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]}

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