Final answer:
The vertex of the quadratic function f(x)=x^2-6x+45 is found using the vertex formula -b/(2a) for the x-coordinate and evaluating f(x) at this x-value for the y-coordinate. The vertex is at the point (3, 36).
Step-by-step explanation:
The question asks to find the vertex of the quadratic function f(x)=x^2-6x+45. To find the vertex of a quadratic function in the form of f(x)=ax^2+bx+c, we can use the vertex formula, which is given by -b/(2a) for the x-coordinate and f(-b/(2a)) for the y-coordinate of the vertex.
In this case, a=1 and b=-6.
Plugging these values into the formula gives us an x-coordinate for the vertex of -(-6)/(2*1) = 3.
The y-coordinate is found by substituting x=3 back into the function: f(3) = (3)^2 - 6*(3) + 45 = 9 - 18 + 45 = 36.
Therefore, the vertex of the function is at the point (3, 36).