Question

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)?

Answer 1

`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]}