Let $\displaystyle g(x) = \log x$. In terms of $\displaystyle g(x)$,

$\displaystyle f(x) =4 g (x- 3)- 2$

$\displaystyle g(x)$, being a logarithmic function, has a graph without a horizontal asymptote. As $\displaystyle f(x)$ represents the result of transformations of $\displaystyle g(x)$, it follows that its graph does not have a horizontal asymptote, either.

Let $\displaystyle g(x) = \log x$. In terms of $\displaystyle g(x)$,

$\displaystyle f(x) =4 g (x- 3)- 2$

The graph of $\displaystyle g(x)$ has as its vertical asymptote the line of the equation $\displaystyle x= 0$. The graph of $\displaystyle f(x)$ is the result of three transformations on the graph of $\displaystyle g(x)$- a right shift of 3 units ( $\displaystyle -3$ ), a vertical stretch ( $\displaystyle 4g$ ), and a downward shift of 2 units ( $\displaystyle -2$ ). Of the three transformations, only the right shift affects the position of the vertical asymptote; the asymptote of $\displaystyle f(x)$ also shifts right 3 units, to $\displaystyle x = 3$.

$\displaystyle g(x) = -f(x-2)$, so the graph of $\displaystyle g(x)$ is the result of performing the following transformations:

1) $\displaystyle f_{1}(x) = f(x-2)$ is the result of translating this graph two units right.

2) $\displaystyle g(x)= -f_{1}(x) = -f(x-2)$ is the result of reflecting the new graph about the $\displaystyle x$-axis.

Reflecting the graph of a function $\displaystyle f(x)$ about the $\displaystyle x$-axis results in the graph of the function

$\displaystyle f_{1}(x)= -f(x)$.

Translating this graph upward $\displaystyle \frac{8}{3}$ results in the graph of the function

$\displaystyle g(x)= f_{1}(x) + \frac{8}{3 }= -f(x)+ \frac{8}{3 }$.

A vertical translation of the graph of a function $\displaystyle f(x)$ by $\displaystyle k$ units yields the graph of the function $\displaystyle g(x)+k$. A translation in an upward direction is a positive translation, so setting $\displaystyle f(x)= \ln x$ and $\displaystyle k = 3$, the resulting graph becomes

$\displaystyle g(x)=( \ln x) +3$

or

$\displaystyle g(x)=3+ \ln x$

Apply properties of logarithms to rewrite this as

$\displaystyle g(x)= \ln e^{3} + \ln x$

$\displaystyle g(x)= \ln (e^{3} x)$.

The reflection of the graph of a function $\displaystyle f(x)$ about the $\displaystyle y$-axis yields the graph of the function $\displaystyle g(x) = f(-x)$. Therefore, set $\displaystyle f(x) = \ln x$ and substitute $\displaystyle -x$ for $\displaystyle x$ to yield the function

$\displaystyle g(x) = \ln (-x)$.

To check for symmetry, we are going to do three tests, which involve substitution. First one will be to check symmetry along the x-axis, replace $\displaystyle -y=y$.

$\displaystyle -y=x^2-6x+10$

This isn't equivilant to the first equation, so it's not symmetric along the x-axis.

Next is to substitute $\displaystyle -x=x$.

$\displaystyle y=(-x)^2-6(-x)+10$

$\displaystyle y=x^2+6x+10$

This is not the same, so it is not symmetric along the y-axis.

For the last test we will substitute $\displaystyle -y=y$, and $\displaystyle -x=x$

$\displaystyle -y=(-x)^2-6(-x)+10$

$\displaystyle -y=x^2+6x+10$

$\displaystyle y=-x^2-6x-10$

This isn't the same as the orginal equation, so it is not symmetric along the origin.

The answer is it is not symmetric along any axis.

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Also, it can be seen to be symmetrical about the origin. Consequently, for each $\displaystyle c$ in the domain, $\displaystyle f(-c) = -f(c)$ - the function is odd.

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Also, it is seen to be symmetric about the origin. Consequently, for each $\displaystyle c$ in the domain, $\displaystyle f(-c) = -f(c)$ - the function is odd.

A function $\displaystyle f(x)$ is even if and only if, for all $\displaystyle x$ in its domain, $\displaystyle f(-x) = f(x)$. It follows that if $\displaystyle f(9) = 10$, then 

$\displaystyle f(-9)= f(9) = 10$.

No restriction is placed on any other value as a result of this information, so the answer is false.