Question #5f4f1

Apr 11, 2015

Leaving the calculations of the table of values to get full marks aside as a clear replacement for hard but obvious work a student must do himself, here is an explanation of the graph transformation
from $f \left(x\right) = {3}^{x}$ to $g \left(x\right) = {3}^{-} \left(x + 1\right) - 2$.

First of all, let's recall the definition of a graph of a function $y = F \left(x\right)$: it's a set of all points on a Cartesian plane with coordinates $\left(a , b\right)$ such that $b = F \left(a\right)$.

We will transform a graph of $f \left(x\right)$ to a graph of $g \left(x\right)$ in the following steps:
Step 1: from $f \left(x\right) = {3}^{x}$ to ${f}_{1} \left(x\right) = {3}^{-} x$.
Step 2: from ${f}_{1} \left(x\right) = {3}^{-} x$ to ${f}_{2} \left(x\right) = {3}^{-} \left(x + 1\right)$.
Step 3: from ${f}_{2} \left(x\right) = {3}^{-} \left(x + 1\right)$ to $g \left(x\right) = {3}^{-} \left(x + 1\right) - 2$.

Here are the transformations in steps.

Step 1: from $f \left(x\right) = {3}^{x}$ to ${f}_{1} \left(x\right) = {3}^{-} x$.
Consider general transformation
from $y = F \left(x\right)$ to $y = F \left(- x\right)$
If a point $\left(a , b\right)$ belongs to a graph of $y = F \left(x\right)$, that is if $b = F \left(a\right)$, then a point $\left(- a , b\right)$ belongs to a graph of $y = F \left(- x\right)$ because $F \left(- \left(- a\right)\right) = F \left(a\right) = b$
Therefore, each point $\left(a , b\right)$ of a graph of $y = F \left(x\right)$ corresponds to a point $\left(- a , b\right)$ of a graph of $y = F \left(- x\right)$. So, the whole graph of $y = F \left(- x\right)$ is symmetrically reflected to a graph of $y = F \left(x\right)$ relatively to the Y-axis.

In our case $F \left(x\right) = {3}^{x}$, so the transformation looks like:
$f \left(x\right) = {3}^{x}$:
graph{3^x [-10, 10, -5, 5]}
${f}_{1} \left(x\right) = {3}^{-} x$:
graph{3^-x [-10, 10, -5, 5]}

Step 2: from ${f}_{1} \left(x\right) = {3}^{-} x$ to ${f}_{2} \left(x\right) = {3}^{-} \left(x + 1\right)$.
Consider general transformation
from $y = F \left(x\right)$ to $y = F \left(x + \epsilon\right)$
If a point $\left(a , b\right)$ belongs to a graph of $y = F \left(x\right)$, that is if $b = F \left(a\right)$, then a point $\left(a - \epsilon , b\right)$ belongs to a graph of $y = F \left(x + \epsilon\right)$ because $F \left(a - \epsilon + \epsilon\right) = F \left(a\right) = b$
Therefore, each point $\left(a , b\right)$ of a graph of $y = F \left(x\right)$ corresponds to a point $\left(a - \epsilon , b\right)$ of a graph of $y = F \left(x + \epsilon\right)$. So, the whole graph shifts to the left by the value $\epsilon$.

In our case $F \left(x\right) = {3}^{-} x$ and $\epsilon = 1$ to transform a function to ${3}^{-} \left(x + 1\right)$. So, the graph of ${3}^{-} x$ is shifted to the left by $\epsilon = 1$.
${f}_{1} \left(x\right) = {3}^{-} x$:
graph{3^-x [-10, 10, -5, 5]}
${f}_{2} \left(x\right) = {3}^{-} \left(x + 1\right)$:
graph{3^-(x+1) [-10, 10, -5, 5]}

Step 3: from ${f}_{2} \left(x\right) = {3}^{-} \left(x + 1\right)$ to $g \left(x\right) = {3}^{-} \left(x + 1\right) - 2$.
Consider general transformation
from $y = F \left(x\right)$ to $y = F \left(x\right) + \delta$
If a point $\left(a , b\right)$ belongs to a graph of $y = F \left(x\right)$, that is if $b = F \left(a\right)$, then a point $\left(a , b + \delta\right)$ belongs to a graph of $y = F \left(x\right) + \delta$ because $F \left(a\right) + \delta = b + \delta$
Therefore, each point $\left(a , b\right)$ of a graph of $y = F \left(x\right)$ corresponds to a point $\left(a , b + \delta\right)$ of a graph of $y = F \left(x\right) + \delta$. So, the whole graph shifts vertically by $\delta$ (up for positive $\delta$ and down for negative).

In our case $F \left(x\right) = {3}^{-} \left(x + 1\right)$ and $\delta = - 2$ to transform a function to ${3}^{-} \left(x + 1\right) - 2$. So, the graph of ${3}^{-} \left(x + 1\right)$ is shifted down by $\delta = - 2$.
${f}_{2} \left(x\right) = {3}^{-} \left(x + 1\right)$:
graph{3^-(x+1) [-10, 10, -5, 5]}
$g \left(x\right) = {3}^{-} \left(x + 1\right) - 2$:
graph{3^-(x+1)-2 [-10, 10, -5, 5]}

This completes the transformation from $f \left(x\right) = {3}^{x}$ to $g \left(x\right) = {3}^{-} \left(x + 1\right) - 2$.

Apr 11, 2015