Explain how to do this with answers!
Explain how to do this with answers!

\( \bf f(x)=log\left( \cfrac{x}{8} \right)\\\\ -\\\\ \textit{x-intercept, setting f(x)=0} \\\\ 0=log\left( \cfrac{x}{8} \right)\implies 0=log(x)-log(8)\implies log(8)=log(x) \\\\ 8=x\\\\ - \)
\( \bf \textit{y-intercept, is setting x=0}\\ \textit{wait just a second! a logarithm never gives 0} \\\\ log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}\qquad\qquad % exponential notation 2nd form {{ a}}^y={{ b}}\iff log_{{ a}}{{ b}}=y \\\\ \textit{now, what exponent for " a" can give you a zero? none}\\ \textit{so, there’s no y-intercept, because " x" is never 0 in }\frac{x}{8}\\ \textit{that will make the fraction to 0, and a}\\ \textit{logarithm will never give that, 0 or a negative}\\\\ \)
\( \bf -\\\\ domain \\\\ \textit{since whatever value " x" is, cannot make the fraction}\\ \textit{negative or become 0, then the domain is }x\ > \ 0\\\\ -\\\\ range \\\\ \textit{those values for " x", will spit out, pretty much}\\ \textit{any " y", including negative exponents, thus}\\ \textit{range is }(-\infty,+\infty) \)
 p, li { white-space: pre-wrap; }


now on 2)

\( \bf f(x)=\cfrac{3}{x^4} \)   if the denominator has a higher degree than the numerator, the horizontal asymptote is y = 0, or the x-axis,

in this case, the numerator has a degree of 0, the denominator has 4, thus y = 0

vertical asymptotes occur when the denominator is 0, that is, when the fraction becomes undefined, and for this one, that occurs at  \( x^4=0\implies x=0 \)  or the y-axis


now on 3)

\( \bf f(x)=\cfrac{1}{x} \)

now, let’s see some transformations templates

\( \bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\ \begin{array}{rllll} left side templates f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ y=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}}\mathbb{R}^{{{ B}}x+{{ C}}}+{{ D}} \end{array}\\ \bf \begin{array}{llll} right side info \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\ \bullet \textit{ vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\ \qquad if\ {{ D}}\textit{ is positive, upwards} \end{array} \)

now, let’s take a peek at g(x)

\( \bf \begin{array}{lcllll} g(x)=&-&\cfrac{1}{x}&+3\\ &\uparrow &&\uparrow \\ &\textit{upside down}&& \begin{array}{llll} \textit{vertical shift up}\\ \textit{by 3 units} \end{array} \end{array} \)