The axis of symmetry for the function f(x) = −x2 − 10x + 16 is x = −5. What are the coordinates of the vertex of the graph? (−5, 41) (−5, 56) (−5, 76) (−5, 91)
The x-coordinate of the axis os symmetry of a quadratic is the x-coorfinate of its vertex. To find the y-coordinate of the vertex here we would substitute -5 for x in:
[tex] f(x) = - {x}^{2} - 10x + 16 \\ so \: we \: have \: \\ f( - 5) = - ( { - 5)}^{2} - 10( - 5) + 16 \\ = - 25 + 50 + 16 = 41[/tex] The coordinates of the vertex are (-5,41)
