How do you sketch the graph of #y=(x+3)^2# and describe the transformation?

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Answer 1 · 2018-05-06T17:05:45

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#y=(x+3)^2#

#y=x^2+6x+9#

we will compare the above function with this below

#y=ax^2+bx+c#

we will get

#a=1,b=6and c==9#

now we will find the vertix

#x_v=-b/(2a)=-6/2=-3#

#y_v=(-3)^2+6(-3)+9=0#

use some points to simplify the sketch

#f(2)=25#

#f(1)=16#

#f(0)=9#

#f(-1)=4#

#f(-2)=1#

now the sketch of our function #y=x^2+6x+9#

graph{x^2+6x+9 [-12.42, 10.08, -2.44, 8.81]}