SOLUTION: Which quadratic equation has maximum at (3, 2)?
Select one:
a. y = - x2 - 6x - 7
b. y = -x2 + 6x - 7
c. y = x2 + 6x - 7
d.
y = - x2 - 6x - 7
y = -x2 + 6x + 7
Algebra.Com
Question 853115: Which quadratic equation has maximum at (3, 2)?
Select one:
a. y = - x2 - 6x - 7
b. y = -x2 + 6x - 7
c. y = x2 + 6x - 7
d.
y = - x2 - 6x - 7
y = -x2 + 6x + 7
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
completing the Square:
y = -x2 + 6x - 7 = -(x-3)^2 + 9 - 7
y = -(x-3)^2 + 2 is a parabola opening downward from V(3,2), it max point
RELATED QUESTIONS
y=x2-6x+11
(answered by Alan3354)
Y=x2-6x+6
(answered by Alan3354)
Which choice is equivalent to the following equation?
y = (x - 3)2 + 36
A. y =... (answered by stanbon)
x2... (answered by jim_thompson5910)
1.
Rewrite the following quadratic equation in standard form: y = (x + 3)2 - 11
A. y (answered by Cromlix)
y=x2-6x+8
y=x-4
(answered by checkley77)
what is... (answered by user_dude2008)
y=x2+6x-2x+2x2-18+4
(answered by ikleyn,josgarithmetic)
Solve for x: y = x2 + 5
6x – y =... (answered by Fombitz)