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Questions: Algebra BusinessCalculus
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Write the function \(\displaystyle f(x)=x^{2} - 4 x + 39\) in vertex form and identify the vertex.
\(\displaystyle h = -\frac{b}{2a}\) and \(\displaystyle k=f(h)\). Using each formula gives \(\displaystyle h= 2\) and \(\displaystyle k = 35\). The vertex is located at \(\displaystyle (2,35)\) and the vertex form is \(\displaystyle f(x) = a(x-h)^2+k=\left(x - 2\right)^{2} + 35\).
\begin{question}Write the function $f(x)=x^{2} - 4 x + 39$ in vertex form and identify the vertex. \soln{9cm}{$h = -\frac{b}{2a}$ and $k=f(h)$. Using each formula gives $h= 2$ and $k = 35$. The vertex is located at $(2,35)$ and the vertex form is $f(x) = a(x-h)^2+k=\left(x - 2\right)^{2} + 35$. } \end{question}
\documentclass{article} \usepackage{tikz} \usepackage{amsmath} \usepackage[margin=2cm]{geometry} \usepackage{tcolorbox} \newcounter{ExamNumber} \newcounter{questioncount} \stepcounter{questioncount} \newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}} \renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}} \newif\ifShowSolution \newcommand{\soln}[2]{% \ifShowSolution% \noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else% \vspace{#1}% \fi% }% \newcommand{\hideifShowSolution}[1]{% \ifShowSolution% % \else% #1% \fi% }% \everymath{\displaystyle} \ShowSolutiontrue \begin{document}\begin{question}(10pts) The question goes here! \soln{9cm}{The solution goes here.} \end{question}\end{document}
<p> <p>Write the function <img class="equation_image" title=" \displaystyle f(x)=x^{2} - 4 x + 39 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3Dx%5E%7B2%7D%20-%204%20x%20%2B%2039%20" alt="LaTeX: \displaystyle f(x)=x^{2} - 4 x + 39 " data-equation-content=" \displaystyle f(x)=x^{2} - 4 x + 39 " /> in vertex form and identify the vertex. </p> </p>
<p> <p> <img class="equation_image" title=" \displaystyle h = -\frac{b}{2a} " src="/equation_images/%20%5Cdisplaystyle%20h%20%3D%20-%5Cfrac%7Bb%7D%7B2a%7D%20" alt="LaTeX: \displaystyle h = -\frac{b}{2a} " data-equation-content=" \displaystyle h = -\frac{b}{2a} " /> and <img class="equation_image" title=" \displaystyle k=f(h) " src="/equation_images/%20%5Cdisplaystyle%20k%3Df%28h%29%20" alt="LaTeX: \displaystyle k=f(h) " data-equation-content=" \displaystyle k=f(h) " /> . Using each formula gives <img class="equation_image" title=" \displaystyle h= 2 " src="/equation_images/%20%5Cdisplaystyle%20h%3D%202%20" alt="LaTeX: \displaystyle h= 2 " data-equation-content=" \displaystyle h= 2 " /> and <img class="equation_image" title=" \displaystyle k = 35 " src="/equation_images/%20%5Cdisplaystyle%20k%20%3D%2035%20" alt="LaTeX: \displaystyle k = 35 " data-equation-content=" \displaystyle k = 35 " /> . The vertex is located at <img class="equation_image" title=" \displaystyle (2,35) " src="/equation_images/%20%5Cdisplaystyle%20%282%2C35%29%20" alt="LaTeX: \displaystyle (2,35) " data-equation-content=" \displaystyle (2,35) " /> and the vertex form is <img class="equation_image" title=" \displaystyle f(x) = a(x-h)^2+k=\left(x - 2\right)^{2} + 35 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20a%28x-h%29%5E2%2Bk%3D%5Cleft%28x%20-%202%5Cright%29%5E%7B2%7D%20%2B%2035%20" alt="LaTeX: \displaystyle f(x) = a(x-h)^2+k=\left(x - 2\right)^{2} + 35 " data-equation-content=" \displaystyle f(x) = a(x-h)^2+k=\left(x - 2\right)^{2} + 35 " /> . </p> </p>