0
$\begingroup$

I'm asking a graph question that has bothered me a while. I'm considering two graphs here:

$$ y=(x+2)^2 $$ $$ y=x^2 + 4 $$

As can be seen, both the difference is a mere $4x$. I would like to know if we were to draw out the graphs, the graph with an additional $4x$ causes the graph to be behind the one without $4x$. Also, for any real value of $x$, why would the difference between the two graphs increase when $x \lt 2$? Does it have to do with rates of change? Thanks!

$\endgroup$

1 Answer 1

0
$\begingroup$

The difference (by which I mean the absolute value of the differences) increases when $x$ moves away from $0$--that's obvious by the difference which is $\left|4x\right|$. And the the two graphs are the same: the first, $y = \left(x + 2\right)^2$, is a regular parabola shifted to the left by two and the second, $y = x^2 + 4$, is a regular parabola shifted to up by four. The only reason the first is "behind" the first is because they are the same and the first one starts to the left and below the second.

$\endgroup$
2
  • $\begingroup$ so is there any relationship to 4x and the moving of the graph to behind? ( like 4 units of x)? So if it is x^2+ 2x+4, how would the difference of 2x to both equations i stated above affect the grpah? Sorry noob here! $\endgroup$
    – Zun Jie
    May 14, 2014 at 12:38
  • $\begingroup$ No, it's already written in terms of a translation of a parabola $y = x^2$: $y = x^2 + 4$ and $y = (x - -2)^2$. Usually you go backwards and are given $y = x^2 + 4x + 4$ and you need to complete the square to see the translation. $\endgroup$
    – Jared
    May 14, 2014 at 17:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .