Question 88331
*[invoke completing_the_square -1, 3, -3]



Since we know the vertex is ({{{3/2}}},{{{-3/4}}}) or (1.5,-0.75), this is one point on the graph. 


Now lets pick any point after {{{x=1.5}}}. Lets evaluate {{{x=2}}}


{{{f(x)=-x^2+3x-3}}} Start with the given polynomial



{{{f(2)=-(2)^2+3(2)-3}}} Plug in {{{x=2}}}



{{{f(2)=-(4)+3(2)-3}}} Raise 2 to the second power to get 4



{{{f(2)=-(4)+6-3}}} Multiply 3 by 2 to get 6



{{{f(2)=-1}}} Now combine like terms


So we get the point (2,-1)


Lets pick another value {{{x=3}}}



{{{f(x)=-x^2+3x-3}}} Start with the given polynomial



{{{f(3)=-(3)^2+3(3)-3}}} Plug in {{{x=3}}}



{{{f(3)=-(9)+3(3)-3}}} Raise 3 to the second power to get 9



{{{f(3)=-(9)+9-3}}} Multiply 3 by 3 to get 9



{{{f(3)=-3}}} Now combine like terms


So another point is (3,-3)



Now since the graph is symmetrical with respect to the axis of symmetry, this means x-values on the other side of the vertex will have the same y-values as their respective counterparts. For instance, the counterpart to {{{x=2}}} is {{{x=1}}} and the counterpart to {{{x=3}}} is {{{x=0}}} (notice they are the same distance away from the vertex along the x-axis)


So here's the table of suitable values



<pre>
<TABLE width=500>

<TR><TD> x</TD><TD>y</TD></TR>
<TR><TD> 0</TD><TD>-3</TD></TR> 
<TR><TD> 1</TD><TD>-1</TD></TR>
<TR><TD> 1.5</TD><TD>-0.75</TD></TR> 
<TR><TD> 2</TD><TD>-1</TD></TR> 
<TR><TD> 3</TD><TD>-3</TD></TR> 
</TABLE>
</pre>


Notice if we graph the equation {{{y=-x^2+3x-3}}} and the table of points we get


{{{drawing(900,900,-10,10,-10,10,
grid( 1 ),
graph(900,900,-10,10,-10,10, -x^2+3x-3),circle(-3,-21,0.05),
circle(0,-3,0.05),
circle(0,-3,0.08),
circle(1,-1,0.05),
circle(1,-1,0.08),
circle(1.5,-0.75,0.05),
circle(1.5,-0.75,0.08),
circle(2,-1,0.05),
circle(2,-1,0.08),
circle(3,-3,0.05),
circle(3,-3,0.08))}}}


Since the points lie on the curve, this verifies our answer.