SOLUTION: Graphing Y=x^2+3x+2 Identify the vertext and the axis of symmetry

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Question 83726: Graphing Y=x^2+3x+2
Identify the vertext and the axis of symmetry
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%2B3+x%2B2 Start with the given equation



y-2=1+x%5E2%2B3+x Subtract 2 from both sides



y-2=1%28x%5E2%2B3x%29 Factor out the leading coefficient 1



Take half of the x coefficient 3 to get 3%2F2 (ie %281%2F2%29%283%29=3%2F2).


Now square 3%2F2 to get 9%2F4 (ie %283%2F2%29%5E2=%283%2F2%29%283%2F2%29=9%2F4)





y-2=1%28x%5E2%2B3x%2B9%2F4-9%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 9%2F4 does not change the equation




y-2=1%28%28x%2B3%2F2%29%5E2-9%2F4%29 Now factor x%5E2%2B3x%2B9%2F4 to get %28x%2B3%2F2%29%5E2



y-2=1%28x%2B3%2F2%29%5E2-1%289%2F4%29 Distribute



y-2=1%28x%2B3%2F2%29%5E2-9%2F4 Multiply



y=1%28x%2B3%2F2%29%5E2-9%2F4%2B2 Now add 2 to both sides to isolate y



y=1%28x%2B3%2F2%29%5E2-1%2F4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-3%2F2, and k=-1%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%2B3x%2B2 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B3x%2B2%29 Graph of y=1x%5E2%2B3x%2B2. Notice how the vertex is (-3%2F2,-1%2F4).



Notice if we graph the final equation y=1%28x%2B3%2F2%29%5E2-1%2F4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x%2B3%2F2%29%5E2-1%2F4%29 Graph of y=1%28x%2B3%2F2%29%5E2-1%2F4. Notice how the vertex is also (-3%2F2,-1%2F4).



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