SOLUTION: I keep finding examples that have the problem squared but none without. I would appreciate any and all help with this. Find the vertex of y=-2(x -3) + 4 and then graph. I think

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I keep finding examples that have the problem squared but none without. I would appreciate any and all help with this. Find the vertex of y=-2(x -3) + 4 and then graph. I think       Log On


   



Question 72735: I keep finding examples that have the problem squared but none without. I would appreciate any and all help with this.
Find the vertex of y=-2(x -3) + 4 and then graph.
I think the vertex is (6,4) and the -2 makes it open downward if im correct. I came up with (5,0) and (-1,0) as the x intercepts and (0,10) as the y intercept. When I plot them on a graph with the vertex, the points are all over the graph so Im doing something wrong. Please please help!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex of y=-2(x -3)^2 + 4 and then graph.
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There are several way to find the vertex.
You can rewrite the equation as y-4 = -2(x-3)^2
This follows the form y-k = a(x-h)^2
where (h,k) is the vertex.
So your vertex is at (3,4).
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To graph you could plot some points:
You know the x=3 is at the vertex, or on the axis of symmetry.
If x=4 you get y=-2*1+4 = 2; Point at (4,2) and by symmetry at (2,2)
If x=5 you get y=-2*5+4 = -6; Point at (5,-6) and by symmetry at (1,-6)
etc.
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C-2%28x-3%29%5E2%2B4%29
Cheers,
Stan H.