Answer:
The coordinates of the vertex are (-5, 41)
Step-by-step explanation:
For a quadratic function of the form
[tex]f(x) = ax ^ 2 + bx + c[/tex]
Where a, b and c are constants and represent the coefficients of the function, then the symmetry of the parabola always passes through its vertex.
In this case we have the following parabola
[tex]f (x) = -x2 - 10x + 16[/tex]
And we know that its axis of symmetry is the line [tex]x = -5[/tex]
Then we know that this axis of symmetry passes through the vertex of the parabola.
Therefore, the x coordinate of the vertex is -5.
To find the coordinate in y of the vertex, we substitute x = -5 in the function.
[tex]f (-5) = -(- 5) ^ 2 -10 (-5) +16\\\\f (-5) = 41[/tex]
Finally, the vertices are in the point (-5, 41).
