How many $y$-intercepts does the graph of the parabola $x = y^2 - 4y - 1$ have?
I know that the $y$-intercept of a graph is the point where it crosses the $y$-axis, which means that the $x$-coordinate of that point is zero. So, to find the $y$-intercepts of this parabola, I need to solve the equation $x = 0$ for $y$. Substituting $x = 0$, I get $0 = y^2 - 4y - 1$, which is a quadratic equation in $y$. I can try to factor this equation, or use the quadratic formula, or complete the square. I'll try factoring first, since it's usually the fastest method. I'm looking for two numbers that multiply to $-1$ and add to $-4$. I see that $-1$ and $1$ are the only factors of $-1$, but they add to zero, not $-4$. I'll use the quadratic formula instead. The quadratic formula says that if I have an equation of the form $ax^2 + bx + c = 0$, then the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In this case, $y$ is playing the role of $x$, and the coefficients are $a = 1$, $b = -4$, and $c = -1$. Plugging these values into the formula, I get $y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)}$. Simplifying, I get $y = \frac{4 \pm \sqrt{20}}{2}$. I can also write this as $y = 2 \pm \sqrt{5}$, by dividing everything by $2$. So, the parabola has two $y$-intercepts, corresponding to the two values of $y$ that I found. They are $(0, 2 + \sqrt{5})$ and $(0, 2 - \sqrt{5})$. 