SOLUTION: How do you solve this graph? Find the vertex and the axis of symmetry of this line y=xsquare-2x+5

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Question 84910This question is from textbook Algebra 1
: How do you solve this graph?
Find the vertex and the axis of symmetry of this line
y=xsquare-2x+5

This question is from textbook Algebra 1

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%5E2-2+x%2B5 Start with the given equation



y-5=1+x%5E2-2+x Subtract 5 from both sides



y-5=1%28x%5E2-2x%29 Factor out the leading coefficient 1



Take half of the x coefficient -2 to get -1 (ie %281%2F2%29%28-2%29=-1).


Now square -1 to get 1 (ie %28-1%29%5E2=%28-1%29%28-1%29=1)





y-5=1%28x%5E2-2x%2B1-1%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1 does not change the equation




y-5=1%28%28x-1%29%5E2-1%29 Now factor x%5E2-2x%2B1 to get %28x-1%29%5E2



y-5=1%28x-1%29%5E2-1%281%29 Distribute



y-5=1%28x-1%29%5E2-1 Multiply



y=1%28x-1%29%5E2-1%2B5 Now add 5 to both sides to isolate y



y=1%28x-1%29%5E2%2B4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=1, and k=4. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-2x%2B5 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-2x%2B5%29 Graph of y=1x%5E2-2x%2B5. Notice how the vertex is (1,4).



Notice if we graph the final equation y=1%28x-1%29%5E2%2B4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-1%29%5E2%2B4%29 Graph of y=1%28x-1%29%5E2%2B4. Notice how the vertex is also (1,4).



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