Sketch the graph of \( \frac{x^3+6x^2+5x}{x^2-4x+4} \).

To sketch the graph of \( \frac{x^3+6x^2+5x}{x^2-4x+4} \), we first need to find the vertical asymptotes, horizontal asymptotes, x-intercepts, and y-intercepts.

First, let's find the vertical asymptotes by setting the denominator equal to zero and solving for x:

\( x^2 - 4x + 4 = 0 \)

This equation factors to:

\( (x-2)^2 = 0 \)

So, we have a vertical asymptote at x = 2.

Next, let's find the horizontal asymptote by looking at the degrees of the numerator and denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the x-intercepts, we set the numerator equal to zero and solve for x:

\( x^3 + 6x^2 + 5x = 0 \)

This equation factors to:

\( x(x^2 + 6x + 5) = 0 \)

\( x(x+1)(x+5) = 0 \)

So, the x-intercepts are at x = 0, x = -1, and x = -5.

To find the y-intercept, we plug in x = 0 into the function:

\( \frac{0^3+6(0)^2+5(0)}{0^2-4(0)+4} = 0 \)

So, the y-intercept is at (0, 0).

Putting all this information together, we can sketch the graph of \( \frac{x^3+6x^2+5x}{x^2-4x+4} \) as shown below:

\[
\begin{array}{c|c}
x & y \\
\hline
-5 & 0 \\
-1 & 0 \\
0 & 0 \\
2 & \text{undefined} \\
\end{array}
\]

The graph will have a hole at x = 2 and will approach the x-axis as x approaches positive or negative infinity.

Step 1: Find the Vertical Asymptotes

Set the denominator to zero:

$$x^2 - 4x + 4 = 0$$

Factor:

$$(x - 2)^2 = 0$$

Solve for \(x\):

$$x = 2$$ (repeated)

Step 2: Find the Horizontal Asymptote

As \(x \to \infty\), the \(x^2\) terms dominate, so the horizontal asymptote is:

$$y = \frac{\infty}{\infty} = 1$$

Step 3: Find the x-intercepts

Set the numerator to zero:

$$x^3 - 6x^2 + 5x = 0$$

Factor:

$$x(x - 1)(x - 5) = 0$$

Solve for \(x\):

$$x = 0, 1, 5$$

Step 4: Find the y-intercept

Set \(x = 0\):

$$y = \frac{0^3 - 6 \cdot 0^2 + 5 \cdot 0}{0^2 - 4 \cdot 0 + 4} = 0$$

Step 5: Find the Local Minimum and Maximum

The first derivative is:

$$y' = \frac{3x^2 - 12x + 5}{(x^2 - 4x + 4)^2}$$

Set \(y' = 0\):

$$3x^2 - 12x + 5 = 0$$

Factor:

$$(x - 1)(3x - 5) = 0$$

Solve for \(x\):

$$x = 1, \frac{5}{3}$$

Evaluate the function at these values:

$$y(1) = \frac{1^3 - 6 \cdot 1^2 + 5 \cdot 1}{1^2 - 4 \cdot 1 + 4} = -\frac{1}{3}$$

$$y\left(\frac{5}{3}\right) = \frac{\left(\frac{5}{3}\right)^3 - 6 \cdot \left(\frac{5}{3}\right)^2 + 5 \cdot \frac{5}{3}}{\left(\frac{5}{3}\right)^2 - 4 \cdot \frac{5}{3} + 4} = \frac{25}{27}$$

The local minimum is \((1, -\frac{1}{3})\) and the local maximum is \(\left(\frac{5}{3}, \frac{25}{27}\right)\).

Step 6: Sketch the Graph

Plot the intercepts, asymptotes, and extrema, and connect them smoothly to obtain the graph:

[Image of the graph of y = (x^3 - 6x^2 + 5x)/(x^2 - 4x + 4)]