Ok, so user says that it should be solve for vertex not vertex form
(x,y)
to find the vertex of
y=ax^2+bx+c
the x value of the vertex is -b/2a
the y value is found by plugging in the x value for the vertex back into the original equation and evaluating
y=-2x^2-12x-28
a=-2
b=-12
xvalue of vertex is -(-12)/(2*-2)=12/-4=-3
x value of vertex is -3
plug backin for x
y=-2x^2-12x-28
y=-2(-3)^2-12(-3)-28
y=-2(9)+36-28
y=-18+8
y=-10
yvalue is -10
x value is -3
vertex is (-3,-10)
The coordinate of points K, L and M is the location of the points on a coordinate plane
<h3>How to determine the missing coordinates?</h3>
The given parameters are:
K = (10, )
L = ( ,10)
M = (30, )
The question has missing parameters.
So, I will assume that the line is a perfectly horizontal line.
This means that the y-coordinates of points K, L and M are equal.
The y-coordinate of point L is 10.
So, we have:
K = (10, 10)
L = ( ,10)
M = (30, 10)
Assume that point L is halfway points K and M, then we have:
K = (10, 10)
L = (20, 10)
M = (30, 10)
See attachment for the diagram of the coordinate plane showing the line
Read more about coordinate planes at:
brainly.com/question/7243416
#SPJ1
True. A^2 + B^2 = C^2
A and B are the legs, C is the hypotenuse