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Interpreting Quadratic Functions in Vertex Form

Interpreting Quadratic Functions in Vertex Form 1.2 - Solution

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Let's start by recalling the vertex form of a quadratic function.

Vertex Formf(x)=a(xh)2+k\begin{gathered} \underline{\textbf{Vertex Form}}\\ f(x) = \textcolor{#FF00FF}{a}(x-{\color{#0000FF}{h}})^2 + {\color{#009600}{k}} \end{gathered} In this format, the vertex is located at (h,k).({\color{#0000FF}{h}},{\color{#009600}{k}}).

Since we want our function to have a vertex at (1,3)({\color{#0000FF}{1}},{\color{#009600}{3}}) we can wirte it in vertex form using h=1{\color{#0000FF}{h}}={\color{#0000FF}{1}} and k=3.{\color{#009600}{k}} = {\color{#009600}{3}}. Note that aa can take any value as it does not affect the location of the vertex. For simplicity, we will use a=1.\textcolor{#FF00FF}{a} = \textcolor{#FF00FF}{1}. f(x)=1(x1)2+3f(x)=(x1)2+3\begin{gathered} f(x) = \textcolor{#FF00FF}{1}(x-\textcolor{#0000FF}{1})^2 +\textcolor{#009600}{3} \quad \Leftrightarrow \quad f(x)=(x-1)^2+3 \end{gathered} Notice that we could have chosen any nonzero value for a.a. Therefore, there are infinitely many quadratic functions whose vertex is located at (1,3).(1,3). This is just one example.