Need help finding vertex form

y = (x-.5)^2 -12.25

Step-by-step explanation:

first, find the vertex

x from (x,y) = -b/2a

ax^2-bx - c

1x^2-1x -12

-1/2(-1) = -1/-2 - 1/2

(1/2,y) then use x value to find y value

y = 1/2^2 - 1/2 - 12

y = .25 -.5 - 12    = -12.25

(.5,-12.25) input values into this formula:

y = a(x-h)^2 + k where h = x-value and k = y-value. use same a value = 1

y = (x-.5)^2 -12.25

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step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: