Graph the function. \[ p(x)=\frac{x^{2}-4 x+4}{x} \]

We start with the function \[ p(x)=\frac{x^2-4x+4}{x}. \] ### Step 1. Factor the Numerator Notice that the quadratic in the numerator is a perfect square: \[ x^2-4x+4=(x-2)^2. \] Thus, the function can be rewritten as \[ p(x)=\frac{(x-2)^2}{x}. \] ### Step 2. Determine the Domain Since the denominator is \( x \), the function is undefined at \( x=0 \). Therefore, the domain is: \[ x\in\mathbb{R}\setminus \{0\}. \] ### Step 3. Rewrite in a Form Suitable for Analyzing Asymptotes Perform polynomial long division or simply rewrite by splitting the fraction: \[ \frac{x^2-4x+4}{x} = \frac{x^2}{x} - \frac{4x}{x} + \frac{4}{x} = x-4+\frac{4}{x}. \] This form, \[ p(x)=x-4+\frac{4}{x}, \] will help us find the asymptotic behavior. ### Step 4. Identify the Asymptotes #### Vertical Asymptote Since the function is undefined at \( x=0 \) and the term \( \frac{4}{x} \) becomes unbounded as \( x \to 0 \), there is a vertical asymptote at: \[ x=0. \] #### Oblique (Slant) Asymptote For large \( |x| \), the term \( \frac{4}{x} \) approaches 0. Therefore, as \( x \to \pm\infty \): \[ p(x) \approx x-4. \] Thus, the oblique asymptote is: \[ y=x-4. \] ### Step 5. Find the Intercepts #### \( x \)-Intercept(s) Set \( p(x) = 0 \): \[ x-4+\frac{4}{x}=0. \] Multiply both sides by \( x \) (noting that \( x\neq 0 \)): \[ x^2-4x+4=0. \] This factors as: \[ (x-2)^2=0, \] so the only solution is: \[ x=2. \] Thus, the \( x \)-intercept is at \( (2,0) \). #### \( y \)-Intercept The \( y \)-intercept would be \( p(0) \), but since \( x=0 \) is not in the domain, there is **no \( y \)-intercept**. ### Step 6. Sketching the Graph - **For \( x>0 \):** - As \( x \to 0^+ \), the term \( \frac{4}{x} \) tends to \( +\infty \), so \( p(x) \to +\infty \). - As \( x \to +\infty \), \( p(x)\approx x-4 \) tends to \( +\infty \) in a linear manner. - The graph passes through the point \( (2, 0) \). - **For \( x<0 \):** - As \( x \to 0^- \), \( \frac{4}{x} \) tends to \( -\infty \), so \( p(x) \to -\infty \). - As \( x \to -\infty \), \( p(x)\approx x-4 \) tends to \( -\infty \) linearly. - Evaluate a sample point: for \( x=-1 \): \[ p(-1)= -1-4+\frac{4}{-1}=-5-4=-9. \] - **Asymptote:** The line \( y=x-4 \) is the oblique asymptote that the graph approaches as \( |x| \) becomes large. ### Step 7. Graph Summary - **Domain:** \( x\in\mathbb{R}\setminus \{0\} \) - **Vertical Asymptote:** \( x=0 \) - **Oblique Asymptote:** \( y=x-4 \) - **\( x \)-Intercept:** \( (2,0) \) - **No \( y \)-Intercept** since \( x=0 \) is excluded. - **Behavior:** - For \( x>0 \), the graph approaches \( +\infty \) as \( x\to0^+ \) and follows the line \( y=x-4 \) for large \( x \). - For \( x<0 \), the graph approaches \( -\infty \) as \( x\to0^- \) and, for large negative \( x \), it tends to follow the line \( y=x-4 \). Using these observations, you can sketch the graph with care around the vertical asymptote \( x=0 \), plot the intercept at \( (2,0) \), and draw the oblique asymptote \( y=x-4 \) guiding the end behavior.