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Questions: Algebra BusinessCalculus
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Use \(\displaystyle f(x) = x^{2} - 5 x + 9\) to evaluate:
Evaluating the function at \(\displaystyle x = 0\) gives \(\displaystyle f(0) = (0)^{2} - 5 (0) + 9 = 9\).
Evaluating the function at \(\displaystyle x = - 2 x\) gives \(\displaystyle f(- 2 x) = (- 2 x)^{2} - 5 (- 2 x) + 9 = 4 x^{2} + 10 x + 9\).
Evaluating the function at \(\displaystyle x = - t - 1\) gives \(\displaystyle f(- t - 1) = (- t - 1)^{2} - 5 (- t - 1) + 9 = t^{2} + 7 t + 15\).
\begin{question}Use $f(x) = x^{2} - 5 x + 9$ to evaluate:\newline
\begin{tabular}{|l|}\hline
$f(0)$ \\ \hline
$f(- 2 x)$ \\ \hline
$f(- t - 1)$ \\ \hline
\end{tabular}\newline
\soln{9cm}{Evaluating the function at $x = 0$ gives $f(0) = (0)^{2} - 5 (0) + 9 = 9$.\newline
Evaluating the function at $x = - 2 x$ gives $f(- 2 x) = (- 2 x)^{2} - 5 (- 2 x) + 9 = 4 x^{2} + 10 x + 9$.\newline
Evaluating the function at $x = - t - 1$ gives $f(- t - 1) = (- t - 1)^{2} - 5 (- t - 1) + 9 = t^{2} + 7 t + 15$.\newline
}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Use <img class="equation_image" title=" \displaystyle f(x) = x^{2} - 5 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20x%5E%7B2%7D%20-%205%20x%20%2B%209%20" alt="LaTeX: \displaystyle f(x) = x^{2} - 5 x + 9 " data-equation-content=" \displaystyle f(x) = x^{2} - 5 x + 9 " /> to evaluate:<br>
</p> </p><p> <p>Evaluating the function at <img class="equation_image" title=" \displaystyle x = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%200%20" alt="LaTeX: \displaystyle x = 0 " data-equation-content=" \displaystyle x = 0 " /> gives <img class="equation_image" title=" \displaystyle f(0) = (0)^{2} - 5 (0) + 9 = 9 " src="/equation_images/%20%5Cdisplaystyle%20f%280%29%20%3D%20%280%29%5E%7B2%7D%20-%205%20%280%29%20%2B%209%20%3D%209%20" alt="LaTeX: \displaystyle f(0) = (0)^{2} - 5 (0) + 9 = 9 " data-equation-content=" \displaystyle f(0) = (0)^{2} - 5 (0) + 9 = 9 " /> .<br>
Evaluating the function at <img class="equation_image" title=" \displaystyle x = - 2 x " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-%202%20x%20" alt="LaTeX: \displaystyle x = - 2 x " data-equation-content=" \displaystyle x = - 2 x " /> gives <img class="equation_image" title=" \displaystyle f(- 2 x) = (- 2 x)^{2} - 5 (- 2 x) + 9 = 4 x^{2} + 10 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20f%28-%202%20x%29%20%3D%20%28-%202%20x%29%5E%7B2%7D%20-%205%20%28-%202%20x%29%20%2B%209%20%3D%204%20x%5E%7B2%7D%20%2B%2010%20x%20%2B%209%20" alt="LaTeX: \displaystyle f(- 2 x) = (- 2 x)^{2} - 5 (- 2 x) + 9 = 4 x^{2} + 10 x + 9 " data-equation-content=" \displaystyle f(- 2 x) = (- 2 x)^{2} - 5 (- 2 x) + 9 = 4 x^{2} + 10 x + 9 " /> .<br>
Evaluating the function at <img class="equation_image" title=" \displaystyle x = - t - 1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-%20t%20-%201%20" alt="LaTeX: \displaystyle x = - t - 1 " data-equation-content=" \displaystyle x = - t - 1 " /> gives <img class="equation_image" title=" \displaystyle f(- t - 1) = (- t - 1)^{2} - 5 (- t - 1) + 9 = t^{2} + 7 t + 15 " src="/equation_images/%20%5Cdisplaystyle%20f%28-%20t%20-%201%29%20%3D%20%28-%20t%20-%201%29%5E%7B2%7D%20-%205%20%28-%20t%20-%201%29%20%2B%209%20%3D%20t%5E%7B2%7D%20%2B%207%20t%20%2B%2015%20" alt="LaTeX: \displaystyle f(- t - 1) = (- t - 1)^{2} - 5 (- t - 1) + 9 = t^{2} + 7 t + 15 " data-equation-content=" \displaystyle f(- t - 1) = (- t - 1)^{2} - 5 (- t - 1) + 9 = t^{2} + 7 t + 15 " /> .<br>
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