<Question>
How many $y$-intercepts does the graph of the parabola $x = y^2 - 4y - 1$ have?
</Question>

<think>
<Plan_1>
Okay, so I need to figure out how many y-intercepts the graph of the parabola given by the equation x = y² - 4y - 1 has. Hmm, let me start by recalling what a y-intercept is. If I remember correctly, a y-intercept occurs where the graph crosses the y-axis. At that point, the value of x should be zero, right? Because the y-axis is where x = 0.

So, to find the y-intercepts, I need to set x = 0 and solve for y. That makes sense. Let me write that down.

Setting x = 0 in the equation:
0 = y² - 4y - 1

Now, this is a quadratic equation in terms of y. The general form of a quadratic equation is ay² + by + c = 0. Here, a = 1, b = -4, and c = -1. To find the solutions for y, I can use the quadratic formula: y = [-b ± √(b² - 4ac)] / (2a). Let me apply that here.

<prompt_1.1> Use the quadratic formula to find the solutions for y in the equation 0 = y² - 4y - 1. </prompt_1.1>

Wait, but let me double-check my steps to make sure I didn't make a mistake. First, identifying the y-intercept correctly by setting x=0.

<prompt_1.2> Verify the steps taken to find the y-intercepts and confirm the number of y-intercepts. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let's compute the discriminant, which is the part under the square root: b² - 4ac. Plugging in the values:

Discriminant = (-4)² - 4 * 1 * (-1) = 16 - (-4) = 16 + 4 = 20.

Wait, hold on. That doesn't seem right. Wait, no, actually, the discriminant is b² - 4ac. So, substituting the values again:

b² = (-4)² = 16

4ac = 4 * 1 * (-1) = -4

So, discriminant = 16 - (-4) = 16 + 4 = 20. Yeah, that's correct. So the discriminant is 20, which is positive, so there are two distinct real roots. That means there are two y-intercepts.
</execution_1.1>

<execution_1.2>
Then solving for y. The equation becomes y² - 4y - 1 = 0. Yes, that's correct. Quadratic in y, so using quadratic formula.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Therefore, the graph crosses the y-axis at two points. So, there are two y-intercepts.
-----

But just to be thorough, let me visualize the graph. The given equation is x = y² - 4y - 1. This is a parabola that opens either to the right or to the left because it's in the form x = ay² + by + c. Since the coefficient of y² is positive (1 in this case), the parabola opens to the right. 

A parabola that opens to the right can indeed intersect the y-axis at two points, one point (if it's tangent), or no points.

<prompt_2.1> Visualize the graph of the parabola and confirm the number of y-intercepts based on the direction it opens and the discriminant. </prompt_2.1>

Alternatively, maybe I can complete the square for the equation x = y² - 4y - 1 to better understand its graph. Let's try that.

<prompt_2.2> Complete the square for the equation x = y² - 4y - 1 and describe the vertex and direction of the parabola. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Since the discriminant is positive (20), it confirms that it intersects the y-axis at two points. Therefore, two y-intercepts. That all lines up. 

Wait, just to make sure I didn't make a mistake in calculating the discriminant. Let's do that part again:

Discriminant D = b² - 4ac = (-4)² - 4*(1)*(-1) = 16 - (-4) = 16 + 4 = 20. Yes, that's correct. So two real roots.
</execution_2.1>

<execution_2.2>
Starting with the right-hand side: y² - 4y - 1.

To complete the square for the y terms:

Take y² - 4y. The coefficient of y is -4. Half of that is -2, and squaring it gives 4. So, we add and subtract 4:

y² - 4y + 4 - 4 - 1 = (y - 2)² - 5.

So, the equation becomes x = (y - 2)² - 5.

This is now in the form x = (y - k)² + h, where (h, k) is the vertex of the parabola. Wait, actually, in standard form, it's x = a(y - k)² + h, so the vertex is at (h, k). Here, h is -5 and k is 2, so the vertex is at (-5, 2). 

Since the parabola opens to the right (as the coefficient a = 1 is positive), the vertex is the leftmost point of the parabola. So, the vertex is at (-5, 2), which is 5 units to the left of the y-axis. Since it opens to the right, it will extend towards the y-axis. 

Now, since the vertex is 5 units left of the y-axis, and the parabola opens to the right, it will intersect the y-axis (x=0) if the distance from the vertex to the y-axis (which is 5 units) is less than or equal to the "width" of the parabola. But actually, for a parabola, the number of intersections with a vertical line depends on the discriminant, which we already calculated as 20, which is positive, so two points.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Therefore, the graph of the parabola x = y² - 4y - 1 has two y-intercepts.
-----

Wait, but just to visualize, the vertex is at (-5, 2).
So, the parabola is symmetric about the line y = 2. When x = 0, solving gives two y-values: y = 2 ± √5. So, those are approximately 2 + 2.236 = 4.236 and 2 - 2.236 = -0.236. So, the two points where it crosses the y-axis are (0, 4.236) and (0, -0.236). That makes sense. Since the vertex is at (-5, 2), the parabola is symmetric around y = 2, so it curves outwards to the right. Therefore, crossing the y-axis at two points.

Based on execution_2.2:
I think that's thorough enough.
The key steps were setting x=0, solving the quadratic, calculating the discriminant, and confirming that it's positive, hence two real solutions. Therefore, two y-intercepts.
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the graph of the parabola $x = y^2 - 4y - 1$ has $\boxed{2}$ $y$-intercepts.
</Final_answer>
</think>