SOLUTION: I'm struggling with this. I need to find the vertex, line of symmetry, and the max or min value of f(x). The equation is f(x)=1/3 (x+7)^2 + 9

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Question 481534: I'm struggling with this. I need to find the vertex, line of symmetry, and the max or min value of f(x).
The equation is f(x)=1/3 (x+7)^2 + 9
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
f(x) = (1/3) * (x+7)^2 + 9
multiply it out to get:
f(x) = (1/3) * (x^2 + 14x + 49) + 9
this becomes:
(1/3)x^2 + (14/3)x + (49/3) + 9 which becomes:
(1/3)x^2 + (14/3)x + (49/3) + (27/3) which becomes:
(1/3)x^2 + (14/3)x + (76/3)
this is a quadratic equation in the form of ax^2 + bx + c
a = (1/3)
b = (14/3)
c = (76/3)
the x value of the vertex is given by the equation x = -b/2a.
that would be (-14/3)/(2/3) which is equal to (-14/3) * (3/2) which is equal to (-7).
the y value of the vertex would be f(-7) which would be equal to:
(1/3)(49) + (14/3)(-7) + (76/3) which becomes:
(49/3) - (98/3) + (76/3) which becomes:
27/3.
the coordinates of the vertex of the equation becomes (x,y) = (-7,27/3).
that's also the max/min point.
the axis of symmetry of the graph is at x = -7.
a graph of your equation looks like this:

that vertical line is at x = -7.
that horizontal line is at y = (27/3).
that vertical line is the axis of symmetry.
the intersection of the vertical line and the horizontal line is the vertex of the graph.
it is also the max/min point of the graph.