Question 943039
The graph of a parabola looks like this {{{graph(100,100,.1,.5,-.5,-.1,-7*(x-.3)^2-.15)}}} or like this {{{graph(100,100,.1,.5,.1,.5,7*(x-0.3)^2+.15)}}} .
First, solve for {{{y}}} :
{{{x^2 +3y=0}}}<-->{{{3y=-x^2}}}<-->{{{y=-(1/3)*x^2}}}
That lat equation allows you to calculate points easily:
you pick a value for {{{x]}}, plug it in, and calculate {{{y}}}
(for example, for {{{x=3}}} , {{{y=-(1/3)*3^2=-(1/3)*9=-3}}} ).
Looking at that equation, you realize that:
for {{{x=0}}} , {{{y=-(1/3)*0=0}}} , and
for {{{x<>0}}} (negative or positive), {{{x^2>0}}} and {{{y=-(1/3)*x^2<0}}} .
That tells you that the graph looks like this:
{{{graph(300,200,-3,3,-2,2,-x^2/3)}}} It has a maximum at {{{O(0,0)}}} , the origin,
and all the other points are below the x-axis, with {{{y<0}}} .
In general a parabola with vertex at {{{V(h,k)}}} has an equation of the form
{{{y=a*(x-h)^2+k}}} <---> {{{y=a(x^2-2hx+h^2)+k}}} <---> {{{y=ax^2-2ahx+ah^2+k}}} .
You may be told that it is a quadratic function of the form {{{y=ax^2+bx+c}}} ,
but that is the same thing,
with {{{b=-2ah}}} and {{{c=ah^2+k}}} .
In any case, you can find {{{h}}}, the x-coordinate of the vertex,
by changing the sign of the coefficient of the term in {{{x}}} ,
to get {{{-b=2ah}}} ,
and dividing by {{{2a}}} , twice the coefficient of the term in {{{x^2}}} ,
to get {{{h=-b/2a=2ah/2a}}} .
After that, you just plug that value of {{{x}}} into the equation and find {{{k}}} ,
the y-coordinate of the vertex.
For example,
{{{y=4x^2+6x+8}}} has {{{a=4}}} and {{{b=6}}} .
The x-coordinate of the vertex of the parabola with that equation is
{{{h=-6/4=-3/2=-1.5}}} .
The y-coordinate of the vertex is
{{{x=4*(-1.5)^2+6*(-1.5)+8=4*2.25-9+8=9-9+8=8}}} .