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.

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