the axis of symmetry for the function $f(x)=-x^{2}-10x + 16$ is $x=-5$. what are the coordinates of the…

the axis of symmetry for the function $f(x)=-x^{2}-10x + 16$ is $x=-5$. what are the coordinates of the vertex of the graph?\n(-5, 41)\n(-5, 56)\n(-5, 76)\n(-5, 91)

the axis of symmetry for the function $f(x)=-x^{2}-10x + 16$ is $x=-5$. what are the coordinates of the vertex of the graph?\n(-5, 41)\n(-5, 56)\n(-5, 76)\n(-5, 91)

Answer

Answer:

A. (-5, 41)

Explanation:

Step1: Recall vertex - axis of symmetry relation

The x - coordinate of the vertex of a parabola is on the axis of symmetry. Given axis of symmetry $x=-5$, so $x_{vertex}=-5$.

Step2: Substitute x - value into function

Substitute $x = - 5$ into $f(x)=-x^{2}-10x + 16$. $f(-5)=-(-5)^{2}-10\times(-5)+16$.

Step3: Calculate the value of f(-5)

First, calculate $-(-5)^{2}=-25$, $-10\times(-5) = 50$. Then $f(-5)=-25 + 50+16=41$. So the vertex is $(-5,41)$.