Curve stretching

Question

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

Answer 1

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 $\sqrt{8}$:

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

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

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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 $\frac{1}{2}.\;\;(\text{This is because}\frac{1}{2}<1)$

Answer 2

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 $\sqrt 8$.