Write the equation of the parabola in standard form with coordinates of points corresponding to P and Q: (-2,3) and (-1,0) and Vertex: (-3,4)?

1 Answer
Mar 27, 2017

Answer:

#y=-x^2-6x-5#

Explanation:

The vertex form of a quadratic equation (a parabola) is #y=a(x-h)^2+v#, where #(h, v)# is the vertex. Since we know the vertex, the equation becomes #y=a(x+3)^2+4#.

We still need to find #a#. To do so, we pick one of the points in the question. I will choose P here. Substituting in what we know about the equation, #3=a(-2+3)^2+4#. Simplifying, we get #3=a+4#. Thus, #a=-1#. The quadratic equation is then #y=-(x+3)^2+4=-x^2-6x-9+4=-x^2-6x-5#. We can substitute the points in to verify this answer.

graph{y=-x^2-6x-5 [-16.02, 16.01, -8.01, 8.01]}