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Factor completely to find the implicit domain of the polynomial or rational function given and all of its roots/zeros. Also identify any $x$-intercepts on the graph of the function, and where this graph is above or below the $x$-axis.
$f(x) = x^2 - 8x + 12$
$f(x) = x^3 - 9x^2 - 4x + 36$
$f(x) = \cfrac{x^2 - 25}{x^2 - 8x + 15}$
$f(x) = \cfrac{36 - 13x + x^2}{x^2-8x+16}$
$x^2 - 8x + 12 = (x-2)(x-6)$.
With no "problem values" of $x$ with regard to computing the above, $f$ has a domain of $\mathbb{R}$.
Solving for where each factor equals zero, we arrive at the roots $x=2$ and $x=6$. These reveal immediately that the graph of $f$ has $x$-intercepts at $(2,0)$ and $(6,0)$.
To answer the question about where the graph of $f$ is above or below the $x$-axis, we note there were two places where we had factors equal to zero, and thus three intervals to consider: $(-\infty,2)$, $(2,6)$, and $(6,\infty)$. Finding the signs of all the factors and their product in each of these intervals reveals that $f$ is positive in the first and third of these intervals and negative in the second. As such, the graph of $f$ is above the $x$-axis when $x \in (-\infty,2) \cup (6,\infty)$ and below the $x$-axis when $x \in (2,6)$.
$\displaystyle{ \begin{array}[t]{rcl} x^3 - 9x^2 - 4x + 36 &=& x^2(x-9) - 4(x-9)\\ &=& (x^2 - 4)(x-9)\\ &=& (x+2)(x-2)(x-9) \end{array}}$
With no "problem values" of $x$ with regard to computing the above, $f$ has a domain of $\mathbb{R}$.
Solving for where each factor equals zero, we arrive at roots $x=\pm 2$ and $x=9$. These reveal immediately that the graph of $f$ has $x$-intercepts at $(2,0)$ and $(9,0)$.
$\cfrac{x^2 - 25}{x^2 - 8x + 15} = \cfrac{(x-5)(x+5)}{(x-3)(x-5)}$
Solving for where each factor equals zero, we note that the denominator is zero (and hence the rational expression is undefined) when $x = 3$ or $x = 5$. As such, the implicit domain of $f$ is $\{x \in \mathbb{R} \ | \ x \neq 3,5\}$.
We also note that the numerator is zero when $x = \pm 5$. Noting that $f$ will only be zero when the numerator is zero and the denominator is not zero, $f$ has only a single zero: $x=-5$. Consequently, the graph of $f$ has only a single $x$-intercept at $(-5,0)$.
To answer the question about where the graph of $f$ is above or below the $x$-axis, we note there were three places where we had factors in either the numerator or denominator equal to zero, and thus four intervals to consider: $(-\infty,-5)$, $(-5,3)$, $(3,5)$, and $(5,\infty)$. Finding the signs of all the factors and their product/quotient in each of these intervals reveals that $f$ is positive for the all but the second of these intervals. As such, the graph of $f$ is above the $x$-axis when $x \in (-\infty,-5) \cup (3,5) \cup (5,\infty)$ and below the $x$-axis when $x \in (-5,3)$.
$\displaystyle{ \begin{array}[t]{rcl} \cfrac{36 - 13x + x^2}{x^2+8x+16} &=& \cfrac{(x-4)(x-9)}{(x-4)^2}\\ \end{array}}$
Solving for where each factor equals zero, we note that the denominator is zero (and hence the rational expressin is undefined) when $x=4$. This tells us that the implicit domain for the related function is $\mathbb{R}_{\neq 4}$.
We also note that the numerator is zero when $x = 4$ or $x = 9$. Noting that the entire rational expression will only be zero when the numerator is zero and denominator is not zero, we have only one zero for this rational expression: $x=9$. The implicit domain of the related function is $\{x \in \mathbb{R} \ | \ x \neq 4\}$ (or, writing things more briefly: $\mathbb{R}_{\neq 4}$).
To answer the question about where the graph of $f$ is above or below the $x$-axis, we note there were two places where we had factors in either the numerator or denominator equal to zero, and thus three intervals to consider: $(-\infty,4)$, $(4,9)$, and $(9,\infty)$. Finding the signs of all the factors and their product/quotient in each of these intervals reveals that $f$ is positive in the first and third of these intervals and negative in the second. As such, the graph of $f$ is above the $x$-axis when $x \in (-\infty,4) \cup (9,\infty)$ and below the $x$-axis when $x \in (4,9)$.
Find all roots/zeros of the equation: $\displaystyle{(2x^4 + 30x^3 + 150x^2 + 250x)(4x^2 - 4x + 1) = 0}$
$\displaystyle{x = 0, -5, \frac{1}{2}}$
Find where (i.e., which $x$-values) the graph of the function $f(x)$ is above the graph of $g(x)$, and where it is below the graph of $g(x)$. Try to use a graphing calculator to verify your conclusions -- note how tricky it is to find the right window on the calculator!
Hint: the graph of $f$ is above the graph of $g$ precisely for those $x$-values where $f(x)-g(x)$ is positive, and $f$ is below the graph of $g$ precisely for those values where $f(x)-g(x)$ is negative.
$f(x) = x^3 + 45; \quad g(x) = 5x^2 + 9x$
$f(x) = \cfrac{24}{x^2-5x+6}; \quad g(x) = \cfrac{x}{x-3}$
As the hint suggested, we need to find out where $(x^3 + 45) - (5x^2 + 9x)$ is positive or negative. Expressing the difference as a polynomial in descending order, we have $$\begin{array}{rcl} x^3 - 5x^2 - 9x^2 + 45 &=& x^2(x-5) -9(x-5)\\ &=& (x^2 - 9)(x-5)\\ &=& (x-3)(x+3)(x-5) \end{array}$$ We then have three $x$ values where the difference is zero: $x=\pm 3$ and $x=5$.
This gives us four intervals to consider: $(-\infty,-3), (-3,3), (3,5), \textrm{ and } (5,\infty)$.
In the first and third intervals the product of the factors is negative. In the second and fourth, the product of factors is positive.
Thus, the graph of $f$ is above the graph of $g$ when $x \in (-3,3) \cup (5,\infty)$ and below the graph of $g$ when $x \in (-\infty,-3) \cup (5,\infty)$.
We need to find out where $\cfrac{24}{x^2-5x+6} - \cfrac{x}{x-3}$ is positive or negative.
To do the needed sign analysis, we want this difference as a single rational expression in completely factored form. Getting common denominators and combining the two rational expressions into one, the difference takes the form of the following rational expression $$\cfrac{-x^2 + 2x + 24}{(x-2)(x-3)}$$ which factored, becomes $$\frac{-(x-6)(x+4)}{(x-2)(x-3)}$$ We then see four $x$ values where the difference is zero or undefined: $x = -4,2,3,\textrm{ and } 6$ (the difference is undefined at $x=2$ and $x=3$).
This gives us five intervals to consider with regard to the sign analysis: $(-\infty,-4), (-4,2), (2,3), (3,6), \textrm{ and } (6,\infty)$.
The first, third, and fifth intervals make the related difference negative, while the rest correspond to a positive difference.
As such, the graph of $f$ is above the graph of $g$ for all $x \in (-\infty,-4) \cup (2,3) \cup (6,\infty)$ and below the graph of $g$ for all $x$ in $(-4,2) \cup (3,6)$.