If f(x)=x^2
Then f(x+3)=(x+3)^2
[tex](x+3)^2=x^2+6x+9[/tex]
Use geogebra to graph the function or calculate the vertex using the equation
x=-b/2a
From the equation we have that
b=6
a=1
x=-6/(2*1)
x=-3
The vertex is on x=-3
Calculate f(-3)=9-18+9=0
The vertex is (-3,0)
y axis cut off point f(0)=0+0+9=9
As "a" is a positive value, parabola open upwards, now you can draw the parabola
This is a sketch, let's use geogebra
