Asked by Rafi
Find b and c so that y=15x2+bx+c has vertex (5,5).
Answers
Answered by
oobleck
from the vertex form, we know that
y = a(x-5)^2 + 5
since a=15, that gives us
y = 15(x-5)^2 + 5 = 15x^2 - 150x + 380
so
b = -150
c = 380
y = a(x-5)^2 + 5
since a=15, that gives us
y = 15(x-5)^2 + 5 = 15x^2 - 150x + 380
so
b = -150
c = 380
Answered by
mathhelper
or...
we know the x of the vertex is -b/(2a)
-b/30 = 5
b = -150
also (5,5) must satisfy
5 = 15(25)-150(5) + c
c = 5 - 375 + 750 = 380
we know the x of the vertex is -b/(2a)
-b/30 = 5
b = -150
also (5,5) must satisfy
5 = 15(25)-150(5) + c
c = 5 - 375 + 750 = 380