To determine the coordinates of the vertex of the function [tex]\( f(x) = -x^2 - 10x + 16 \)[/tex], we start with the information that the axis of symmetry is at [tex]\( x = -5 \)[/tex]. The vertex of a parabola given by a quadratic function [tex]\( ax^2 + bx + c \)[/tex] lies on its axis of symmetry. Here, since we already know that the axis of symmetry is [tex]\( x = -5 \)[/tex], the x-coordinate of the vertex is [tex]\(-5\)[/tex].
Next, we need to find the y-coordinate of the vertex by substituting [tex]\( x = -5 \)[/tex] into the function [tex]\( f(x) \)[/tex].
[tex]\[
f(-5) = -(-5)^2 - 10(-5) + 16
\][/tex]
Now calculate each term step-by-step:
[tex]\[
(-5)^2 = 25
\][/tex]
[tex]\[
-(-5)^2 = -25
\][/tex]
[tex]\[
10(-5) = -50 \quad \text{so} \quad -10(-5) = 50
\][/tex]
Putting it all together:
[tex]\[
f(-5) = -25 + 50 + 16
\][/tex]
Simplify the result:
[tex]\[
-25 + 50 = 25
\][/tex]
[tex]\[
25 + 16 = 41
\][/tex]
As a result, the y-coordinate of the vertex is [tex]\( 41 \)[/tex]. Therefore, the coordinates of the vertex are [tex]\( (-5, 41) \)[/tex].
From the given options, the correct answer is:
[tex]\[
(-5, 41)
\][/tex]
