cafdfc…: “Out of the bright tree”

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The best way to analyze a polynomial is with its factored form.

Analyze p(x)

\[ p(x) = 3 x^2 + 5 x^2 - 107 x + 35 \, \text { with its natural domain } \]

Algebraic Language

$\blacktriangleright $ Domain: We are given that the domain is the natural domain and the natural domain of all polynomials is $(-\infty , \infty )$.

$\blacktriangleright $ Continuity: All polynomials are continuous on their domain. So, $Q$ has no discontinuities. Since, the domain is all real numbers, there can be no singularities.

$\blacktriangleright $ Zeros:

There are several approaches to factoring. Identifying zeros or roots is one. For this we need a graph.

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The DESMOS graph is suggesting that perhaps $-7$ and $5$ might be roots and then there is a third root. Let’s check:

\[ p(-7) = 3 (-7)^2 + 5 (-7)^2 - 107 (-7) + 35 = 0 \]

\[ p(5) = 3 (5)^2 + 5 (5)^2 - 107 (5) + 35 = 0 \]

We can now factor $Q$.

We know that $(x+7)$ and $x-5$ are both factors of $p$. We just need to figure out the third factor.

\[ p(x) = 3 x^2 + 5 x^2 - 107 x + 35 = (x+7)(x-5)(A x + B) \]

\[ p(x) = 3 x^2 + 5 x^2 - 107 x + 35 = A x^3 + (2 A + B) x^2 + (-35 A + 2 B) x - 35 B \]

Comparing the constant terms tells us that $B = -1$

Comparing leading terms tells us that $A = 3$.

\[ p(x) = 3 x^2 + 5 x^2 - 107 x + 35 = (x+7)(x-5)(3 x - 1) \]

The zeros or roots are $-7$, $\frac {1}{3}$, and $5$. All three have multiplicity of $1$, which is odd. Therefore, $p$ changes signs across all three.

$\blacktriangleright $ End-Behavior: Polynomials with odd degree (like cubics) have the different end-behavior on either side. Since $p$ has a positive leading coefficient, $Q$ is unbounded negatively as $w$ tends to $-\infty $. and $Q$ is unbounded positively as $w$ tends to $\infty $.

\[ \lim \limits _{x \to -\infty } p(x) = -\infty \]

\[ \lim \limits _{x \to \infty } p(x) = \infty \]

There is no global maximum or global minimum.

$\blacktriangleright $ Behavior: The end-behavior for this cubic tells us that

$p$ increases on $(-\infty , -7)$
$p$ increases and then decreases on $\left ( -7, \frac {1}{3} \right )$
$p$ decreases and then increases on $\left ( \frac {1}{3}, \infty \right )$

$\blacktriangleright $ Extrema: We’ll need Calculus to know how to get a derivative for cubic polynomials. So, we cannot get the criticval numbers for $p$. That means we cannot get exact information about the local maximum or minimum. However, we can establish some information.

Since the cubic polynomial $p$ changes sign from negative to positive across $-7$, $p$ must be increasing around $-7$. Since the cubic polynomial $p$ changes sign from positive to negative across $\frac {1}{3}$, $p$ must be decreasing around $\frac {1}{3}$.

We also know that $p(-7) = 0$ and $p\left ( \frac {1}{3} \right ) = 0$ and that $p$ is continuous on $\left [-7, \frac {1}{3} \right ]$. There must be a local maximum somewhere on this interval.

Similar reasoning tells us that must be a local minimum somewhere on $\left [ \frac {1}{3}, 5 \right ]$.

At this point, we can only turn to the graph for approximations.

$p$ has a local maximum of approximately $351.072$ at approximately $-4.048$. $-4.048$ is a critical number.

$p$ has a local minimum of approximately $-160.126$ at approximately $2.937$. $2.937$ is a critical number.

$\blacktriangleright $ Range:

$p$ is continuous and $\lim \limits _{x \to -\infty } p(x) = -\infty $ and $\lim \limits _{x \to \infty } p(x) = \infty $. This tells us that the range is $(-\infty , \infty )$.

Graphical Language

$\blacktriangleright $ The graph of $y = p(x)$ is that of a cubic with one hill and one valley. It has no holes or breaks.

$\blacktriangleright $ The graph has three intercepts:

\[ \left ( -7, 0 \right ) \, \left ( \frac {\sqrt {1}}{3}, 0 \right ) \, \text { and } \, (5, 0) \]

$\blacktriangleright $ The graph slopes up to the right until it hits the top of a hill at approximately the point, $(-4.048, 351.072)$, then it slopes down to the right until it hits the bottom of a valley at approximately the point, $(2.937, -160.126)$, then is slopes up tot he right.

The graph has no highest or lowest points.

2025-01-09 20:19:51