step 3: Asymptotes

  • Vertical asymptotes: These occur where the denominator is zero and the numerator is non-zero. From the factored form of the denominator, we have vertical asymptotes at:

x=1andx=6x = -1 \quad \text{and} \quad x = 6

  • Horizontal asymptote: Since the degree of the denominator (2) is greater than the degree of the numerator (1), the horizontal asymptote is at:

y=0y = 0

Step 4: Graph the function

The graph will show:

  • Vertical asymptotes at x=1x = -1 and x=6x = 6,
  • A horizontal asymptote at y=0y = 0,
  • An xx-intercept at (1,0)(1, 0),
  • A yy-intercept at (0,16)\left(0, \frac{1}{6}\right).

Let me know if you'd like me to graph this function for you.


Would you like further details or clarifications? Here are 5 related questions that can deepen your understanding:

  1. How do vertical asymptotes affect the shape of the graph near these values of xx?
  2. What happens to the function's value as xx approaches infinity or negative infinity?
  3. How can you determine if there are slant (oblique) asymptotes in a rational function?
  4. What is the significance of horizontal asymptotes in terms of the function's end behavior?
  5. How can the factored form of the denominator help in identifying intercepts and asymptotes more quickly?

Tip: When analyzing rational functions, always check both the numerator and denominator for factors to simplify identifying intercepts and asymptotes!

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Intercepts
Asymptotes
Domain

Formulas

Rational function: f(x) = (numerator)/(denominator)
Quadratic factorization: x^2 - 5x - 6 = (x - 6)(x + 1)

Theorems

Factoring Quadratics
Asymptote Behavior in Rational Functions

Suitable Grade Level

Grades 10-12