How do you graph # (y-3)^2=8(x-3)#?

Assumption: #y# is the dependant variable (answer)

#color(blue)("Deriving an equation that is simpler to plot")#

Multiply out the RHS bracket

#(y-3)^2=8x-24#

Square root both sides

#y-3=sqrt(8x-24)#
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Consider this point demonstrated by example

#(-3)xx(-3)=+9#
#(+3)xx(+3)=+9#

So depending on the conditions the square root of any number, say #a# is #+-sqrt(a)#
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Using the above principle we have:

#y-3=+-sqrt(8x-24)#

Add 3 to both sides

#y=3+-sqrt(8x-24)->" graph shape of "sub#
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#color(blue)("Determine the critical points on the plot")#

For the values to remain in the 'real' set of numbers ( #RR# )
#8x-24# must remain positive. The trigger point is when #x<3#.

So for #yinRR# we have #{x: x>=3}#

Thus there is not any plot for #x<3#

#color(brown)("y-intercept")#

Set #x=0# this does not comply with #{x: x>=3}#
so there is no y-intercept.

#color(brown)("x-intercept")#

Set #y=0=3+sqrt(8x-24)#

Thus for this to work #sqrt(8x-14)# needs to be -3. This is not acceptable so this condition does not exist.

Set #y=3-sqrt(8x-24)#

So this means that #8x-24=+9" " =>" " 8x=35#

#x=35/8=4.375#

#color(brown)("Determine "y_("vertex")" at "x_("vertex")=3)#

This is the turning point ( vertex ) of #sub#

#y=3+-sqrt(24-24)=3#

Vertex#->(x,y)=(3,3)#
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#color(blue)("Some people would say that the plot should be:")#