Question 91926
{{{y=-x^2+6x-2}}} Start with the given quadratic



{{{y+2=-x^2+6x}}} Add {{{2  }}} to both sides



{{{y+2=-1*(x^2-6*x)}}} Factor out the leading coefficient {{{-1  }}}. This step is important since we want the {{{x^2}}} coefficient to be 1. This is where you made your mistake



Take half of the x coefficient {{{-6}}} to get {{{-3}}} (ie {{{-6/2=-3}}})


Now square  {{{-3}}} to get {{{9}}}  (ie {{{(-3)^2=9}}})




{{{y+2=-1*(x^2-6*x+9)}}} Add this number ({{{9}}}) to the expression inside the parenthesis.



{{{y+2+(-1)(9)=-1*(x^2-6*x+9)}}} Since we added {{{9}}} inside the parenthesis, we need to add {{{(-1)(9)}}} to the other side (remember, we factored out the leading coefficient {{{-1}}}).



{{{y+2+(-1)(9)=-1*(x-3)^2}}} Factor {{{x^2-6*x+9}}} into {{{(x-3)^2}}}. Now the right side is a perfect square  (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)



{{{y+2-9=-1*(x-3)^2}}} Multiply {{{-1}}} and {{{9}}} to get {{{-9}}}.





{{{y-7=-1*(x-3)^2}}} Combine like terms




{{{y=-1*(x-3)^2+7}}}Add {{{7  }}} to both sides to isolate y



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Now the quadratic {{{y=-1*(x-3)^2+7}}} is in vertex form {{{y=a(x-h)^2+k}}} where {{{a=-1}}},  {{{h=3}}}, and {{{k=7}}}



Remember, for any quadratic in vertex form {{{y=a(x-h)^2+k}}} 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


<pre>
<TABLE width=50>

<TR><TD><TABLE frame=below><TR><TD> x</TD></TR></TABLE></TD><TD>|</TD><TD><TABLE frame=below><TR><TD> y</TD></TR></TABLE></TD></TR>

<TR><TD> 1</TD><TD>|</TD><TD> 3</TD></TR> 
<TR><TD> 2</TD><TD>|</TD><TD> 6</TD></TR> 
<TR><TD> 3</TD><TD>|</TD><TD> 7</TD></TR> 
<TR><TD> 4</TD><TD>|</TD><TD> 6</TD></TR> 
<TR><TD> 5</TD><TD>|</TD><TD> 3</TD></TR> 
</TABLE>
</pre>


Now plot the points

{{{drawing(900,900,-10,10,-10,10,
  grid( 1 ),circle(1,3,0.05),
circle(1,3,0.08),
circle(2,6,0.05),
circle(2,6,0.08),
circle(3,7,0.05),
circle(3,7,0.08),
circle(4,6,0.05),
circle(4,6,0.08),
circle(5,3,0.05),
circle(5,3,0.08))}}}



Now connect the points to graph {{{y=-x^2+6x-2}}}

{{{drawing(900,900,-10,10,-10,10,
grid( 1 ),
graph(900,900,-10,10,-10,10, -x^2+6x-2),circle(1,3,0.05),
circle(1,3,0.08),
circle(2,6,0.05),
circle(2,6,0.08),
circle(3,7,0.05),
circle(3,7,0.08),
circle(4,6,0.05),
circle(4,6,0.08),
circle(5,3,0.05),
circle(5,3,0.08))}}}