SOLUTION: For the following parabola, find the x-axis intercepts and the coodinates of the vertex: -x2 + 11x - 18 I have not a clue on how to start this,,please help me!!! ThAnKs!

Question 71739: For the following parabola, find the x-axis intercepts and the
coodinates of the vertex:
-x2 + 11x - 18

I have not a clue on how to start this,,please help me!!!
ThAnKs!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find the x-axis intercepts and the
coodinates of the vertex:
y = -x^2 + 11x - 18
Rearrange:
x^2-11x+? =-y-18+?
x^2-11x+(11/2)^2 = -y-(72/4)+(121/4)
(x-(11/2))^2 = -y+49/4
Rearrange:
y-(49/4) =-(x-(11/2))^2
Vertex at (11/2 , 49,4)
-------
To find x intercepts, let y=0
-x^2+11x-18 = 0
(-x+-9)(x+2)=0
x=-9 or x=-2
============
Cheers,
Stan H.
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