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Suppose f(x) = x². Find the graph off(x+3).???

Sagot :

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

View image BezalelM300503
View image BezalelM300503
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