What is the minimum point of the graph of the equation y=(2x^2)+8x+9
A- (2,33)
B- (2,17)
C- (-2,-15)
D- (-2,1)
Best answer gets 10 pts!
2006-12-19 13:15:21 · 4 answers · asked by fachizzzzle 3 in Education & Reference ➔ Homework Help
the Derived of the eq. is y=4x+8, if 4x+8=0 then X=-2 if you substitute -2 in the original eq. then you get the minimum point that is
D: X=-2 Y=1
2006-12-19 13:18:28 · answer #1 · answered by oposites2 2 · 0⤊ 0⤋
(1) Y = 2X^2 + 8X + 9 => Y' = 4X + 8
The graph passes the minimum point when 4X + 8 = 0 or X = -2
Plug X = -2 into (1) => Y = 1
The minimum point is D(-2, 1)
2006-12-19 21:28:46 · answer #2 · answered by lloyd w 1 · 0⤊ 0⤋
the derivative of the equation is :dy/dx=4x+8. At the minimum point ,which is also the turning point ,the derivative is zero(0).So when you solve dy/dx=4x+8=0, the value of x comes -2.When you put x=-2 in"y" the value of y comes 1.Hence the minimum point is (-2,1)
2006-12-20 11:42:47 · answer #3 · answered by no one 1 · 0⤊ 0⤋
um
2006-12-19 21:17:19 · answer #4 · answered by Myhappylife 2 · 0⤊ 0⤋