SOLUTION: For the function y= x2 - 6x + 8, put the function in the form y = a (x - h) squared + k. I was able to solve it by factoring and using the quadratic formula, but I can't figure out

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: For the function y= x2 - 6x + 8, put the function in the form y = a (x - h) squared + k. I was able to solve it by factoring and using the quadratic formula, but I can't figure out      Log On


   

Question 77475: For the function y= x2 - 6x + 8, put the function in the form y = a (x - h) squared + k. I was able to solve it by factoring and using the quadratic formula, but I can't figure out how to use this formula.
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%5E2-6+x%2B8 Start with the given equation



y-8=1+x%5E2-6+x Subtract 8 from both sides



y-8=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-8=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-8=1%28%28x-3%29%5E2-9%29 Now factor x%5E2-6x%2B9 to get %28x-3%29%5E2



y-8=1%28x-3%29%5E2-1%289%29 Distribute



y-8=1%28x-3%29%5E2-9 Multiply



y=1%28x-3%29%5E2-9%2B8 Now add 8 to both sides to isolate y



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




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=3, and k=-1. 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%2B8 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-6x%2B8%29 Graph of y=1x%5E2-6x%2B8. Notice how the vertex is (3,-1).



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


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



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





So the quadratic y=x%5E2+-+6x+%2B+8 can be converted to the standard form y=a+%28x+-+h%29%5E2+%2B+k which is
y=%28x-3%29%5E2-1 where a=1, h=3, and k=-1
Now we let y=0
0=%28x-3%29%5E2-1
1=%28x-3%29%5E2 Add 1 to both sides
sqrt%281%29=sqrt%28%28x-3%29%5E2%29

Which means

x-3=1 or x-3=-1
Now add 3 to both sides for each case to solve for x:
So our solutions are
x=4 or x=2