How do you identify the important parts of #y = 1/2(x-3)(x+1)# to graph it?

1 Answer
Nov 25, 2015

See stepwise explanation

Explanation:

Tony B

#color(green)("Given: "y=1/2(x-3)(x+1).................................(1)#

#color(green)("Same as: "y=1/2(x^2-2x-3))#

#color(blue)(x^2 " is positive so" underline(" upwards U-shape")).................(2)#

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#color(brown)("To find "x_("vertex")" Consider the -2 of "-2x)#

#color(blue)(x_("vertex")=-1/2(-2)=+1)......................(3)#
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#color(brown)("To find "y_("vertex")" Substitute (3) into (1)")#

#y=1/2(1-3)(1+1) = 1/2(-2)(2)=-2#

#color(blue)(y_("vertex")=-2)....................................(4)#
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#color(brown)("To find y-intercept")#

Substitute #x=0# in (1)

#y=1/2(0-3)(0+1) =1/2(-3)(1) = -3/2#

#color(blue)(y_("intercept") = -3/2)#
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#color(brown)("To find x-intercept")#

Substitute #y=0# in (1)

#0=1/2(x-3)(x+1)#

#(x-3)=0 -> x=+3#
#(x+1)=0-> x=-1#

#color(blue)(x_("intercept")-> (x ,y) -> (-1 , 0) " and "(+3 , 0))#
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