Given y-k=a (x-h) ^2, a=-1, h>0, k<0 in which quadrant is the vertex

Since the equation is a quadractic, the graph would be a parabola.

With a being - 1, the parabola will represent a reflection of the y values. In other words, the parabola will be upside down and the vertex will be a maximum value. Ultimately, the a in the function doesn't determine the location of the vertex.

Since the k value is negative, that means the equation begins y - (-k). The K value being negative restricts the transformation of the parabola to being down k number of units. The location of moving the parabola down places the vertex in the third or fourth quadrant.

The h value being positive means that the parabola is shifted to the right h number of units. For example, if the parabola f (x) = x² has a vertex at (0,0), the parabola f (x) = (x-2) ² must have a vertex at (2,0) because 2 - 2 = 0. Shifting right places the vertex of the parabola in the first or fourth quadrant.

Therefore the k value and h value restrictions must overlap in the fourth quadrant.