SOLUTION: What is the standard form for y=1/2x^2+4x-3 ?

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Question 119297: What is the standard form for y=1/2x^2+4x-3 ?
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=%281%2F2%29+x%5E2%2B4+x-3 Start with the given equation



y%2B3=%281%2F2%29+x%5E2%2B4+x Add 3 to both sides



y%2B3=%281%2F2%29%28x%5E2%2B8x%29 Factor out the leading coefficient %281%2F2%29



Take half of the x coefficient 8 to get 4 (ie %281%2F2%29%288%29=4).


Now square 4 to get 16 (ie %284%29%5E2=%284%29%284%29=16)





y%2B3=%281%2F2%29%28x%5E2%2B8x%2B16-16%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 16 does not change the equation




y%2B3=%281%2F2%29%28%28x%2B4%29%5E2-16%29 Now factor x%5E2%2B8x%2B16 to get %28x%2B4%29%5E2



y%2B3=%281%2F2%29%28x%2B4%29%5E2-%281%2F2%29%2816%29 Distribute



y%2B3=%281%2F2%29%28x%2B4%29%5E2-8 Multiply



y=%281%2F2%29%28x%2B4%29%5E2-8-3 Now add %2B3 to both sides to isolate y



y=%281%2F2%29%28x%2B4%29%5E2-11 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1%2F2, h=-4, and k=-11. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=%281%2F2%29x%5E2%2B4x-3 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C%281%2F2%29x%5E2%2B4x-3%29 Graph of y=%281%2F2%29x%5E2%2B4x-3. Notice how the vertex is (-4,-11).



Notice if we graph the final equation y=%281%2F2%29%28x%2B4%29%5E2-11 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C%281%2F2%29%28x%2B4%29%5E2-11%29 Graph of y=%281%2F2%29%28x%2B4%29%5E2-11. Notice how the vertex is also (-4,-11).



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