To find the coordinates of the vertex of the quadratic function y = -x^2 + 6x - 11, we first need to find the x-coordinate of the vertex.
The x-coordinate of the vertex of a quadratic function in the form y = ax^2 + bx + c is given by x = -b/2a.
In this case, the coefficient of x^2 is a = -1 and the coefficient of x is b = 6. Plugging these values into the formula, we get x = -6 / (2 * -1) = 3.
Now that we have the x-coordinate of the vertex, we can substitute this back into the original function to find the y-coordinate.
y = -(3)^2 + 6(3) - 11
y = -9 + 18 - 11
y = -2
Therefore, the coordinates of the vertex of the graph of the function y = -x^2 + 6x - 11 are (3, -2).
Quadratic Functions
what are the coordinates of the vertex of the graph of the function y=-x^2+6x-11?
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