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

Algebra ->  Quadratic-relations-and-conic-sections -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2-5+x%2B3 Start with the given equation



y-3=1+x%5E2-5+x Subtract 3 from both sides



y-3=1%28x%5E2-5x%29 Factor out the leading coefficient 1



Take half of the x coefficient -5 to get -5%2F2 (ie %281%2F2%29%28-5%29=-5%2F2).


Now square -5%2F2 to get 25%2F4 (ie %28-5%2F2%29%5E2=%28-5%2F2%29%28-5%2F2%29=25%2F4)





y-3=1%28x%5E2-5x%2B25%2F4-25%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 25%2F4 does not change the equation




y-3=1%28%28x-5%2F2%29%5E2-25%2F4%29 Now factor x%5E2-5x%2B25%2F4 to get %28x-5%2F2%29%5E2



y-3=1%28x-5%2F2%29%5E2-1%2825%2F4%29 Distribute



y-3=1%28x-5%2F2%29%5E2-25%2F4 Multiply



y=1%28x-5%2F2%29%5E2-25%2F4%2B3 Now add 3 to both sides to isolate y



y=1%28x-5%2F2%29%5E2-13%2F4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=5%2F2, and k=-13%2F4. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-5x%2B3 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-5x%2B3%29 Graph of y=1x%5E2-5x%2B3. Notice how the vertex is (5%2F2,-13%2F4).



Notice if we graph the final equation y=1%28x-5%2F2%29%5E2-13%2F4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-5%2F2%29%5E2-13%2F4%29 Graph of y=1%28x-5%2F2%29%5E2-13%2F4. Notice how the vertex is also (5%2F2,-13%2F4).



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 (5%2F2,-13%2F4) or (2.5,-3.25), this is one point on the graph.

Now lets pick any point after x=2.5. Lets evaluate x=3

f%28x%29=x%5E2-5x%2B3 Start with the given polynomial


f%283%29=%283%29%5E2-5%283%29%2B3 Plug in x=3


f%283%29=%289%29-5%283%29%2B3 Raise 3 to the second power to get 9


f%283%29=%289%29-15%2B3 Multiply 5 by 3 to get 15


f%283%29=-3 Now combine like terms

So we get the point (3,-3)


Lets pick another point x=4

f%28x%29=x%5E2-5x%2B3 Start with the given polynomial


f%284%29=%284%29%5E2-5%284%29%2B3 Plug in x=4


f%284%29=%2816%29-5%284%29%2B3 Raise 4 to the second power to get 16


f%284%29=%2816%29-20%2B3 Multiply 5 by 4 to get 20


f%284%29=-1 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 x=4 is x=1 and the counterpart to x=3 is x=2 (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 y=x%5E2-5x%2B3 and the table of points we get

graph of y=x%5E2-5x%2B3 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.