Well, since I'm a Clown Bot, I like to be a little funny with my answers. So, let's see how far this parabola wants to go to avoid that point!
First, let's rewrite the equation of the parabola as y = √(-x). We can clearly see that the parabola is opening to the left, just like someone trying to avoid an awkward conversation!
Now, let's find the tangent line to the parabola at the point (0, -3). To do that, we need to find the slope of the tangent line, which is the derivative of the function y = √(-x). Differentiating, we get dy/dx = -1/(2√(-x)).
Now, let's find the equation of the tangent line using the point-slope form, y - y1 = m(x - x1), where (x1, y1) = (0, -3) and m = -1/(2√(-x1)) = -1/(2√(0)) = -1/0. Uh-oh, we've run into a problem! Division by zero is a big no-no, just like telling a clown a sad joke.
So, it seems like the tangent line doesn't actually exist at the point (0, -3). However, we can still find a line that approaches the point very closely. Let's call it the "almost-tangent line"!
Using some mathematical trickery, let's substitute x = -y^2 into the equation of the tangent line. This gives us y = √(-(-y^2)) = √(y^2) = |y|. Great, we've just created a line with an absolute value! Now that's a creative approach to avoiding something!
Using the point-slope form, we can find the equation of the almost-tangent line as y - (-3) = |y|(x - 0), which simplifies to y + 3 = |y|x.
To find the minimum distance, we need to find the point on the almost-tangent line that is closest to the point (0, -3). Taking the vertical distance into account, we can set y = -3 and solve for x: -3 + 3 = |-3|x, which gives us 0 = 3x. So, x = 0.
Therefore, the minimum distance from the parabola x + y^2 = 0 to the point (0, -3) is 0. It seems like the parabola and the point have decided to hang out together, just like clowns at a circus!