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's 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 x-coordinate is 0, right? So, to find the y-intercepts, I should set x equal to 0 and solve for y. That makes sense. 

Let me write that down. If x = 0, then the equation becomes 0 = y² - 4y - 1. So, the quadratic equation is y² - 4y - 1 = 0. Now, I need to solve this quadratic for y. Since it's a quadratic equation, there can be zero, one, or two real solutions, which would correspond to the number of y-intercepts. 

To solve y² - 4y - 1 = 0, I can use the quadratic formula. The quadratic formula is y = [4 ± √(b² - 4ac)] / (2a). Here, a is 1, b is -4, and c is -1. Let me plug those values into the formula. 

First, calculate the discriminant: b² - 4ac. That would be (-4)² - 4*1*(-1). So, 16 - (-4) which is 16 + 4 = 20. The discriminant is 20, which is positive. Since the discriminant is positive, there are two distinct real roots. That means there are two y-intercepts. 

Wait, but let me make sure I didn't make a mistake in calculating the discriminant. Let's double-check. The general form is ax² + bx + c, so here, a=1, b=-4, c=-1. So discriminant D = (-4)² - 4*1*(-1) = 16 - (-4) = 16 + 4 = 20. Yep, that's correct. So two real solutions. 

Therefore, the parabola should cross the y-axis at two points. But let me think again. The equation given is x = y² - 4y - 1. This is a parabola that opens either to the right or to the left. Since the coefficient of y² is positive (1), the parabola opens to the right. Typically, a parabola that opens to the right can have a maximum of two y-intercepts because it's a function of y. Wait, but actually, in this case, since x is expressed in terms of y, it's a sideways parabola. So normally, for a sideways parabola, each x value can correspond to two y values. But in terms of y-intercepts, which are points where x=0, there can be two points where the parabola crosses the y-axis. 

But let's visualize this. If the parabola opens to the right, its vertex is at the minimum x-value. If the vertex's x-coordinate is less than 0, then the parabola would cross the y-axis twice. If the vertex's x is equal to 0, it would touch the y-axis at one point (a tangent). If the vertex's x is greater than 0, then the parabola wouldn't cross the y-axis at all. 

So, let's find the vertex of this parabola. The vertex of a parabola given by x = ay² + by + c is at y = -b/(2a). Here, a=1, b=-4, so y = -(-4)/(2*1) = 4/2 = 2. So the y-coordinate of the vertex is 2. Then, the x-coordinate is x = (2)² - 4*(2) - 1 = 4 - 8 -1 = -5. So the vertex is at (-5, 2). Since the vertex is at x = -5, which is to the left of the y-axis (x=0), and the parabola opens to the right, it should cross the y-axis twice. That matches with the discriminant being positive, so two real solutions. 

Therefore, the graph of the parabola has two y-intercepts. 

But just to be thorough, let's solve the quadratic equation again to find the exact y-values. 

The equation is y² - 4y - 1 = 0. Using the quadratic formula: y = [4 ± √(16 + 4)] / 2 = [4 ± √20]/2. √20 simplifies to 2√5, so y = [4 ± 2√5]/2 = 2 ± √5. So the two solutions are y = 2 + √5 and y = 2 - √5. Both are real numbers, so the parabola crosses the y-axis at these two points. 

Therefore, the answer is 2 y-intercepts. 

Wait, but let me check if there's any other way this could be approached. For example, maybe completing the square? Let's try that. 

Starting with y² - 4y -1 = 0. Completing the square for y: 

y² - 4y = 1. 

Take half of the coefficient of y, which is -2, square it to get 4. Add that to both sides: 

y² - 4y + 4 = 1 + 4 => (y - 2)² = 5. 

Taking square roots: y - 2 = ±√5 => y = 2 ± √5. 

Same result as before. So that confirms the solutions. 

Therefore, there are two y-intercepts. 

I think that's solid. The discriminant is positive, so two real solutions, and solving confirms that. The vertex is to the left of the y-axis, and since the parabola opens to the right, it must cross the y-axis twice. 

So, the final answer is 2. The graph has two y-intercepts.
Therefore, the graph of the parabola $x = y^2 - 4y - 1$ has $\boxed{2}$ $y$-intercepts.
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