How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. $y = - x^{2} + 6 x - 9$ $y = -x^2 + 6x -9$ ?

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Answer 1 · 2015-05-04T15:14:59

Coordinate of vertex:

x = -b/2a = -6/-2 = 3
y = f(3) = -9 + 18 - 9 = 0.

Axis of symmetry: x = -b/2a = 3
$x - intercepts: - (x^{2} - 6 x + 9) = 0 = - {(x - 3)}^{2}$ $x-"intercepts: "-(x^2 - 6x + 9) = 0 = -(x -3)^2$

There is double root at x = 3.

How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. y = -x^2 + 6x -9y=−x2+6x−9y = -x^2 + 6x -9?