In my grade 11 math course the function $f(x)$ is said to have domain $[-4,\infty)$ and range $(-\infty,-1)$. The question asks that range of the function $y=2f(x)$ be determined. The textbook says "this is a vertical stretch by a factor of 2 so it expands the upper bounds of the range by $2$" and so the answer is $(-\infty,-2)$.

I can see how this works for a linear function for example, but I believe that a vertical stretch would have no impact on the range of several other functions such as a parabola, a square root function, or an inverse function. Thus I am a little bit confused.

  • $\begingroup$ Why do you believe that scaling a parabola by a factor of $2$ would have no effect on the range of that parabola? $\endgroup$
    – Xander Henderson
    Sep 21 at 22:41
  • $\begingroup$ Draw any function. You don't even need a formula. Now consider that every $y$ goes to $2y$. If $f(x)=-5$ then $2f(x)=-10$ etc. If you do this for every $x$ value and every $y$ value gets doubled, what happens to the curve? $\endgroup$
    – John Douma
    Sep 21 at 22:46
  • 1
    $\begingroup$ Sorry, I've been incorrectly thinking that the coefficient 2 merely represents a vertical stretch as in $2f(x-h)+k$ when it obviously applies to $f(x)$ as a whole. I also failed to realize that the given range means that $f(x)$ has a vertical translation of 1 unit down; I've been thinking of parabolas whose vertex sit on the x-axis i.e., $x^2$ for which a vertical stretch would not affect the range. Thanks for your help. $\endgroup$ Sep 21 at 23:47

1 Answer 1


It's not the shape of the function that matters, but where the range sits relative to the $x$-axis.

Yes, $y=x^2$ would have the same range if "stretched" by a factor of 2, but $y=x^2+1$ would not.

The $x$-axis, i.e. the line $y=0$, is the set of fixed points of the transformation "vertical stretch by a factor of 2". So for your average "nice" function, where the range is some connected set, if you want the range to be unchanged you can't have an "edge" of the range be anywhere but at $y=0$. (I'm using "edge" and "nice" rather loosely here, but I hope it helps you develop your intuition.)


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