\documentclass{article} \usepackage[fleqn]{amsmath} \usepackage[pdf,forpaper,cfg=hw,nopoints,useforms, % Try compiling this file under one of these options: % nosolutions, % answerkey, vspacewithsolns, ]{eqexam} \subject[CA]{College Algebra} \title[HW2]{HW \#2} \author{Dr.\ D. P. Story} \date{Spring 2011} \duedate{03/05/11} \keywords{Homework due \theduedate} % Make the end of solution label blank \def\exrtnlabelformat{} \def\exrtnlabelformatwp{} \solAtEndFormatting{\eqequesitemsep{3pt}} \begin{document} \maketitle \begin{exam}{HW} \begin{instructions}[] In preparation for the quiz on Thursday, solve each of these short problems in the space provided before looking at their solutions at the end of the document. \[ \text{\url{http://faculty.nwfsc.edu/web/math/storyd}} \] All class assignments and other announcements will be posted on this web site. \end{instructions} \begin{problem*}[4ea] Let $P(-4,2)$ and $Q(2,-3)$ be two points in the plane. \begin{parts} \item Find the distance $d(P,Q)$ between $P$ and $Q$. \begin{solution}[1in] We use the distance formula \[ d(P,Q)=\sqrt{(2+4)^2+(-3-2)^2}=\sqrt{61} \] to obtained the required answer. \end{solution} \item Find the midpoint $M$ between $P$ and $Q$. \begin{solution}[\sameVspace] We use the midpoint formula \[ M=\left(\frac{-4+2}{2},\frac{2+(-3)}{2}\right)=\left(-1,-\dfrac{1}{2}\right) \] to obtained the required answer. \end{solution} \end{parts} \end{problem*} \begin{problem*}[3ea] Complete each of the two sentences below with correct entries. \begin{parts} \item The function $ g(x) = | x+2 | $ can be graphed from the library function $ f(x) = |x| $ by shifting it \fillin[u]{.25in}{2} units \fillin[u]{1.25in}{horizontally} (horizontally/vertically) \fillin[u]{.75in}{left} (left/right/up/down). % Make the solution label blank. \ifkeyalt \begin{solution}[] % We want this to be shown at the end of the file, but not if the author changes the option % to answerkey The function $ g(x) = | x+2 | $ can be graphed from the library function $ f(x) = |x| $ by shifting it \fillin[u]{.25in}{2} units \fillin[u]{1.25in}{horizontally} (horizontally/vertically) \fillin[u]{.75in}{left} (left/\penalty0right/\penalty0up/\penalty0down). % Note: \penalty0 is inserted to help TeX break the line after the forward slash / \end{solution} \fi \item The function $ g(x) = 5 - x^2 $ can be graphed from the library function $ f(x) = x^2 $ by first reflecting it with respect to the \fillin[u]{.25in}{$x$} axis, then shifting it \fillin[u]{.25in}{5} units \fillin[u]{1.25in}{vertically} (horizontally/vertically) \fillin[u]{.75in}{upward} (left/right/up/down). \ifkeyalt \begin{solution}[] % We want this to be shown at the end of the file, but not if the author changes the option % to answerkey. The function $ g(x) = 5 - x^2 $ can be graphed from the library function $ f(x) = x^2 $ by first reflecting it with respect to the \fillin[u]{.25in}{$x$} axis, then shifting it \fillin[u]{.25in}{5} units \fillin[u]{1.25in}{vertically} (horizontally/vertically) \fillin[u]{.75in}{upward} (left/right/up/down). \end{solution} \fi \end{parts} \end{problem*} \begin{problem}[5] The circle $ x^2 + y^2 = 25 $ passes through the point $P(3,4)$. Let $\ell$ be the line passing though the origin and the point $P$. Find the equation of the line perpendicular to line $\ell$ and passing through point $P$. \begin{solution}[1in] The slope of the line perpendicular to $\ell$ is $ m = -\tfrac{3}{4} $, the line must pass through $(3,4)$; thus, the line is $ y-4 = -\tfrac{3}{4}(x-3) \implies y = -\tfrac{3}{4} x + \tfrac{25}{4} $. Thus, \[ \text{Ans:}\quad\boxed{ y = -\tfrac{3}{4} x + \tfrac{25}{4} } \] This is the equation of the line tangent to the circle at $P(3,4)$. \end{solution} \end{problem} \begin{problem}[3] If the slope the a line is negative, then the line is \begin{answers}{4} \bChoices \Ans0 increasing\eAns \Ans1 decreasing\eAns \Ans0 constant\eAns \Ans0 none of these\eAns \eChoices \end{answers} % The \texttt{answers} and \texttt{manswers} environments can now % be copied and pasted into the solutions environment as well. % \ifkeyalt \begin{solution}[.25in] If the slope the a line is negative, then the line is \begin{answers}{4} \bChoices \Ans0 increasing\eAns \Ans1 decreasing\eAns \Ans0 constant\eAns \Ans0 none of these\eAns \eChoices \end{answers} \end{solution} \fi \end{problem} \end{exam} \end{document}