Question 124293
To find h, use this formula:


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


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


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



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




{{{h=-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+1}}} Start with the given polynomial



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



{{{f(-2)=(4)+4(-2)+1}}} Raise -2 to the second power to get 4



{{{f(-2)=(4)+-8+1}}} Multiply 4 by -2 to get -8



{{{f(-2)=-3}}} Now combine like terms



So the vertex is (-2,-3)




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



Lets evaluate {{{f(-4)}}}


{{{f(x)=x^2+4x+1}}} Start with the given polynomial



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



{{{f(-4)=(16)+4(-4)+1}}} Raise -4 to the second power to get 16



{{{f(-4)=(16)+-16+1}}} Multiply 4 by -4 to get -16



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



So our 1st point is (-4,1)




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




Lets evaluate {{{f(-3)}}}


{{{f(x)=x^2+4x+1}}} Start with the given polynomial



{{{f(-3)=(-3)^2+4(-3)+1}}} Plug in {{{x=-3}}}



{{{f(-3)=(9)+4(-3)+1}}} Raise -3 to the second power to get 9



{{{f(-3)=(9)+-12+1}}} Multiply 4 by -3 to get -12



{{{f(-3)=-2}}} Now combine like terms



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



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

This means the y-value for {{{x=-3}}} is equal to the y-value of {{{x=-1}}}. So when {{{x=-1}}}, {{{y=-2}}}.

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



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

Now let's plot these points to graph {{{y=x^2+4x+1}}}

{{{drawing(900,900,-12,8,-13,7,
grid( 1 ),
graph(900,900,-12,8,-13,7, x^2+4x+1),
circle(-4,1,0.05),
circle(-4,1,0.08),
circle(-3,-2,0.05),
circle(-3,-2,0.08),
circle(-2,-3,0.05),
circle(-2,-3,0.08),
circle(-1,-2,0.05),
circle(-1,-2,0.08),
circle(0,1,0.05),
circle(0,1,0.08)
)}}}