For each point $(a,b)$ on the graph $y = f(x)$ the point $(3a-1,\frac{b}{2})$ is plotted forming the graph of another function $y = g(x)$. The problem asks us to find the function $g(x)$ in terms of $f(x)$.

Also, we need to describe the transformation.

Since $a$ goes to $3a-1$, the inverse of the function will be $x \to \frac{x+1}{3}$.

Hence the required function is $$g(x) = \frac{1}{2}f\left( \frac{x+1}{3}\right).$$

Note that $$g(3a-1) = \frac{1}{2}f\left( \frac{3a-1+1}{3}\right) = \frac{1}{2}f(a) = \frac{b}{2}.$$

Description of the transformation:

$\bullet$ Horizontal stretch by a factor of $3$.

$\bullet$ Vertical stretch by a factor of $\frac{1}{2}$.

$\bullet$ Horizontal translation by $-1$.

Is the solution correct?


1 Answer 1


Let $$X=3a-1\;\text{ and }\; g(X)=\frac b2$$


$$b=f(a) $$ or

$$2g(X)=f(\frac{X+1}{3})$$ So

$$g(X)=\frac 12f(\frac{X+1}{3})$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.