Find the vertex for these parabolas?

Question

x^2-2x-24=0

and

3m^2-6m+1=0

thanks so much

Answer 1

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

Answer 2

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)

Answer 3

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.

Answer 4

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.

Answer 5

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