SOLUTION: what is the vertex of f(x) = x^2 -6x + 3

Question 884576: what is the vertex of f(x) = x^2 -6x + 3
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
since the quadratic equation is in the standard form of ax^2 + bx + c = 0, you can determine the value of a, b, and c.
you get:
a = 1
b = -6
c = 3
the formula for the line of symmetry is x = -b/2a which becomes x = 6/2 which becomes 3.
the vertex has the coordinates of (-b/a2a, f(-b/2a)
to find f(-b/2a), you solve for f(3) to get f(3) = 3^2 - 6*3 + 3 which results in f(3) = 9 - 18 + 3 which results in f(3) = -6.
your vertex is at (3,-6)
the graph of your equation is shown below:

draw a vertical line at x = 3 and a horizontal line as y = -6 and the intersection of those lines will be the minimum point on the graph of y = x^2 - 6x + 3.
the vertex of a quadratic equation is either the minimum point on the graph or the maximum point on the graph.
for this graph, the vertex is the minimum point on the graph because the coefficient of the x^2 term is positive.

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