SOLUTION: Could someone please show me step by step how to figure the axis of symmetry with a x,y chart. The problem is y=x^2-5x+3. I know how to graph it, but I need to know how to identify

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Question 87726: Could someone please show me step by step how to figure the axis of symmetry with a x,y chart. The problem is y=x^2-5x+3. I know how to graph it, but I need to know how to identify the axis of symmetry, and to create a suitable table of values for the above problem. Thank you
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


Start with the given equation



Subtract from both sides



Factor out the leading coefficient



Take half of the x coefficient to get (ie ).


Now square to get (ie )





Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of does not change the equation




Now factor to get



Distribute



Multiply



Now add to both sides to isolate y



Combine like terms




Now the quadratic is in vertex form where , , and . Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation we get:


Graph of . Notice how the vertex is (,).



Notice if we graph the final equation we get:


Graph of . Notice how the vertex is also (,).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.






Since we know the vertex is (,) or (2.5,-3.25), this is one point on the graph.

Now lets pick any point after . Lets evaluate

Start with the given polynomial


Plug in


Raise 3 to the second power to get 9


Multiply 5 by 3 to get 15


Now combine like terms

So we get the point (3,-3)


Lets pick another point

Start with the given polynomial


Plug in


Raise 4 to the second power to get 16


Multiply 5 by 4 to get 20


Now combine like terms

So another point is (4,-1)




Now since the graph is symmetrical with respect to the axis of symmetry, this means x-values on the other side of the vertex will have the same y-values as their respective counterparts. For instance, the counterpart to is and the counterpart to is (notice they are the same distance away from the vertex along the x-axis)

So here's the table of suitable values 




 xy


 1-1
 

 2-3
 

 2.5-3.25
 

 3-3
 

 4-1


Notice if we graph the equation and the table of points we get

graph of with the points (1,1),(2,-3),(2.5,-3.25),(3,-3),(4,-1)


Since the points lie on the curve, this verifies our answer.
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