find the maximum value of y=-x^2+6x?

Answer 1

don't know if you've taken Calc I or not, but if you have, differentiate your equation. If not, you can look at the next paragraph.You'll get y'=-2x+6. Remember that deriviatives tell the slope of a function at the x input. Also remember that quadriatics are U shaped with the co-efficient of the squared term telling if the equation opens up or down. (positive up, negative down) Since it's negative, the equation will open down and have a maximum point. (the slope of the max point will be 0) so y'= -2x+6=0 (x=3) will give max value (3,9)

Set your equation equal to zero and solve.
(-x^2+6x=0)
-x(x-6)=0
x=0,6
since quadriatics are symetrical around the max/min point, find the point half way between 0 and 6 (x=3) and this will yeild your max/min point. (3,9)

Answer 2

y= -x^2 + 6x
= 9 - 9 + 6x - x^2
= 9 - (x^2 - 6x + 9)
= 9 - (x - 3)^2

Minimum value of (x - 3)^2 = 0 and that occurs when x = 3
So maximum value of y = 9 and occurs when x = 3

Answer 3

y = -x^2+6x

y' = -2x+6 = 0
-2x = -6
x = (-6)/(-2)
x = 3

y = -3^2+6*3 = 9

So the max value is 9

Answer 4

it is 3

Answer 5

y = - x^2 + 6x
y' = -2x + 6 = 0 for maximum, or
x = 3
y = -9 + 18 = 9

check:
-2.999(-2.999 + 6) = -8.999999
-3.001(-3.001 + 6) = -8.999999

Answer 6

graph it out but i think it the answer is 0

Answer 7

i suck at math.
so im not even gonna try.