Graph of $y = G(h)$, the original (basic) function.

• The domain of $G$ is $[-7,-4) \cup \{-3\} \cup [-2,7)$.
• $G$ has no global maximum
• $G$ has a global (and local) minimum of $\answer {-9}$, which occurs at $\answer {-2}$.
• $G$ has a local maximum of $1$, which occurs at $\answer {-7}$.
• $G$ has both a local minimum and local maximum of $2$, which occurs at $-3$.

$\blacktriangleright$ Now to define a new function based on $G$.

Define $T(p) = -2G\left ( \frac {1}{2} p - 4 \right ) + 1$.

How do we construct the graph of $z = T(p) = -2G\left (\frac {1}{2} p - 4\right ) + 1$?

First, let’s map over all of the endpoints in the graph of $y = G(h)$.

Remember: If $h = \frac {1}{2} p - 4$, then $p = \answer {2(h+4)}$.

• Domain: $2 (h+4)$
  • (1) add $4$
  • (2) multiply by $2$
• Range: $-2 G + 1$
  • (1) multiply by $-2$
  • (2) add $1$

First, we’ll transform the domain coordinate, which is the left coordinate. Then we’ll transform the range coordinate, which is the right coordinate.

$$\begin {array}{c|l|l|l|l|l} & & \text {domain} & \text {domain} & \text {range} & \text {range} \\ \text {Point} & \text {Original} & +4 & \times 2 & \times (-2) & + 1 \\ \hline A & (-7, 1) & (-3, 1) & (-6, 1) & (-6, -2) & (-6,-1) \\ B & (-4, -2) & (0, -2) & (0, -2) & (0,4) & (0,5) \\ C & (-3,2) & (1,2) & (2,4) & (2,-8) & (2,-7) \\ D & (-2,-9) & (2, -9) & (4,-9) & (4, 18) & (4,19) \\ E & (1,-6) & (5,-6) & (10,-6) & (10,12) & (10,13) \\ F & (1,6) & (5, 6) & (10,6) & (10,-12) & (10,-11) \\ H & (7,0) & (11,0) & (22,0) & (22,0) & (22,1) \end {array}$$

Now, we’ll graphically map the important points from $y=G(h)$ to $t = T(p)$:

Linear transformations cannot change the shape of the graph, and they didn’t. The graph may have flipped vertically, but relatively speaking its shape is the same.

The table above shows that the transformations follow the order of operations applied to the original function and graph. The only thing to keep in mind is that you have to phrase the arithmetic of the transformation as being applied to the orginal function notation. To accomplish this, we solve the new domain expression, $h = \frac {1}{2} p - 4$, for the original domain variable. That rephrases it back to what happened to the original domain numbers. Then you can just apply the arithmetic to the coordinates of each point - following the order of operations.

The graph is filled-in exactly as it was connected in the original graph.

• Line segment between $A$ and $B$.
• Line segment between $D$ and $E$.
• Line segment between $F$ and $H$.
• $C$ is an isolated point.

What Happened?

Our new function was defined by

$$T(p) = -2G\left ( \frac {1}{2} p - 4 \right ) + 1$$

On the inside of the domain parentheses, we have $\frac {1}{2} p - 4$. Remember, $h$ is the variable representing domain numbers of $G$. The new function has $h = \frac {1}{2} p - 4$. Solving for $p$, gives us $p = 2(h+4)$. Now we can read off the arithmetic according to the order of operations.

When looking at $p = 2(h+4)$, the parentheses are grouping symbols here. We perform the arithmetic inside these parentheses first.

First, add $4$. The graph should shift to the right left $4$.
Second, multiplication by $2$. The graph should stretch horizontally by a factor of $2$.

On the outside of $G(h)$, we have multiplication by $-2$ and then addition of $1$. These are applied in the order of operations.

First, multiplication by $-2$. This is negative, so the graph should flip vertically over the horizontal axis.

• All positive second coordinates become negative.
• All negative second coordinates become positive.
• All $\answer {0}$ second coordinates remain $\answer {0}$.

The graph should also stretch vertically by a factor of $2$. Adding $1$ shifts the whole graph up down $1$.

Let $B(r) = \frac {1}{2}|3r-1| + 4$.

$B$ appears to be a linear transformation of the basic absolute value function: $|x|$.

Thinking Ahead

The basic absolute value function has a formula like $|x|$, with $\mathbb {R}$ as its domain. The graph has a corner at $(0,0$) and a “V” shape. This shape can’t change from a linear transformation.

The graph of $B$ is also a “V”, with a corner. The corner occurs where $\answer {3r-1}=0$, which is when $r=\frac {1}{3}$. The corner is at $\left ( \frac {1}{3}, \answer {4} \right )$

The outside coefficient of the transformation is $\frac {1}{2}$, which is positive. Therefore the “V” will again open up for the graph of $B$.

The “V” is made of two rays emanating from the corner. Therefore, we just need another point on each ray to plot the graph.

We can randomly choose any numbers on either side of $\frac {1}{3}$, like $r=-5$ and $r=5$.

$B(-5) = 12$ and $B(5) = 11$, giving us the points $\left (\answer {-5}, \answer {12}\right )$ and $\left (\answer {5}, \answer {11}\right )$, respectively.

We can now draw the graph of $y = B(r)$.

What are the graphical affects?

$\blacktriangleright$ Horizontally

The original inside was just $x$. This became $3r-1$ in the new function.

$x = 3r - 1$

But let’s rephrase this to show the arithmetic performed on the original variable, $x$.

$\frac {1}{3} (x + 1)$

The order of operations tell is to add $1$ first. That shifts the graph to the right by $1$. Then multiplication by $\frac {1}{3}$ compresses the graph horizontally.

The orginal corner was at $x=0$. This moved right to $1$ and then was compressed to $\frac {1}{3}$.

$\blacktriangleright$ Vertically

The original vertical measurements are given by the whole function: $|x|$. On the outside, we multiplied by $\frac {1}{2}$, which compressed the graph vertically. Then we added $4$, which sifted the graph vertically by $4$.

The original corner was at a height of $0$. This was compressed to $0$ and then raised to $4$.

The new corner is at $\left ( \frac {1}{3}, 4 \right )$

Here is a graph of $y=k(A)$

The graph of $k(A)$ has a shape with important or strategic points.

• The domain of $k$ is the union of two intervals: $[-9,-3) \cup (3,6]$.
• The longer interval is twice as long as the shorter interval.
• The inner endpoints are hollow on the graph.
• The outer endpoints are solid on the graph.

A linear transformation will maintain this shape.

$\blacktriangleright$ Define $f(m)$ by

$$f(m) = -2 \, k\left (-\frac {1}{2} m + 1\right ) - 5$$

What will the graph of $z = f(m) = -2 k\left (-\frac {1}{2} m + 1\right ) - 5$ look like?

$A = -\frac {1}{2} m + 1$ gives us $m = \answer {-2(A-1)}$.

Reading the order of operations, for $-2(A-1)$, the graph will shift left right $\answer {1}$. Then, the graph will reflect about the vertical axis, then the graph will stretch horizontally by a factor of $\answer {2}$.

The order of operations can be read directly on the outside. Reflect vertically, then stretch by a factor of $2$, then shift down $\answer {5}$.

$k$ only had two values, $8$ and $4$. For $f(m)$, these values will become $-2 \cdot 8 - 5 = \answer {-21}$ and $-2 \cdot 4 - 5 = \answer {-13}$.

The horizontal endpoints of the intervals will become

• $-2(-9-1) = \answer {20}$
• $-2(-3-1) = \answer {8}$
• $-2 \left ( \answer {3}-1 \right ) = -4$
• $-2 \left (\answer {6}-1 \right ) = -10$

The graph of $z = f(m)$.

The length of the longer interval is $\answer {12}$. The length of the shorter interval is $\answer {6}$. Twice as long.

The inner endpoints are hollow solid on the graph.

The outer endpoints are hollow solid on the graph.

The longer interval is now on the right, because multiplication by $-\frac {1}{2}$ reflected the graph horizontally vertically .

The longer interval is now on the bottom, because multiplication by $-2$ reflected the graph horizontally vertically .