Math Problem Statement

Find the intercepts and asymptotes, then sketch a graph of the function. x-2/x^2-4x+4

Solution

Let's analyze the given function step by step:

Function:

f(x)=x2x24x+4f(x) = \frac{x - 2}{x^2 - 4x + 4}

Step 1: Simplify the function

Factor the denominator:

x24x+4=(x2)2x^2 - 4x + 4 = (x - 2)^2

So, the function simplifies to:

f(x)=x2(x2)2f(x) = \frac{x - 2}{(x - 2)^2}

This can be rewritten as:

f(x)=1x2f(x) = \frac{1}{x - 2}

Now, we have the simplified form of the function: f(x)=1x2f(x) = \frac{1}{x - 2}.

Step 2: Find the intercepts

a) x-intercept:

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

1x2=0\frac{1}{x - 2} = 0

This equation has no solution because the numerator is 1, and it can never equal 0. Thus, there is no x-intercept.

b) y-intercept:

To find the y-intercept, set x=0x = 0 and evaluate f(0)f(0):

f(0)=102=12=12f(0) = \frac{1}{0 - 2} = \frac{1}{-2} = -\frac{1}{2}

So, the y-intercept is (0,12)(0, -\frac{1}{2}).

Step 3: Asymptotes

a) Vertical Asymptote:

The vertical asymptote occurs when the denominator equals zero. Since the function is f(x)=1x2f(x) = \frac{1}{x - 2}, set the denominator equal to zero:

x2=0x=2x - 2 = 0 \quad \Rightarrow \quad x = 2

So, there is a vertical asymptote at x=2x = 2.

b) Horizontal Asymptote:

For large values of xx, f(x)=1x2f(x) = \frac{1}{x - 2} approaches zero because the numerator stays constant and the denominator grows larger. Thus, the horizontal asymptote is:

y=0y = 0

Step 4: Sketching the graph

  • The function has a vertical asymptote at x=2x = 2.
  • The horizontal asymptote is at y=0y = 0.
  • The y-intercept is at (0,12)(0, -\frac{1}{2}).
  • As x2x \to 2^-, the function f(x)f(x) \to -\infty.
  • As x2+x \to 2^+, the function f(x)+f(x) \to +\infty.
  • As xx \to \infty or xx \to -\infty, f(x)0f(x) \to 0.

Would you like me to generate a sketch of this graph?


Here are 5 related questions:

  1. What are the steps to find vertical asymptotes for more complex functions?
  2. How do you determine if a function has a slant asymptote?
  3. What is the behavior of rational functions near vertical asymptotes?
  4. How do horizontal asymptotes differ from limits at infinity?
  5. What happens when both the numerator and denominator have the same degree?

Tip: Always check for common factors in rational functions before finding asymptotes, as they can reveal removable discontinuities.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Intercepts
Asymptotes
Simplification

Formulas

f(x) = (x - 2) / (x^2 - 4x + 4)
f(x) = 1 / (x - 2)

Theorems

Finding Vertical Asymptotes: Set the denominator equal to zero
Finding Horizontal Asymptotes: Analyze the end behavior of the function

Suitable Grade Level

Grades 10-12