Question 105679
For your first problem, read my lesson---Shortcut in finding the vertex of any parabola 

Then, go back to this page.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

{{{y=x^2+x-2}}}
{{{dy/dx=2x+1=0}}}

Solving {{{2x+1=0}}}

{{{2x+1=0}}}
{{{2x+1-1=0-1}}}
{{{2x+0=-1}}}
{{{2x=-1}}}
{{{x=-1/2}}}

Solving for y,

{{{y=x^2+x-2}}}
{{{y=(-1/2)^2+(-1/2)-2}}}
{{{y=1/4-1/2-2}}}
{{{4y=4(1/4)-4(1/2)-4(2)}}}
{{{4y=1-2-8}}}
{{{4y=(1-2)-8}}}
{{{4y=-1-8}}}
{{{4y=-9}}}
{{{y=-9/4}}}

Since we obtained {{{x=-1/2}}} and {{{y=-9/4}}}, thus, the vertex is located at ({{{-1/2}}},{{{-9/4}}}).


Proof? Look below!


{{{graph(1000,1000,-20,20,-20,20,x^2+x-2,-9/4)}}}



Power up,
HyperBrain!