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**Soroban** Hello, mathhelpplease!

Your work seems to be correct,

. . but I'm not sure where (or why) you are stuck.

I'll walk through it for you.

$\displaystyle y \:=\:a(x-h)^2+k$

$\displaystyle y\:=\:a(x^2-2hx + h^2) + k$

$\displaystyle y\:=\:ax^2 - 2ahx + ah^2 + k$

$\displaystyle y\:=\:ax^2 - 2ahx + (ah^2+k)$ .[1]

$\displaystyle y \:=\:a(x-p)(x-q)$

$\displaystyle y\:=\:a(x^2 - px - qx + pq)$

$\displaystyle y\:=\:ax^2 - apx - aqx + apq$

$\displaystyle y\:=\:ax^2 - a(p+q)x + apq$ .[2]

[1] .$\displaystyle y\:=\:ax^2 - 2ahx + (ah^2+k) \quad\Rightarrow\quad \begin{Bmatrix}a &=& a \\ b &=& \text{-}2ah \\ c &=& ah^2+k \end{Bmatrix}$

Vertex: .$\displaystyle x \;=\;\frac{\text{-}b}{2a} \;=\;\frac{\text{-}(\text{-}2ah)}{2(a)} \;=\;\frac{2ah}{2a} \;=\;h$

$\displaystyle \text{The vertex occurs at }\,x \,=\, h.$

[2] .$\displaystyle y \:=\:ax^2 - a(p+q)x + apq \quad\Rightarrow\quad \begin{Bmatrix}a &=& a \\ b &=& \text{-}a(p+q) \\ c &=& apq \end{Bmatrix}$

Vertex: .$\displaystyle x \;=\;\frac{\text{-}b}{2a} \;=\;\frac{\text{-}[\text{-}a(p+q)]}{2(a)} \;=\;\frac{a(p+q)}{2a} \;=\;\frac{p+q}{2}$

$\displaystyle \text{The vertex occurs at }\,x \,=\,\tfrac{p+q}{2}$