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x^2-2x-24=0

and

3m^2-6m+1=0

thanks so much

2007-05-19 04:47:48 · 5 answers · asked by Irish 3 in Science & Mathematics Mathematics

5 answers

1)
vertex(r,k)

r=-b/(2a)
=-(-2)/(2*1)=1

k=(4ac-b^2)/(4a)
=(4*1*(-24))/(4*1)
=-24

2)
r=-(-6)/(2*3)=-1

k=(4*3*1-(-6)^2)/(4*3)
=(12-36)/12
=-2

2007-05-19 04:58:02 · answer #1 · answered by iyiogrenci 6 · 0 0

They should be equal to y not 0.
vertex = turning point?
in that case...
x^2-2x-24=0
(x-1)^2-25=0
turning point
(1,-25)

and
3m^2-6m+1=y
dy/dx=6m-6
0=6m-6
6=6m
=> m=1
(1,-2)

2007-05-19 11:55:43 · answer #2 · answered by Anonymous · 0 0

y=x^2-2x-24=(x-6)(x+4)
x=(6-4)/2 = 1, y=-25
vertex(1, -25)

y=3m^2-6m+1
m=-b/(2a)=6/(2*3)=1, y=-2
vertex(1, -2)
----------
I showed you two different ways to find the vertex point for a parabolas.

2007-05-19 11:52:27 · answer #3 · answered by sahsjing 7 · 0 0

ok in the quadratic forumla u hvae -b + - sqrt b2 - 4ac / 2a
now the -b/2a part is the equation to ur vertex
so in the first one ..
u shud have x = 1
and the second one x = 1

hope it helps u out.
BTW if u are wondering the b2 - 4ac is also known as the discriminant and it tells u how many solutions ur quadratic equation has.. less than 0 it means u have 2 imaginary roots,
0 means u have 1 root.. greater than 0 means u have 2 real roots.

2007-05-19 12:12:13 · answer #4 · answered by sudhi_kandi 3 · 0 0

Is the first one x=6, x=-4?

2007-05-19 11:52:10 · answer #5 · answered by Brent F 2 · 0 1

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