Questions: Find the vertex, focus, and directrix of the parabola. Graph the equation. (y+1)^2=4(x-1) The vertex of the parabola is (Type an ordered pair.) The focus of the parabola is (Type an ordered pair.) The directrix of the parabola is (Simplify your answer. Type an equation.)

Solution Steps

The given equation is in the form $(y-k)^2 = 4p(x-h)$, where $(h, k)$ is the vertex. Comparing this with $(y+1)^2 = 4(x-1)$, we have $h=1$ and $k=-1$. Thus, the vertex is $(1, -1)$.

Since the parabola is in the form $(y-k)^2 = 4p(x-h)$, it opens horizontally. Here, $4p = 4$, so $p=1$. Since $p>0$, the parabola opens to the right. The focus is at $(h+p, k)$. Substituting $h=1$, $k=-1$, and $p=1$, the focus is at $(1+1, -1) = (2, -1)$.

The equation of the directrix is $x = h-p$. Substituting $h=1$ and $p=1$, we have $x = 1-1 = 0$. Thus, the directrix is the equation $x=0$.

Final Answer: