SOLUTION: can some one help me with this i have never seen this before and i am very lost and confused. identify the axis of symmetry, create a suitable table of values, and sketch the gr

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: can some one help me with this i have never seen this before and i am very lost and confused. identify the axis of symmetry, create a suitable table of values, and sketch the gr      Log On


   

Question 80089: can some one help me with this i have never seen this before and i am very lost and confused.
identify the axis of symmetry, create a suitable table of values, and sketch the graph (including the axis of symmetry).
y=-x^2+6x-2
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
In order to find the vertex, we must complete the square and place the quadratic into y=a%28x-h%29%5E2%2Bk form

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


y=-1+x%5E2%2B6+x-2 Start with the given equation



y%2B2=-1+x%5E2%2B6+x Add 2 to both sides



y%2B2=-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%2B2=-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%2B2=-1%28%28x-3%29%5E2-9%29 Now factor x%5E2-6x%2B9 to get %28x-3%29%5E2



y%2B2=-1%28x-3%29%5E2%2B1%289%29 Distribute



y%2B2=-1%28x-3%29%5E2%2B9 Multiply



y=-1%28x-3%29%5E2%2B9-2 Now add %2B2 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%2B6x-2 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-1x%5E2%2B6x-2%29 Graph of y=-1x%5E2%2B6x-2. 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%2C-1%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.






So we can see that the vertex is (3,7)

Now we know one point on the graph, and it is (3,7). So lets go one unit to the right to x=4. Plug in x=4
f%284%29=-%284%29%5E2%2B6%284%29-2
f%284%29=6
So another point is (4,6). Since the graph is symmetrical around x=3, then x=2 will have the same y value as x=4. So another point is (2,6). If we continue this, we can create an adequate table. Anyways, here's what we have so far: 

x  |   y
2  |   6
3  |   7
4  |   6

Look above for the sketched graph. To sketch the axis of symmetry, simply draw a vertical line through the vertex at x=3.