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To do this problem I substituted the $$(x+2)$$ as the x in the equation

$$ 3f(x+2)=2(x+2)-4$$ $$3f(x+2)=2x+4-4$$ $$3f(x+2)=2x$$ $$3f(x+2)/3=2x/3$$

when i graphed the equation I did not get the answer stated as the answer choice. What did i do wrong?

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    $\begingroup$ Vertical stretch by factor of $3$ and move $2$ units over. $\endgroup$ – MathMajor Jun 5 '16 at 20:41
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    $\begingroup$ $3f(x+2)=3\left[2(x+2)-4\right]$ $\endgroup$ – Semiclassical Jun 5 '16 at 20:42
  • $\begingroup$ @MathMajor In this case, it's fine, but I must point out that in general, one should be careful about the order of operations, when it comes to transformations. $\endgroup$ – pjs36 Jun 5 '16 at 20:50
  • $\begingroup$ @semiclassical So multiply 2 first and than by 3? I thought i should've divided it by 3? $\endgroup$ – MATH ASKER Jun 5 '16 at 21:14
  • $\begingroup$ @pjs36 Yes, agreed. $\endgroup$ – MathMajor Jun 5 '16 at 21:17
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Begin with the graph of $f$.

Move it two to the left to get the graph of $y= f(x + 2)$.

Now stretch vertically by a factor of 3 to get the graph of $y = 3 f(x + 2)$.

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  • $\begingroup$ so left two units and from there up 3 units and 1 units over? $\endgroup$ – MATH ASKER Jun 5 '16 at 20:58
  • $\begingroup$ No. Read what I said. $\endgroup$ – ncmathsadist Jun 6 '16 at 1:10
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The graph of $f(x)=2x-4$ is a straight line; if I stretch/shift things (as I do by multiplying by a constant overall and replacing $x\to x+2$) the graph will still be that of a straight line. But when $x=-2$ one has $3f(0)=3(-4)=-12$, and for $x=0$ one has $3f(2)=3(0)=0$. Hence the new graph will be a line through the points $(-2,-12)$ and $(0,0)$, corresponding to the graph of $y=6x$.

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  • $\begingroup$ how did u get x=-2? $\endgroup$ – MATH ASKER Jun 5 '16 at 21:01
  • $\begingroup$ @MATHASKER I picked a point arbitrarily and tried it. I could've instead taken, say, $x=131$ and gotten instead $3f(133)=3(262)=786$. So the point $(131,768)$ should also be on the line. (And it is!) $\endgroup$ – Semiclassical Jun 5 '16 at 21:04

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