SOLUTION: Hello, Would you please help me with this problem: Identify the axis of symetry, create a suitable table of values, and sketch th graph. y=-x^2+6x-2 This is how far I have

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello, Would you please help me with this problem: Identify the axis of symetry, create a suitable table of values, and sketch th graph. y=-x^2+6x-2 This is how far I have       Log On


   

Question 91926This question is from textbook Begining Algebra
: Hello,
Would you please help me with this problem:
Identify the axis of symetry, create a suitable table of values, and sketch th graph. y=-x^2+6x-2
This is how far I have gotten, but I am not too sure if I have gotten this part right.
y=-x^2+6x-2
y+2=-x^2+6x
y+2+9=x^2+6x+9
y+11=(x+3)^2
Can you please walk me through how to complete this type of problem? This question is from textbook Begining Algebra

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
y=-x%5E2%2B6x-2 Start with the given quadratic


y%2B2=-x%5E2%2B6x Add 2++ to both sides


y%2B2=-1%2A%28x%5E2-6%2Ax%29 Factor out the leading coefficient -1++. This step is important since we want the x%5E2 coefficient to be 1. This is where you made your mistake


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

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



y%2B2=-1%2A%28x%5E2-6%2Ax%2B9%29 Add this number (9) to the expression inside the parenthesis.


y%2B2%2B%28-1%29%289%29=-1%2A%28x%5E2-6%2Ax%2B9%29 Since we added 9 inside the parenthesis, we need to add %28-1%29%289%29 to the other side (remember, we factored out the leading coefficient -1).


y%2B2%2B%28-1%29%289%29=-1%2A%28x-3%29%5E2 Factor x%5E2-6%2Ax%2B9 into %28x-3%29%5E2. Now the right side is a perfect square (note: if you need help with factoring, check out this solver)


y%2B2-9=-1%2A%28x-3%29%5E2 Multiply -1 and 9 to get -9.




y-7=-1%2A%28x-3%29%5E2 Combine like terms



y=-1%2A%28x-3%29%5E2%2B7Add 7++ to both sides to isolate y


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Now the quadratic y=-1%2A%28x-3%29%5E2%2B7 is in vertex form y=a%28x-h%29%5E2%2Bk where a=-1, h=3, and k=7


Remember, for any quadratic in vertex form y=a%28x-h%29%5E2%2Bk the vertex is (h,k) and the axis of symmetry is x=h

So the vertex is (3,7) and the axis of symmetry is x=3



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In order to create a table of suitable values, simply plug in x-values that are close to the axis of symmetry.


For instance, plug in x=1 (which is two units away from the axis of symmetry) to get y=3. So our first point is (1,3)

Now plug in x=2 (which is one unit away from the axis of symmetry) to get y=6. That makes our second point (2,6).


Since we know the vertex is (3,7), our third point is (3,7)


Since the graph is symmetric with respect to the axis of symmetry, the x-value that is one unit to the right of the axis of symmetry (x=4) will have the same y-value as x=2 (since x=2 is also one unit away from the axis of symmetry). So our fourth point is (4,6)


Also the x-value that is two units to the right of the axis of symmetry (x=5) will have the same y-value as x=1 (since x=1 is also two units away from the axis of symmetry). This makes our fifth point at (5,3)


So our table looks like 




 x
|
 y



 1| 3
 

 2| 6
 

 3| 7
 

 4| 6
 

 5| 3


Now plot the points



Now connect the points to graph y=-x%5E2%2B6x-2