To get the vertex of the parabola we proceed as follows;
y=-7(x-4)^2-5
The above can be written as:
y=-7x^2+56x-117
The values of a,b and c are:
a=-7, b=56 and c=-117
x=-b/(2a)
x=-56/(-7*2)=4
but;
y=-7x^2+56x-117
y=-7(4)^2+56(4)-117
y=-5
Thus;
x=4 and y=-5
The vertex will be at point (4,-5)
Discussion
The discriminate is b^2 - 4*a*c
The general equation for a quadratic is ax^2 + bx + c
In this equation's case
a = 1
b= -5
c = - 3
Solve
(-5)^2 - 4*(1)*(-3)
25 - (-12)
25 + 12
37
Note
Since the discriminate is > 0, the roots are real and different. The roots do exist and there are 2 of them.
Adding the two equations
3x + 6y + 3x - 6y = 36+0
6x = 36
x = 6
.
subtracting second equation from first
(3x + 6y ) - (3x-6y) =36 -0
12y = 36
y = 3
.
therefore
x = 6 and y = 3
(6,3)
I don’t know if this is correct but about 4 times because
3/4 -> 18/24 and 1/6 -> 4/24
0.0000021 you need to divide 2.1 by 10^6