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

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) (Show Source): You can put this solution on YOUR website!
Start with the given quadratic

Add to both sides

Factor out the leading coefficient . This step is important since we want the coefficient to be 1. This is where you made your mistake

Take half of the x coefficient to get (ie )

Now square to get (ie )

Add this number () to the expression inside the parenthesis.

Since we added inside the parenthesis, we need to add to the other side (remember, we factored out the leading coefficient ).

Factor into . Now the right side is a perfect square (note: if you need help with factoring, check out this solver)

Multiply and to get .

Combine like terms

Add to both sides to isolate y

--------------------------------------------------------------------------------

Now the quadratic is in vertex form where , , and

Remember, for any quadratic in vertex form the vertex is (h,k) and the axis of symmetry is x=h

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

-----------------------------------------------------------------------

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

Now plot the points

Now connect the points to graph

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