78.8k views
5 votes
Given y-k=a(x-h)^2, a=-1, h>0, k<0 in which quadrant is the vertex

User Artier
by
7.9k points

1 Answer

5 votes

Answer:

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.

Explanation:


User DadyByte
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories