Given:#" "y=x^2-x-2#
#color(blue)("General shape of the graph")#
The coefficient of #x^2# is +1 ie positive. So the general shape is:# color(blue)(uu)#
If it had been negative the general shape would have been #nn#
#color(blue)("The vertex is a minimum")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("y-intercept")#
For the general equation form of #y=ax^2+bx+c# the y-intercept is #c#
#color(blue)("y-intercept" = -2)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the vertex")#
For the general equation form of #y=a(x^2+b/ax)+c#
#x_("vertex")->"axis of symmetry"->(-1/2)xx(b/a)#
In your case #a=+1#
#color(blue)(x_("vertex")=(-1/2)xx(-1)=+1/2)#
#color(blue)(y_("vertex")=(1/2)^2-(1/2)-2 = -2 1/4 -> -9/4)#
#color(blue)("Vertex "->(x,y)->(1/2,-9/4)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the x-intercepts")#
#y=x^2-x-2 -> y=(x+1)(x-2) = 0#
#color(blue)(=> x= -1" and "x=+2)#
