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this question was given on a previous exam:

Below is a graph of a function y = f(x) with a few key points. graph of function

Draw a graph of y = 3f(2x) + 4, with the corresponding points shown in the graph above.

And the answer key is given as the points:

(-1/2,10), (1/2, 10), (1/4), (3/2,10)

for the new points on the new graph; but, I am confused as to how these answers were computed. It would be great if someone can explain to me how these answers were computed.

Thank you!

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    The graph of $y=f(2x)$ is scaling of $y=f(x)$ in $x$ directions by a factor of $1/2$; ie. it "shrinks" the graph of $y=f(x)$. So, for example the point $(-1,2)$ is mapped to $(-1/2,2)$. Then $y=3f(2x)$ is a shrinking followed by scaling in direction of $y$ axes by a factor of $3$. So $(-1/2,2)$ maps to $(-1/2,6)$. Finally, $y=3f(2x)+4$ translates this graph in positiv e $y$ direction by $4$, so $(-1/2,6)$ goes to $(-1/2,10)$. Same thing applies to other points – 2017-01-31

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Suppose you have a function defined by some equation $y=f(x)$ as in this problem and suppose each $y$ in the equation is replaced by a linear expression $ay+b$ where $a$ and $b$ are constants, $a\ne0$. Suppose also that each $x$ in the equation is replaced by a linear expression $cx+d$ where $c\ne0$ and $d$ are constants.

Then one will have a new functional equation which maps each straight line connecting any two points of the original graph to a straight line connecting two points of the new, transformed graph. Such a transformation is called a linear transformation.

In order to use this information to graph $y=3f(2x)+4$ we must first solve for $f(2x)$ which we find is

$$\frac{1}{3}y-\frac{4}{3}=f(2x)$$

which we compare to

$$y=f(x)$$

to see that the linear transformation which is being applied to the given graph is

$$x\,\to\,2x,\qquad y\,\to\,\frac{1}{3}y-\frac{4}{3}$$

Since the graph of $y=f(x)$ consists of straight line segments and since this is a linear transformation, we just have to find out where the four endpoints of $y=f(x)$ are mapped and connect them with straight line segments.

We can do this with each of the four points as follows:

  1. $2x=-1$ so $x=-\frac{1}{2}.\qquad$ $\frac{1}{3}y-\frac{4}{3}=2$ so $y=10$. Therefore $(-1,2)\,\to\,\left(-\frac{1}{2},10\right)$
  2. $2x=1$ so $x=\frac{1}{2}.\qquad$ $\frac{1}{3}y-\frac{4}{3}=2$ so $y=10$. Therefore $(1,2)\,\to\,\left(\frac{1}{2},10\right)$
  3. $2x=2$ so $x=1.\qquad$ $\frac{1}{3}y-\frac{4}{3}=0$ so $y=4$. Therefore $(2,0)\,\to\,(1,4)$
  4. $2x=3$ so $x=\frac{3}{2}.\qquad$ $\frac{1}{3}y-\frac{4}{3}=2$ so $y=10$. Therefore $(3,2)\,\to\,\left(\frac{3}{2},10\right)$

The original graph is mapped to the new graph shown in red.

Below the graph is a second way to work the problem which requires finding the piece-wise equations for the given graph.

Graph of solution

There is a second way to work this kind of problem if you are given the equation. This particular function consists of three line segments "glued" together.

$y=\begin{cases} 2 & \text{for }-1\le x<1\\ -2x+4 & \text{for }\phantom{-}1\le x<2\\ 2x-4 & \text{for }\phantom{-}2\le x\le3\\ \end{cases}$

Now apply the transformation $x\,\to\,2x,\quad y\,\to\,\frac{1}{3}y-\frac{4}{3}$

$\frac{1}{3}y-\frac{4}{3}=\begin{cases} 2 & \text{for }-1\le 2x<1\\ -4x+4 & \text{for }\phantom{-}1\le 2x<2\\ 4x-4 & \text{for }\phantom{-}2\le 2x\le3\\ \end{cases}$

Remove fractions by multiplying all three equations by $3$ to obtain

$y-4=\begin{cases} 6 & \text{for }-1\le 2x<1\\ -12x+12 & \text{for }\phantom{-}1\le 2x<2\\ 12x-12 & \text{for }\phantom{-}2\le 2x\le3\\ \end{cases}$

Then add four to each side of the three equations.

$y=\begin{cases} 10 & \text{for }-1\le 2x<1\\ -12x+16 & \text{for }\phantom{-}1\le 2x<2\\ 12x-12 & \text{for }\phantom{-}2\le 2x\le3\\ \end{cases}$

Finally, solve the three inequalities for $x$ to express the domain correctly.

$y=\begin{cases} 10 & \text{for }-\frac{1}{2}\le x<\frac{1}{2}\\ -12x+16 & \text{for }\phantom{-}\frac{1}{2}\le x<1\\ 12x-12 & \text{for }\phantom{-}1\le x\le\frac{3}{2}\\ \end{cases}$

This gives the piece-wise equations for the transformed graph.

Note that the actual reason for learning this is so that you can do the process in reverse. That is when the translating, reflecting and scale changes comes into play.

For example, an equation such as

\begin{equation} y=3+\sqrt{x+1} \end{equation}

can be re-written as

\begin{equation} y-3=\sqrt{x+1} \end{equation}

so that we recognize it as the graph of $y=\sqrt{x}$ transformed by $x\,\to\,x+1,\quad y\,\to\,y-3$ which translates the graph graph of $y=\sqrt{x}$ by $-1$ units horizontally and $+3$ units vertically.

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    Thank you so much! This was explained mathematically great. I was also wondering, do you think this new graph can be generated from using just transformations (e.g. translations, reflection, magnifications)? – 2017-02-02
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    Yes it can be done in terms of translations, scale changes (magnification & shrinking) but that approach is easier when you have the equation of the function. I will add that to my answer so you will see how to do it when you have the equation of the graph. In this case it will be a piece-wise defined function so first I will re-write it in that form. Give me a few minutes to write that up. – 2017-02-02