Question

question #1
everett factored the rational equation in order to determine the domain and point(s) of discontinuity.
y = \frac{3x^{2}+5x - 2}{x^{2}+5x + 6}
everetts work is shown.
did everett make a mistake? if so, in which step was the mistake made? what was the mistake?
step 1: \frac{(3x - 1)(x + 2)}{x^{2}+5x + 6}
step 2: \frac{(3x - 1)(x + 2)}{(x + 3)(x + 2)}
step 3: domain: (-\infty,-3)\cup(-3,-2)\cup(-2,\infty)
step 4: point of discontinuity: (-2,7)
step 3; everett stated the domain incorrectly.
step 1; everett has the wrong factors for the numerator.
step 4, everett determined the point of discontinuity incorrectly.
step 2; everett has the wrong factors for the denominator.

Explanation:

Step1: Check numerator factoring

The numerator $3x^{2}+5x - 2=(3x - 1)(x + 2)$ by factoring techniques (product - sum method: product of $3\times(-2)=-6$ and sum of $5$, factors of $-6$ that add to $5$ are $6$ and $-1$, then rewrite middle term and factor by grouping). Everett's Step 1 for numerator is correct.

Step2: Check denominator factoring

The denominator $x^{2}+5x + 6=(x + 3)(x+ 2)$ since the product of the constant term is $6$ and the sum of the coefficient of the linear term is $5$, and $3\times2 = 6$ and $3 + 2=5$. Everett's Step 2 for denominator is correct.

Step3: Check domain calculation

The domain of a rational function $y=\frac{f(x)}{g(x)}$ is all real numbers except where $g(x)=0$. Setting $(x + 3)(x + 2)=0$, we get $x=-3$ and $x=-2$. So the domain $(-\infty,-3)\cup(-3,-2)\cup(-2,\infty)$ is correct as in Everett's Step 3.

Step4: Check point - of - discontinuity

A removable discontinuity occurs at a value of $x$ that makes both the numerator and denominator zero. After canceling out the common factor $(x + 2)$ (since $x
eq - 2$ in the domain of the original rational function), we have $y=\frac{3x - 1}{x + 3}$. To find the $y$ - value of the removable discontinuity, we substitute $x=-2$ into $\frac{3x - 1}{x + 3}$. So $y=\frac{3(-2)-1}{-2 + 3}=\frac{-6-1}{1}=-7$. Everett's Step 4 has an incorrect $y$ - value for the point of discontinuity.

Answer:

Step 4; Everett determined the point of discontinuity incorrectly.