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**Soroban** Hello, Jools!

With the given points of the parabola: .$\displaystyle (-5,0),(0,-5), (1,0)$

. . we can determine its equation: .$\displaystyle g'(x) \:=\:x^2+4x-5$

Then: .$\displaystyle g(x)\;=\;\int(x^2+4x-5)\,dx \;=\;\tfrac{1}{3}x^3 + 2x^2 - 5x + C$

The extreme values occur where $\displaystyle g'(x) = 0 \quad\Rightarrow\quad x \:=\:-5,\:1$

The second derivative is: .$\displaystyle g''(x) \:=\:2x + 4$

. . $\displaystyle \begin{array}{cccccccc}g''(\text{-}5) &=& 2(\text{-}5)+4 &=& -6 & \text{concave down: }\cap & \text{maximum} \\ g''(1) &=& 2(1)+4 &=& +6 & \text{concave up: } \cup & \text{minimum}\end{array}$

Therefore: .$\displaystyle \begin{Bmatrix}\text{maximum at:}& \left(-5,\:C + \frac{100}{3}\right) \\ \\[-3mm] \text{minimum at:} & \left(1,\:C-\frac{8}{3}\right) \end{Bmatrix}$

Inflection points: .$\displaystyle g''(x) \:=\:0 \quad\Rightarrow\quad 2x + 4 \:=\:0 \quad\Rightarrow\quad x \:=\:-2$

Therefore: . $\displaystyle \begin{Bmatrix}g''(x) > 0 & \text{when }x > -2 & \text{concave up} \\ \\[-3mm] g''(x) < 0 & \text{when }x < -2 & \text{concave down} \end{Bmatrix}$