SOLUTION: It says to find the vertex, but the fraction is throwing me off. 1/2x^2+6x+7

Question 487757: It says to find the vertex, but the fraction is throwing me off.
1/2x^2+6x+7
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
this is a quadratic equation in standard form.
the standard form of a quadratic equation is ax^2 + bx + c = 0
set your equation to 0 and it becomes:
(1/2)x^2 + 6x + 7 = 0
this means that:
a = (1/2)
b = 6
c = 7
the formula for finding the x value of the vertex is:
x = -b/2a
substitute for b and a in this formula to get:
x = -6 / (2 * (1/2)) which becomes:
x = -6 / 1 which becomes:
x = -6
to find the value of y for the vertex, then substitute -6 in the equation to get:
y = (1/2)x^2 + 6x + 7 becomes:
y = (1/2)*(-6)^2 + 6*(-6) + 7 which becomes:
y = (1/2)*36 - 36 + 7 which becomes:
y = 18 - 36 + 7 which becomes:
y = -11
the vertex should be at the point (x,y) = (-6,-11)
a graph of your equation looks like this:

i drew a horizonal line at y = -11 and a vertical line at x = -6 to show your where the vertex would be.
it would be at the intersection of the vertical and horizontal line.
the vertical line is not perfectly vertical, but it should be vertical enough for you to get the idea.

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