Question 127398
To find the axis of symmetry, use this formula:


{{{x=-b/(2a)}}}


From the equation {{{y=x^2-4x}}} we can see that a=1 and b=-4


{{{x=(--4)/(2*1)}}} Plug in b=-4 and a=1



{{{x=4/(2*1)}}} Negate -4 to get 4



{{{x=(4)/2}}} Multiply 2 and 1 to get 2




{{{x=2}}} Reduce



So the axis of symmetry is  {{{x=2}}}



So the x-coordinate of the vertex is {{{x=2}}}. Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(2)}}}


{{{f(x)=x^2-4x}}} Start with the given polynomial



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



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



{{{f(2)=-4}}} Now combine like terms



So the vertex is (2,-4)



Since the x-coordinate is 2, this means the axis of symmetry is {{{x=2}}}


Now lets find 2 other points to the left of the vertex



Lets evaluate {{{f(0)}}}


{{{f(x)=x^2-4x}}} Start with the given polynomial



{{{f(0)=(0)^2-4(0)}}} Plug in {{{x=0}}}



{{{f(0)=(0)-4(0)}}} Raise 0 to the second power to get 0



{{{f(0)=(0)-0}}} Multiply 4 by 0 to get 0



{{{f(0)=0}}} Remove any zero terms



So our 1st point is (0,0)




----Now lets find another point----




Lets evaluate {{{f(1)}}}


{{{f(x)=x^2-4x}}} Start with the given polynomial



{{{f(1)=(1)^2-4(1)}}} Plug in {{{x=1}}}



{{{f(1)=(1)-4(1)}}} Raise 1 to the second power to get 1



{{{f(1)=(1)-4}}} Multiply 4 by 1 to get 4



{{{f(1)=-3}}} Now combine like terms



So our 2nd point is (1,-3)



Now remember, the parabola is symmetrical about the axis of symmetry (which is {{{x=2}}})

This means the y-value for {{{x=1}}} is equal to the y-value of {{{x=3}}}. So when {{{x=3}}}, {{{y=-3}}}.

Also, the y-value for {{{x=0}}} is equal to the y-value of {{{x=4}}}. So when {{{x=4}}}, {{{y=0}}}.



Now lets make a table of the values we have calculated

<pre>
<TABLE width=500>

<TR><TD> x</TD><TD>y</TD></TR>
<TR><TD> 0</TD><TD>0</TD></TR> 
<TR><TD> 1</TD><TD>-3</TD></TR> 
<TR><TD> 2</TD><TD>-4</TD></TR> 
<TR><TD> 3</TD><TD>-3</TD></TR> 
<TR><TD> 4</TD><TD>0</TD></TR> 
</TABLE>
</pre>


Now plot the points 

{{{drawing(900,900,-8,12,-14,6,
grid( 1 ),
graph(900,900,-8,12,-14,6, 0),
circle(0,0,0.05),
circle(0,0,0.08),
circle(1,-3,0.05),
circle(1,-3,0.08),
circle(2,-4,0.05),
circle(2,-4,0.08),
circle(3,-3,0.05),
circle(3,-3,0.08),
circle(4,0,0.05),
circle(4,0,0.08)
)}}}


Now connect the points to graph {{{y= x^2-4x}}}. Also, to draw the axis of symmetry, simply draw a vertical line through the vertex

{{{drawing(900,900,-8,12,-14,6,
grid( 1 ),
graph(900,900,-8,12,-14,6, x^2-4x),
blue(line(2,-20,2,20)),
circle(0,0,0.05),
circle(0,0,0.08),
circle(1,-3,0.05),
circle(1,-3,0.08),
circle(2,-4,0.05),
circle(2,-4,0.08),
circle(3,-3,0.05),
circle(3,-3,0.08),
circle(4,0,0.05),
circle(4,0,0.08)
)}}} Graph of {{{y= x^2-4x}}} (red) and the axis of symmetry (blue)