SOLUTION: Complete the square in each quadratic function, placing the final answer in the fom of y=a(x-h)^2+k: 1. y=x^2 -6x+16 2.y=3x^2 -6x+7 . Could any one help with these?

Algebra ->  Rational-functions -> SOLUTION: Complete the square in each quadratic function, placing the final answer in the fom of y=a(x-h)^2+k: 1. y=x^2 -6x+16 2.y=3x^2 -6x+7 . Could any one help with these?      Log On


   

Question 160593: Complete the square in each quadratic function, placing the final answer in the fom of y=a(x-h)^2+k: 1. y=x^2 -6x+16 2.y=3x^2 -6x+7 . Could any one help with these?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Complete the square in each quadratic function, placing the final answer in the fom of y=a%28x-h%29%5E2%2Bk
1. y=x%5E2+-6x%2B16

Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2-6+x%2B16 Start with the given equation



y-16=1+x%5E2-6+x Subtract 16 from both sides



y-16=1%28x%5E2-6x%29 Factor out the leading coefficient 1



Take half of the x coefficient -6 to get -3 (ie %281%2F2%29%28-6%29=-3).


Now square -3 to get 9 (ie %28-3%29%5E2=%28-3%29%28-3%29=9)





y-16=1%28x%5E2-6x%2B9-9%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 9 does not change the equation




y-16=1%28%28x-3%29%5E2-9%29 Now factor x%5E2-6x%2B9 to get %28x-3%29%5E2



y-16=1%28x-3%29%5E2-1%289%29 Distribute



y-16=1%28x-3%29%5E2-9 Multiply



y=1%28x-3%29%5E2-9%2B16 Now add 16 to both sides to isolate y



y=1%28x-3%29%5E2%2B7 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=3, and k=7. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-6x%2B16 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-6x%2B16%29 Graph of y=1x%5E2-6x%2B16. Notice how the vertex is (3,7).



Notice if we graph the final equation y=1%28x-3%29%5E2%2B7 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-3%29%5E2%2B7%29 Graph of y=1%28x-3%29%5E2%2B7. Notice how the vertex is also (3,7).



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






2.y=3x%5E2+-6x%2B7
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=3+x%5E2-6+x%2B7 Start with the given equation



y-7=3+x%5E2-6+x Subtract 7 from both sides



y-7=3%28x%5E2-2x%29 Factor out the leading coefficient 3



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-7=3%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-7=3%28%28x-1%29%5E2-1%29 Now factor x%5E2-2x%2B1 to get %28x-1%29%5E2



y-7=3%28x-1%29%5E2-3%281%29 Distribute



y-7=3%28x-1%29%5E2-3 Multiply



y=3%28x-1%29%5E2-3%2B7 Now add 7 to both sides to isolate y



y=3%28x-1%29%5E2%2B4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=3, 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=3x%5E2-6x%2B7 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C3x%5E2-6x%2B7%29 Graph of y=3x%5E2-6x%2B7. Notice how the vertex is (1,4).



Notice if we graph the final equation y=3%28x-1%29%5E2%2B4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C3%28x-1%29%5E2%2B4%29 Graph of y=3%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.