## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 1 - Functions - 1.2 Representing Functions - 1.2 Exercises: 83

#### Answer

$x = \frac{{{x_1} + {x_2}}}{2}$

#### Work Step by Step

$\begin{gathered} The\,equation\,of\,the\,parabola\,having\,zeros\,at\,x = {x_1} \hfill \\ and\,x = {x_2}\,\,is. \hfill \\ \hfill \\ f\,\left( x \right) = c\,\left( {x - {x_1}} \right)\,\left( {x - {x_2}} \right) \hfill \\ \hfill \\ differentiating \hfill \\ \hfill \\ f\left( x \right) = c\,\left( {x - {x_1} + x - {x_2}} \right) \hfill \\ \hfill \\ add \hfill \\ \hfill \\ f\left( x \right) = c\,\left( {2x - {x_1} - {x_2}} \right) \hfill \\ \hfill \\ for\,vertex\,,\,we\,must\,have\,f\left( x \right) = 0\,\,the\,gives \hfill \\ \hfill \\ c\,\left( {2x - \,\left( {{x_1} + {x_2}} \right)} \right) = 0 \hfill \\ \hfill \\ solve\,for\,x \hfill \\ \hfill \\ x = \frac{{{x_1} + {x_2}}}{2} \hfill \\ \end{gathered}$

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