pre-cal

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what is the vertex and directrix of the parabola y=-1/2x^2 + 10 - 46

• pre-cal -

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.

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