# 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)?

Mar 27, 2017

$y = - {x}^{2} - 6 x - 5$
The vertex form of a quadratic equation (a parabola) is $y = a {\left(x - h\right)}^{2} + v$, where $\left(h , v\right)$ is the vertex. Since we know the vertex, the equation becomes $y = a {\left(x + 3\right)}^{2} + 4$.
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, $3 = a {\left(- 2 + 3\right)}^{2} + 4$. Simplifying, we get $3 = a + 4$. Thus, $a = - 1$. The quadratic equation is then $y = - {\left(x + 3\right)}^{2} + 4 = - {x}^{2} - 6 x - 9 + 4 = - {x}^{2} - 6 x - 5$. We can substitute the points in to verify this answer.