Use the following equation to perform the operations below (do not graph the equation yet):

y ={x + 5}{x + 1}\{(x + 3)(x - 2)}{(x - 4)(x + 1)}) - {1}/{x - 2}

What graphical feature occurs at {x} = -1? Use the "trace" function, zoom in on the graph, or look at a table of values to verify what is happening at {x} = -1. Why isn't there a vertical asymptote at {x} = -1?
c. Solve the equation for {x}when y = 0}
find the roots of the equation. Show all of your work and explain the strategies you use in each step.

$f(x) = x^{5} - 3x^{4} - 5x^{3} + 5x^{2} - 6x + 8$

b. Identify all possible rational zeros.
c. How many possible positive real zeros are there? How many possible negative real zeros? How many possible complex zeros?
d. Graph the polynomial to approximate the zeros. What are the rational zeros? Use synthetic division to verify these are correct.
e. Write the polynomial in factor form.
f. What are the complex zeros?

To determine the graphical feature at x = -1, we will substitute x = -1 into the given equation and evaluate it.

y = ((x + 5)(x + 1))/((x + 3)(x - 2)((x - 4)(x + 1))) - 1/(x - 2)

Substituting x = -1 into the equation:

y = ((-1 + 5)(-1 + 1))/((-1 + 3)(-1 - 2)((-1 - 4)(-1 + 1))) - 1/(-1 - 2)
y = (4)(0)/(2)(-3)(-5)(-3)
y = 0/(-30)
y = 0

Therefore, at x = -1, the equation gives a y-coordinate of 0. The graph of the equation will intersect the x-axis at x = -1.

Now let's move on to solving the equation for x when y = 0.

0 = ((x + 5)(x + 1))/((x + 3)(x - 2)((x - 4)(x + 1))) - 1/(x - 2)

To simplify the equation, we will get rid of the fractions by multiplying through by the denominators:

0*((x + 3)(x - 2)((x - 4)(x + 1))) = (x + 5)(x + 1) - (x - 2)
0 = (x + 5)(x + 1) - (x - 2)

Expanding the brackets:

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

Combining like terms:

0 = x^2 + 4x + 12

Now we need to solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In the given equation, a = 1, b = 4, and c = 12. Substituting these values into the quadratic formula:

x = (-4 ± √(4^2 - 4(1)(12))) / (2(1))
x = (-4 ± √(16 - 48)) / 2
x = (-4 ± √(-32)) / 2
x = (-4 ± √(-1*32)) / 2
x = (-4 ± √(-1)√(32)) / 2
x = (-4 ± 4i√2) / 2
x = -2 ± 2i√2

Therefore, the roots of the equation are x = -2 + 2i√2 and x = -2 - 2i√2.

To answer these questions, we will first need to simplify and graph the given equation in order to analyze its features and solve for the roots.

1. Simplifying the Equation:
The given equation is:
y = (x + 5)(x + 1)/((x + 3)(x - 2)(x - 4)(x + 1)) - 1/(x - 2)

To proceed, we can simplify the equation by canceling out common factors if they exist. Simplifying further, we get:
y = (x + 5)/(x + 3)(x - 4)

2. Graphical Feature at x = -1:
To determine the graphical feature at x = -1, we can plot the graph of the equation using a graphing calculator or software. Alternatively, we can analyze the behavior of the equation.

To trace the graph or analyze the behavior of the equation, we can substitute x = -1 into the equation and calculate the corresponding y-value.

Substituting x = -1 into the simplified equation, we get:
y = (-1 + 5)/(-1 + 3)(-1 - 4)
y = 4/(-2)(-5)
y = 4/10
y = 2/5

Therefore, at x = -1, we have y = 2/5 or y = 0.4. This indicates that the graph intersects the y-axis at the point (x, y) = (-1, 0.4).

3. Absence of Vertical Asymptote at x = -1:
There is no vertical asymptote at x = -1 because it cancels out as a factor in the denominator of the equation. Additionally, the factor (x + 1) appears in both the numerator and denominator and gets canceled out. As a result, there is no vertical asymptote at x = -1.

4. Solving the equation for x when y = 0:
To find the roots of the equation, we need to solve for x when y = 0, as stated.

The equation can be rewritten as:
0 = (x + 5)/(x + 3)(x - 4)

Now, set the numerator equal to zero:
x + 5 = 0
x = -5

Next, set the denominator equal to zero:
(x + 3)(x - 4) = 0

Using the zero product property, we can set each factor equal to zero and solve for x:

x + 3 = 0
x = -3

x - 4 = 0
x = 4

Hence, the roots or solutions to the equation when y = 0 are x = -5, x = -3, and x = 4.

Note: In solving for the roots, we used the strategy of setting both the numerator and denominator equal to zero to ensure we capture all potential values of x that make the equation equal to zero.