Classify the two turning points exactly

The printed self-reading algebraic method uses the quotient rule:

$\begin{aligned}&\frac{dy}{dx}\\ & =\frac{(x+1)(2x-3)-(x^2-3x)(1)}{(x+1)^2}\\ & =\frac{x^2+2x-3}{(x+1)^2}.\end{aligned}$

Retain $x\ne-1$, where the original curve is undefined.

The source sets $dy/dx=0$, so $x^2+2x-3=0$. Factor $(x+3)(x-1)=0$ to obtain $x=-3$ or $1$. Both inputs belong to the domain. Substitution gives $y(-3)=-9$ and $y(1)=-1$.

Applying the quotient rule again, the source gives

$\begin{aligned}&\frac{d^2y}{dx^2}\\ & =\frac{(x+1)^2(2x+2)-(x^2+2x-3)(2)(x+1)}{(x+1)^4}.\end{aligned}$

At $x=-3$, $y=-9$ and $y^{\prime\prime}=-1<0$, so $(-3,-9)$ is a maximum point. At $x=1$, $y=-1$ and $y^{\prime\prime}=1>0$, so $(1,-1)$ is a minimum point. Independently the derivative simplifies to $8/(x+1)^3$; these are local extrema on separate branches, not global bounds on the whole curve.

The source gives six steps. 1. Enter $(x^2-3x)/(x+1)$ in Y1. 2. Press 2nd → TRACE to access CALCULATE, then choose 4:maximum. Both copies say “to find the x-intercepts” here; this is a wording slip—the selected routine finds the maximum. 3. Move left of the maximum with the arrows and press ENTER, or enter a left-bound number. The next screen asks for the right bound; the source’s red note marks this transition.

4. Move right of the maximum and press ENTER, or enter a right-bound number. The source shows the Right Bound? screen before confirming and the Guess? screen afterwards. Its numerical-method remark says to round appropriately and verify the answer. 5. Move close to the maximum for the guess, then press ENTER. The calculator display is approximately $(-3,-9)$; the exact calculation above verifies it. Independently, keep the bracket on the same continuous branch, away from the vertical asymptote $x=-1$.

6. Repeat the process for the minimum. Use 3:minimum in the displayed CALCULATE menu for that search. The source’s final display and conclusion give minimum $(1,-1)$, alongside maximum $(-3,-9)$. The following source remark directs exact turning-point calculus to Applications of Differentiation. The reused reconstruction shows the same curve and printed turning-point coordinates; its asymptotes were established in Example 3.