Question 163607: Find the answers to a, b,c for for the quadratic function y= -x^2 + 6x -5
a) Find the co-ordinates of the vertex of the parabola
b) Find the equation of the axis of symmetry
c) Find the maximum or minimum value
Here is what I have - I must say my concern is the - (negative) in front of the -x^2.
a) x = - b/2a = - 6/2(-1) = 3 and y = (-3^2) + 6(3) - 5 = -9 +18 - 5 = 4
b) ? is this x = 3
c) x | y
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0 -5
1 0
2 3
3 4
4 3
5 0
6 -5
Based on this maximum is 4
Thanks for your help.
Found 3 solutions by Earlsdon, Fombitz, gonzo:
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Congratulations for making an attempt at solving this problem!
Given:
Find:
a) The vertex of the parabola:
The x-coordinate is given by:
 In your equation, a = -1 and b = 6, so...
 You got this part correct! Now substitute this into the original equation to find the y-coordinate.
 You also got this correct!
The vertex is located at (3, 4) but please note, in your calculations, you show... and this is not correct.
Remember that:  = but,  =
The axis of symmetry is the vertical line that bisects the parabola and, of course, it runs right through the vertex, so its equation would be:
The maximum or minimum value?
Well, notice that the coefficient of the  term is negative, so this indicates that the parabola opens downward, hence, the vertex will be a maximum, and its location is, as you have already found, (3, 4).
Let's see what the graph looks like for this equation:
Answer by Fombitz(32388)  (Show Source):
You can put this solution on YOUR website! Yes, you're correct for all of your answers.
Also, the min or max (depending on the sign of x^2 coefficient, positive min, negative max) is the y coordinate of the vertex.
Vertex is (3,4), x^2 coefficient is -1, 4 is the max. value.
.
.
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Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website! vertex is (3,4)
answer to a is correct.
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axis of symmetry looks like x = 3.
answer to b is correct based on viewing the graph.
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since there is a minus in front of the x^2, this is a parabola that has a maximum at the vertex, so your answer of 4 is correct.
following are two graphs.
the first one shows the turning point and the maximum clearly.
the second one shows the axis of symmetry clearly.
first graph:
second graph:
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