algebra
posted by drew .
Find the maximum or minimum of the following quadratic function: y = x2  x  56.
A. 56
B. 56
C. 225/4
D. 225/4
d

1. y=x2  16
for maximum or minumum, dy/dx=0
hence 2x=0 or x=0
minimum value of function occurs when x=0 which is y=16
2.using same principle as above,
dy/dx=10x+5
when dy/dx=0 => 10x+5=0 => x=1/2
minimum value of y= 5/4  5/2 + 11
for 3, 4 and 5 use the same principle as 1 and 2 
max: infinity (at x=inf)
min: x=1/2, y= you figure it. check sign. 
Not sure if you know Calculus, as done by Me
so let's complete the square
y = x^2  x  56
= x^2  x + 1/4  1/4  56
= (x 1/2)^2  225/4
so we have a min of 225/4 , when x = 1/2
(min because the parabola opens upwards)
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