We are given the equation of the parabola:

$x^2 + 2x = -y + 1$

Step 1: Rearrange the equation

To better analyze this equation, let's first rearrange it so that the $y$-terms are on one side and the $x$-terms are on the other:

$x^2 + 2x + y = 1$

Now, move the constant to the other side:

$x^2 + 2x = -y + 1$

$y = -x^2 - 2x + 1$

Step 2: Complete the square

To rewrite the equation in a form that reveals the vertex, we need to complete the square for the $x$-terms. The $x$-terms in this equation are $x^2 + 2x$.

1. Take half of the coefficient of $x$, which is $2$, then square it: $\left(\frac{2}{2}\right)^2 = 1$.
2. Add and subtract this number inside the equation:

$y = -(x^2 + 2x + 1 - 1) + 1$

Now simplify:

$y = -(x + 1)^2 + 1$

So the equation becomes:

$y = -(x + 1)^2 + 1$

This is now in vertex form, where the vertex is at $(-1, 1)$.

Step 3: Identify the vertex, focus, and directrix

1. Vertex: From the equation $y = -(x + 1)^2 + 1$, the vertex is $(-1, 1)$.

2. Focus: The equation $y = a(x - h)^2 + k$ has the general form of a parabola that opens up or down. For $y = -(x + 1)^2 + 1$, the parabola opens downward because the coefficient of $(x + 1)^2$ is negative.

   The standard form of a parabola is $y = a(x - h)^2 + k$, and the distance from the vertex to the focus is given by:

   $\text{Distance} = \frac{1}{4a}$

   Here, $a = -1$, so:

   $\text{Distance} = \frac{1}{4(-1)} = -\frac{1}{4}$

   Since the parabola opens downward, the focus will be at a distance of $\frac{1}{4}$ units below the vertex. The vertex is at $(-1, 1)$, so the focus is at:

   $\left(-1, 1 - \frac{1}{4}\right) = \left(-1, \frac{3}{4}\right)$

3. Directrix: The directrix is a horizontal line located at a distance of $\frac{1}{4}$ units above the vertex. Since the vertex is at $(-1, 1)$, the directrix will be the line:

   $y = 1 + \frac{1}{4} = \frac{5}{4}$

Step 4: Graph the parabola

The graph will have the following characteristics:

• The vertex is at $(-1, 1)$.
• The focus is at $(-1, \frac{3}{4})$.
• The directrix is the line $y = \frac{5}{4}$.

I can help you graph it, or if you'd prefer, I can generate a visual for you. Let me know!

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Would you like more details on how to plot the parabola or any other aspect of this problem?

Here are some related questions:

1. How do I find the vertex of a parabola from its standard equation?
2. How do you find the distance to the focus and the directrix?
3. What is the difference between a parabola that opens upward versus downward?
4. How do you complete the square when the coefficient of $x^2$ is negative?
5. How can I apply this to other types of parabolas (sideways or vertically shifted)?

Tip: Completing the square is a key step when dealing with quadratic equations and helps convert a general equation into vertex form, making it easier to identify the vertex, focus, and directrix.