Question 414468: please help me find the roots, lines of symmetry, and vertex for y=-x2+5x-4
Found 3 solutions by stanbon, rfer, MathLover1:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! please help me find the roots, lines of symmetry, and vertex for
y=-x^2+5x-4
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-x^2+5x+? = y+4+?
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Complete the square on the x-terms:
-(x^2-5+(5/2)^2) = y+4-(5/2)^2
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Factor:
-(x-(5/2))^2 = y + 16/4 - (25/4)
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-(x-(5/2))^2 = y -9/4
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Symmetry line : x = 5/2
Vertex: (5/2 , (9/4))
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Roots of y = -x^2+5x-4
Solve -x^2+5x-4 = 0
x^2-5x+4 = 0
(x-4)(x-1) = 0
x = 1 or x = 4
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Cheers,
Stan H.
Answer by rfer(16322) (Show Source):
Answer by MathLover1(20849)  (Show Source):
You can put this solution on YOUR website!  ....factor
This is a quadratic equation (degree of 2), so
First, find x-intercepts,
Let and solve for  :
 ........->.... and
 ........->....
So, the  (roots) are (4,0) and (1,0)
Now, find the  , let  and solve for  ,
 , so the  is (0,-4)
The axis of  is just the average of the two ,
So, the axis of symmetry is 
Since the sign of  is negative then this function will have a maximum.
Now, to find it's maximum (vertex).
Formula for maximum value is 
 ,  ,  ( Coefficients of  )
this is exactly  , which is not surprising because the   on the axis of .
Now, find the value corresponding to the value,
Substitute, into the equation and solve for ,
Summarize:
You should have more than enough now for a graph,
: (4,0) and (1,0)
: (0,-4)
maximum value (vertex) : (2.5,2.25)
axis of :
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