Question 16 of 25 What are the coordinates of the vertex of the parabola described by the
equation below?
y=-7(x-4)^2-5
A. (-4,5)
B. (5,4)
C. (4,-5)
D. (-5, -4)

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)

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Thanks

-1/2 +2

step-by-step explanation:

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length