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Let $f(x) = x^2$ Let's try to transform that to $8x^2$.

First question, is this a vertical stretch scale factor $8$?

That would be $8f(x)$ then? But, could it also be $f(\frac{x}{\sqrt{8}})$ because that would equate to a horizontal stretch $\frac{1}{a}$ (a being the multiple, in this case $\frac{1}{\sqrt{8}}$) which would give $(1/1/\sqrt{8} *x)^2 = (\sqrt{8}x)^2 = 8x^2$, the same thing as before. This is my second question

Cobbles
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2 Answers2

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First question, yes. It is a vertical translation since it takes it multiplies the output of $f(x)$ by $8$.

I'm not entirely clear what you are asking in the second question, but note: $$f(x) = x^2 \implies f(x/\sqrt 8) = \left(\frac x{\sqrt 8}\right)^2 = \dfrac {x^2}{8} \neq 8x^2$$

That flattens (vertically) the graph of $f(x) = x^2$ by a factor of $\frac{1}{8}$

What you can do is as follows:if you evaluate $f(\sqrt 8 x)$, you obtain output as follows: $$(\sqrt 8 x)^2 = 8x^2$$ as desired, which squeezes the graph horizontally by a factor of $\sqrt 8$.

amWhy
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  • I thought that if a is inside (f(ax)) then it becomes (1/a x)^2 – Cobbles Apr 17 '14 at 15:15
  • Note that if you evaluate $f(\sqrt 8 x)$, you obtain output as follows: $(\sqrt 8 x)^2 = 8x^2$, as desired. – amWhy Apr 17 '14 at 15:20
  • I was taught that a stretch f(ax) means scale factor 1/a, is this correct? If so, why is it not $1/\sqrt{8}x^2$? – Cobbles Apr 17 '14 at 15:27
  • Because it would compress by a factor of $\sqrt 8$ (to the left), it would stretch horizontally by the factor of $\frac 1{\sqrt 8}$. The point is, it would move x values in to the left. So you can think of stretching as the inverse of shrinking. – amWhy Apr 17 '14 at 15:31
  • I see, but you said that $f(\sqrt{8}x)$ = 8x^2 and is a horizontal stretch/shrink, yet it is a vertical stretch? – Cobbles Apr 17 '14 at 15:32
  • I am replacing $x$ in f(x) with $\sqrt 8 x$. The function squares this value, to get $8x^2$ – amWhy Apr 17 '14 at 15:34
  • Thank you, i was confusing horizontal stretch with vertical shrink – Cobbles Apr 17 '14 at 15:37
  • The behaviors are reversed between "horizontal" and "vertical" transformations: $ \ af(x) \ $ for $ \ a > 1 \ $ produces a "vertical stretch", $ \ 0 < a < 1 \ $ , a "vertical compression"; $ \ f(ax) \ $ for $ \ a > 1 \ $ produces a "horizontal compression", $ \ 0 < a < 1 \ $ , a "horizontal stretch". For horizontal transformations, we are taking the value of $ \ x \ $ , multiplying it by $ \ a \ $ , evaluating the function there, and then "plotting" the result back at $ \ x \ $ . (continued) – colormegone Apr 17 '14 at 15:42
  • So, for $ \ a > 1 \ $ , the value of the function at larger values of $ \ x \ $ get plotted at smaller values of $ \ x \ $ , "squeezing" the curve toward the $ \ y-$ axis; for $ \ 0 < a < 1 \ $ , the function at smaller values of $ \ x \ $ is "taken out" to larger values of $ \ x \ $ , producing the "horizontal stretch" away from the $ y-$ axis. – colormegone Apr 17 '14 at 15:44
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Transforming $f(x)=x^2$ into $g(x)=8x^2$ can both be interpreted as a vertical stretch by a factor of $8$, or it can also be interpreted as a horizontal compression, by a factor of $\sqrt{8}$:

$8f(x)=8\cdot x^2=8x^2 \;\;(\text{This is a vertical stretch by factor of 8})$

$f(\sqrt{8}x)=((\sqrt{8}x))^2=8x^2\;\; (\text{This is a horizontal compression by a factor of } \sqrt{8})$

Maybe you are confusing horizontal stretches and compressions. Recall that for $f(ax)$:

  • If $a$ is between $0$ and $1$, then the graph is stretched horizontally. This means that $a$ is a fraction. If this happens, we say that the graph is stretched by a factor $a$.
  • If $a$ is bigger than $1$, then the graph is compressed horizontally. If this happens, we say that the graph is compressed by a factor of $a$.

For example, say I have a function $g(x)$. Then:

  • $g(5x)$ is a horizontal compression by a factor of $5$. $(\text{This is because 5>1})$

  • $g\left(\frac{1}{2}x\right)$ is a horizontal stretch by a factor of $\frac{1}{2}.\;\;(\text{This is because}\frac{1}{2}<1)$

Sujaan Kunalan
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