#color(blue)("Determine "x_("intercepts")#
The graph crosses the x-axis at #y=0# thus:
#x_("intercept")" at " y=0#
Thus we have #color(brown)(y=2(x+1)(x-4))color(green)(->0=2(x+1)(x-4))#
Thus #color(blue)(x_("intercept") -> (x,y)-> (-1,0)" and "(+4,0))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "x_("vertex"))#
If you multiply out the right hand side you get:
#" "y=2(x^2-3x-4) -> #
From this we have two options to determine #x_("vertex")
#color(brown)("Option 1:")# This is the allowed format to apply:
#color(blue)(" "x_("vertex")=(-1/2)xx(-3) = +3/2)#
#color(brown)("Option 1:")# Take the mean of #x_("intercepts")" "(x" values only)"#
#color(blue)(" "x_("vertex")= ((-1)+(+4))/2 = +3/2)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "y_("vertex"))#
Substitute for #x# in original equation using #x_("vertex")" to find "y_("vertex")#
#color(blue)(=>y_("vertex")=2(3/2+1)(3/2-4) = -12 1/2 = -25/2)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "y_("intercept"))#
The graph crosses the y-axis at x=0. Substituting x=0 giving:
#color(blue)(y_("intercept")=2(0+1)(0-4)=-8)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine general shape of the graph")#
If you totally multiply out the right hand side and look at the highest order you have:
#y=2x^2-.....#
The coefficient of #x^2# is positive (+2)
#color(green)("So the general shape of the graph is: "uu)#
#color(blue)("Thus we have a "underline("minimum")->(x,y)->(3/2,-24/2 ))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
