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)
Answer:
why does it say answer doesn't exist but still lemme put a answer
The strategy is following
Smaller number subtract from the greater. Result is distance between two numbers
For example you have 3 and 8
8-3-5
5 is the distance between 3 and 8
It also works with the negative numbers, for example 3 and -5
3-(-5)=8
