what is the vertex and directrix of the parabola y=-1/2x^2 + 10 - 46
I'm assuming your equation is this:
y = -1/2x^2 + 10x - 46
The above equation would then be in the format:
y = ax^2 + bx + c
...where a, b, and c are the coefficients of each term in this format.
Therefore, in your equation, a = -1/2, b = 10, and c = -46
To find the vertex, use this:
-b/2a for x
(4ac - b^2)/4a for y
You can also find y by substituting the value for x into the equation and solving for y.
Vertex will be in the format (x,y).
I'll let you take it from here.