Explain how these answers were computed from this graph and equation?

Question

this question was given on a previous exam:

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

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!

Answer 1

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.

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.