SOLUTION: Any ideas here? Graph the function using the equation in part a (x2 - 6x + 8 = 0). Explain why it is not necessary to plot points to graph when using y = a (x - h) 2 + k. Also,

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Any ideas here? Graph the function using the equation in part a (x2 - 6x + 8 = 0). Explain why it is not necessary to plot points to graph when using y = a (x - h) 2 + k. Also,       Log On


   

Question 85410: Any ideas here?
Graph the function using the equation in part a (x2 - 6x + 8 = 0). Explain why it is not necessary to plot points to graph when using y = a (x - h) 2 + k. Also, how would this graph compares to the graph of y = x2?
Thanks ahead of time
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming you want to convert y=x%5E2-6x%2B8 into vertex form right?

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.






If a=1, which it does, then we only need the vertex coordinates to plot the graph of y=%28x-3%29%5E2-1. Simply plot the vertex at (3,-1)



Starting at the vertex, go to the right one unit and up 1 unit and plot (4,0). Go back to the vertex and go to the left one unit and up 1 unit and plot (2,0).



Now that you have 3 points, you can draw a parabola through them.


This is the quick and easy way to graph parabolas


Since a=1, the graph of y=%28x-3%29%5E2-1 has the same shape as y=x%5E2. The only difference is the vertex of y=%28x-3%29%5E2-1 has been shifted 3 units to the right and one unit down. In other words, y=x%5E2 has a vertex at (0,0) and to get y=%28x-3%29%5E2-1, simply move the graph y=x%5E2 to the point (3,-1)

graph of y=x%5E2 (red) and the graph of y=%28x-3%29%5E2-1(green) that has been shifted from the origin