Find an equation of parabola that has the indicated vertex and whose graph passes through the given point.
Vertex (-3,5)
Point (-6,1)
****Point(-6,-1)
from the vertex, we know that
y = a(x+3)^2+5
Now plug in the values to get a:
1 = a(-6+3)^2 + 5
9a = -4
a = -4/9
y = -4/9 (x+3)^2 + 5
See
http://www.wolframalpha.com/input/?i=y+%3D+-4%2F9+%28x%2B3%29^2+%2B+5
To find an equation of a parabola with a given vertex and a point on the graph, you can use the general form of the equation for a parabola, which is (x - h)^2 = 4p(y - k), where (h, k) is the vertex of the parabola.
We are given the vertex (-3, 5), so h = -3 and k = 5. Substituting these values into the equation, we have (x + 3)^2 = 4p(y - 5).
Next, we need to find the value of p, which is the distance from the vertex to the focus of the parabola. To find p, we can use the given point (-6, 1) on the parabola.
Since the vertex is (-3, 5), the focus will be p units closer to the directrix, which is a horizontal line parallel to the x-axis. In this case, the directrix will be 5 units below the vertex.
The distance from the directrix to the vertex is 5 units, so the focus will be 5 units above the vertex. Therefore, p = 5.
Now we can substitute the values of h, k, and p into the equation to get the final equation of the parabola:
(x + 3)^2 = 4(5)(y - 5)
Simplifying,
(x + 3)^2 = 20(y - 5)
This is the equation of the parabola with the given vertex (-3, 5) and the point (-6, 1).