SOLUTION: The problem asks to find the vertex and graph the following quadratic function.
y=-x+6x
My professor says I need only turn in the coordinates. This is what I have tried... x=-b/
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Question 320535: The problem asks to find the vertex and graph the following quadratic function.
y=-x+6x
My professor says I need only turn in the coordinates. This is what I have tried... x=-b/2a = -6/2(-1) = -6/-2 =3 x=3
y=-2(3)^2+6(3)= -2(9) + 18 = -18 + 18=0 y=0 My book says the vertex is (3,9) so I am totally confused. I come up with (0,0) for the y intercept but I have no idea how that relates to me finding the coordinates for this graph. I have a very hard time grasping this concept. Thanks for your time.
Tracy Jones
Found 2 solutions by scott8148, solver91311:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
y = -x^2 + 6x
you found the axis of symmetry OK (x = -b / 2a)
substituting back into the function ___ y = -(3^2) + 6(3) = -9 + 18 = 9
not sure where the x^2 coefficient of 2 comes from in your substitution?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You did absolutely everything correctly except for one thing.
You began with the function:
You calculated the 
-coordinate of the vertex by:
}\ =\ 3)
And then you proceeded to evaluate the function at this value of . Here is where you made your error.
The function you started with was:
But you evaluated:
^2\ +\ 6(3)\ =\ 0)
Your evaluation calculation is spot on, but where did you get that extra factor of 2 in the lead coefficient?
What you should have done:
^2\ +\ 6(3)\ =\ -9\ +\ 18\ =\ 9)
and therefore the vertex is at the point )
John
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