2006-07-29 08:18:38 · 4 answers · asked by Brandon ツ 3 in Science & Mathematics ➔ Mathematics
. (3, –136)
b. (–6, –136)
c. (6, –244)
d. (–6, 108)
2006-07-29 08:31:11 · update #1
Above are the answers for y = 4x^2 + 48x + 8
2006-07-29 08:31:46 · update #2
Vertex: -6 and -136
2006-07-29 08:50:42 · answer #1 · answered by gcfnl 1 · 3⤊ 0⤋
The vertex is the only point on the parabola where the slope = 0. So find the derivative:
y' = 8x + 48
Where is y' = 0?
0 = 8x + 48
-48 = 8x
- 6 = x
At x = -6 is the vertex. The y-coordinate at this point on the curve is:
y = 4(-6)^2 + 48(-6) + 8
y = 4*36 - 288 + 8
y = 144 - 288 + 8
y =-136
So your coordinates are (-6, -136)
2006-07-29 11:41:02 · answer #2 · answered by jimbob 6 · 0⤊ 0⤋
y = 4x^2 + 48x + 8
y = 4(x^2 + 12x + 2)
x = (-b/2a)
x = (-12)/(2(1))
x = -12/2
x = -6
y = 4((-6)^2 + 12(-6) + 2)
y = 4(36 - 72 + 2)
y = 4(-36 + 2)
y = 4(-34)
y = -136
Vertex = (-6,-136)
ANS : b.
2006-07-30 05:58:26 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
its in the format y=ax^2 +bx +c. so x=-b/2a. plug in a=4 and b=48. and then you get teh x value for the vertex. take that and plug it back into the original equation to get the y value. HA theres your point.
good luck!
2006-07-29 08:22:56 · answer #4 · answered by sun34529 2 · 0⤊ 0⤋