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Luv2Teach
[tex]-1( x^{2} +10x+25)=-41[/tex]Do this by completing the square.  Start by setting the function equal to 0 and at the same time moving the constant over to the other side of the equals sign. [tex]- x^{2} -10x=-16[/tex].  The rules for completing the square is the leading coefficient has to be a positive 1, but ours is -1, so we have to factor out the negative, and when we do that, we have this: [tex]-1( x^{2} +10x)=-16[/tex].  10x is our linear term.  We will take half of tha 10 to get 5, square the 5 to get 25 and add 25 to both sides.  HOWEVER, that -1 is still hanging around out front, so what we actually have "added" in is -1(25) = -25.  So now what we have is this: [tex]-1( x^{2} +10x+25)=-16-25.  Simplifying we have [tex]-1( x^{2} +10x+25)=-41[/tex].  In this process we have created a perfect square binomial on the left: [tex]-1(x+5) ^{2}=-41 [/tex].  Now we will move the -41 by adding to get [tex]-1(x+5) ^{2} +41=y[/tex].  Our vertex, then, is (-5, 41). 

Answer:

(-5,41)

Step-by-step explanation:

The axis of symmetry for the function [tex]f(x) = -x^2 - 10x + 16[/tex] is x = -5

Axis of symmetry lies at the x coordinate of vertex

axis of symmetry is x=-5, so the x coordinate of vertex is -5

vertex is (x,y) that is (-5,y)

To find out y we plug in -5 for x in f(x)

[tex]f(-5) = -(-5)^2 - 10(-5)+ 16[/tex]

[tex]f(-5) = -25+50+ 16=41[/tex]

so the value of y is 41

Vertex is (-5,41)

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