What are the important points needed to graph #y= 2 (x + 1) (x - 4)#?

#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))#
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#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)#
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#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)#
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#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)#

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#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 ))#

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