Question 102742
To find the axis of symmetry, use this formula:


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


From the equation {{{y=x^2-5x+4}}} we can see that a=1 and b=-5


{{{x=(--5)/(2*1)}}} Plug in b=-5 and a=1



{{{x=5/(2*1)}}} Negate -5 to get 5



{{{x=(5)/2}}} Multiply 2 and 1 to get 2



So the axis of symmetry is  {{{x=5/2}}}



So the x-coordinate of the vertex is {{{x=5/2}}} (which is {{{x=2.5}}} in decimal form). Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(2.5)}}}


{{{f(x)=x^2-5x+4}}} Start with the given polynomial



{{{f(2.5)=(2.5)^2-5(2.5)+4}}} Plug in {{{x=2.5}}}



{{{f(2.5)=(6.25)-5(2.5)+4}}} Raise 2.5 to the second power to get 6.25



{{{f(2.5)=(6.25)-12.5+4}}} Multiply 5 by 2.5 to get 12.5



{{{f(2.5)=-2.25}}} Now combine like terms



So the vertex is (2.5,-2.25)



Since the x-coordinate is 2.5, this means the axis of symmetry is {{{x=2.5}}}


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



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


{{{f(x)=x^2-5x+4}}} Start with the given polynomial



{{{f(0)=(0)^2-5(0)+4}}} Plug in {{{x=0}}}



{{{f(0)=(0)-5(0)+4}}} Raise 0 to the second power to get 0



{{{f(0)=(0)-0+4}}} Multiply 5 by 0 to get 0



{{{f(0)=4}}} Remove any zero terms



So our 1st point is (0,4)




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




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


{{{f(x)=x^2-5x+4}}} Start with the given polynomial



{{{f(1)=(1)^2-5(1)+4}}} Plug in {{{x=1}}}



{{{f(1)=(1)-5(1)+4}}} Raise 1 to the second power to get 1



{{{f(1)=(1)-5+4}}} Multiply 5 by 1 to get 5



{{{f(1)=0}}} Now combine like terms



So our 2nd point is (1,0)



Now remember, the parabola is symmetrical about the axis of symmetry (which is {{{x=2.5}}})

This means the y-value for {{{x=1}}} is equal to the y-value of {{{x=4}}}. So when {{{x=4}}}, {{{y=0}}}.

Also, the y-value for {{{x=0}}} is equal to the y-value of {{{x=5}}}. So when {{{x=5}}}, {{{y=4}}}.



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>4</TD></TR> 
<TR><TD> 1</TD><TD>0</TD></TR> 
<TR><TD> 2.5</TD><TD>-2.25</TD></TR> 
<TR><TD> 4</TD><TD>0</TD></TR> 
<TR><TD> 5</TD><TD>4</TD></TR> 
</TABLE>
</pre>

Notice if you graph the equation {{{y=x^2-5x+4}}} with the axis of symmetry you get

{{{drawing(900,900,-7.5,12.5,-12.25,7.75,
grid( 1 ),
graph(900,900,-7.5,12.5,-12.25,7.75, x^2-5x+4),
circle(0,4,0.05),
circle(0,4,0.08),
circle(1,0,0.05),
circle(1,0,0.08),
circle(2.5,-2.25,0.05),
circle(2.5,-2.25,0.08),
circle(4,0,0.05),
circle(4,0,0.08),
circle(5,4,0.05),
line(2.5,-20,2.5,20),
circle(5,4,0.08)
)}}} Graph of {{{y=x^2-5x+4}}} through the given points. Notice the axis of symmetry is the equation {{{x=2.5}}} which is the vertical line through the vertex of the parabola.