SOLUTION: I need to take this equation from standard form to vertex form can someone please help me the problem is x^2-18x+50

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Question 112938: I need to take this equation from standard form to vertex form can someone please help me the problem is
x^2-18x+50
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%5E2-18+x%2B50 Start with the given equation



y-50=1+x%5E2-18+x Subtract 50 from both sides



y-50=1%28x%5E2-18x%29 Factor out the leading coefficient 1



Take half of the x coefficient -18 to get -9 (ie %281%2F2%29%28-18%29=-9).


Now square -9 to get 81 (ie %28-9%29%5E2=%28-9%29%28-9%29=81)





y-50=1%28x%5E2-18x%2B81-81%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 81 does not change the equation




y-50=1%28%28x-9%29%5E2-81%29 Now factor x%5E2-18x%2B81 to get %28x-9%29%5E2



y-50=1%28x-9%29%5E2-1%2881%29 Distribute



y-50=1%28x-9%29%5E2-81 Multiply



y=1%28x-9%29%5E2-81%2B50 Now add 50 to both sides to isolate y



y=1%28x-9%29%5E2-31 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=9, and k=-31. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-18x%2B50 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-18x%2B50%29 Graph of y=1x%5E2-18x%2B50. Notice how the vertex is (9,-31).



Notice if we graph the final equation y=1%28x-9%29%5E2-31 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-9%29%5E2-31%29 Graph of y=1%28x-9%29%5E2-31. Notice how the vertex is also (9,-31).



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