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Question

Write the equation of the parabola in standard form, and find the vertex of its graph. y = x2^{2} - 8x + 3

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y=x28x+3Write original functionf(x)=x28x+3Complete the squaref(x)=x28x+1616+3Group termsf(x)=(x28x+16)+(16+3)Factor the trinomial and simplifyf(x)=(x4)213Standard form f(x)=a(xh)2+k, sof(x)=(x4)213f(x)=a(xh)2+ka=1, h=4, k=13Vertex (h,k)Vertex (4,13)\begin{gathered} y = {x^2} - 8x + 3 \\ \textcolor{#4257b2}{{\text{Write original function}}} \\ f\left( x \right) = {x^2} - 8x + 3 \\ \textcolor{#4257b2}{{\text{Complete the square}}} \\ f\left( x \right) = {x^2} - 8x + 16 - 16 + 3 \\ \textcolor{#4257b2}{ {\text{Group terms}}} \\ f\left( x \right) = \left( {{x^2} - 8x + 16} \right) + \left( { - 16 + 3} \right) \\ \textcolor{#4257b2}{ {\text{Factor the trinomial and simplify}}} \\ f\left( x \right) = {\left( {x - 4} \right)^2} - 13 \\ \textcolor{#4257b2}{{\text{Standard form }}f\left( x \right) = a{\left( {x - h} \right)^2} + k,{\text{ so}}} \\ \underbrace {f\left( x \right) = {{\left( {x - 4} \right)}^2} - 13}_{f\left( x \right) = a{{\left( {x - h} \right)}^2} + k} \Rightarrow a = 1,{\text{ }}h = 4,{\text{ }}k = - 13 \\ {\text{Vertex }}\left( {h,k} \right) \Rightarrow {\text{Vertex }}\left( {4, - 13} \right) \\ \end{gathered}

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