Question 87726
*[invoke completing_the_square 1, -5, 3]


Since we know the vertex is ({{{5/2}}},{{{-13/4}}}) or (2.5,-3.25), this is one point on the graph. 


Now lets pick any point after {{{x=2.5}}}. Lets evaluate {{{x=3}}}


{{{f(x)=x^2-5x+3}}} Start with the given polynomial



{{{f(3)=(3)^2-5(3)+3}}} Plug in {{{x=3}}}



{{{f(3)=(9)-5(3)+3}}} Raise 3 to the second power to get 9



{{{f(3)=(9)-15+3}}} Multiply 5 by 3 to get 15



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


So we get the point (3,-3)



Lets pick another point {{{x=4}}}


{{{f(x)=x^2-5x+3}}} Start with the given polynomial



{{{f(4)=(4)^2-5(4)+3}}} Plug in {{{x=4}}}



{{{f(4)=(16)-5(4)+3}}} Raise 4 to the second power to get 16



{{{f(4)=(16)-20+3}}} Multiply 5 by 4 to get 20



{{{f(4)=-1}}} Now combine like terms


So another point is (4,-1)





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=4}}} is {{{x=1}}} and the counterpart to {{{x=3}}} is {{{x=2}}} (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> 1</TD><TD>-1</TD></TR> 
<TR><TD> 2</TD><TD>-3</TD></TR> 
<TR><TD> 2.5</TD><TD>-3.25</TD></TR> 
<TR><TD> 3</TD><TD>-3</TD></TR> 
<TR><TD> 4</TD><TD>-1</TD></TR> 

</TABLE>
</pre>


Notice if we graph the equation {{{y=x^2-5x+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^2-5x+3),circle(-5,53,0.05),
circle(1,-1,0.05),
circle(1,-1,0.08),
circle(2,-3,0.05),
circle(2,-3,0.08),
circle(2.5,-3.25,0.05),
circle(2.5,-3.25,0.08),
circle(3,-3,0.05),
circle(3,-3,0.08),
circle(4,-1,0.05),
circle(4,-1,0.08)
)}}} graph of {{{y=x^2-5x+3}}} with the points (1,1),(2,-3),(2.5,-3.25),(3,-3),(4,-1)



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