SOLUTION: i need help finding the equation of the axis of symmetry for 5(x-6)^2+12

Question 839837: i need help finding the equation of the axis of symmetry for 5(x-6)^2+12
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
here's a link that will help you to find it.
http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php
the axis of symmetry is the vertical line that goes through the vertex.
your equation is in the vertex form of the equation of a parabola.
that form is:
y = a*(x-h)^2 + k
(h,k) is the vertex.
that makes your vertex at the point (6,12)
that means that your axis of symmetry is a vertical line at x = 6.
you can also find it by converting the vertex form of your equation to the standard form of the equation and then looking for x = -b/2a which gives you the x-coordinate of the vertex.
the vertex is also the minimum / maximum point of the quadratic equation.
the x-value of the vertex is at x = -b/2a
the y-value of the vertex is at f(-b/2a).
since you're already in vertex form, the easiest thing to do is just take if from that form.
the graph of your equation is shown below:

i drew your axis of symmetry as best i could at x = 6. it may not be perfectly vertical but it should be very close to vertical.
you can see that the graph of the equation is symmetrical about that line.

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