color(brown)("Formatting tip - write "color(white)("d")" hash y = a root(3)(x)+b hash ")
color(brown)( "giving: ............."color(white)("ddddddddddddd") y=a root(3)(x)+b
Have a look at: https://socratic.org/help/symbols
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Set point 1 as P_1->(x_1,y_1)=(-1,4)
Set point 2 as P_2->(x_2,y_2)=(1,-8) color(red)( larr" +8 corrected to -8")
Given: color(white)("d") y=aroot(3)(x)+b
P_1->+4=aroot(3)(-1)+b" "...............Equation(1)" checked"
P_2->color(white)(".")color(red)(-8)=aroot(3)(+1)+b" "...............Equation(2)" corrected"
color(blue)("Determine the value of "b)
Note that:
(-1)xx(-1)xx(-1)color(white)("d")=color(white)("d")(-1)" so "root(3)(-1)=-1
Note that root(3)(+1)=1
So we have:
color(white)("d")+4=-a+b" "....................Equation(1_a)
ul(color(white)("d")-8=+a+b)" "....................Equation(2_a)
color(green)(color(white)("d")-4=color(white)(".dd")0+2b) larr" "Eqn(1_a)+Eqn(2_a)
Divide both sides by color(red)(2)
color(green)(-4/color(red)(2)=(2b)/color(red)(2))
-2=bcolor(white)("d") more formally presented as: b=-2
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color(blue)("Determine the value of "a)
Choosing Equation(2_a) substitute for b where color(red)(b=-2)
color(green)(-8=+a+color(red)(b) color(white)("ddd") -> color(white)("ddd")-8=a+(color(red)(-2)))
Add 2 to both sides
color(green)(color(white)("dddddddddddddd")->color(white)("ddd")-6=a)
More formally a=-6
color(white)("d")