What are the important points needed to graph #y=3(x+3)^2 - 3#?

1 Answer
Mar 20, 2017

#y_("intercept")=24#
#x_("intercepts")=-4 and -2#
Vertex #->(x,y)=(-3,-3)#

Explanation:

This is the vertex form of the standardised#" " y=ax^2+bx+c#
Multiplying out the brackets the we have #" "y=3x^2+18x+24#

As the #3x^2# is positive the graph is of general shape #uu#

#color(blue)("Determine the vertex using the questions equation format")#

Given:#" "y=3(xcolor(red)(+3))^2color(green)(-3)#

#x_("vertex")=(-1)xxcolor(red)((+3)) = -3#
#y_("vertex")=color(green)(-3)#

Vertex#->(x,y)=(-3,-3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Determine the x-intercepts")#

set #y=0#

#=>0=3(x+3)^2-3#

#3=3(x+3)^2#

#(x+3)^2=1#

#x+3=+-sqrt((1)#

#x=-3+-1#

#x=-4 and -2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine y-intercept")#

Set #x=0#

#y= 3(0+3)^2-3#

#y=27-3=24#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Foot note")#

Using the format #y=3x^2+18x+24#

#x_("vertex")=(-1/2)xx(18/3)=-3#

Tony B