Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.10 - Modeling with Functions - Exercise Set - Page 294: 37

Answer

The distance from a point $P\left( x,y \right)$ to $\left( 1,0 \right)$ in terms of the point’s $x\text{-coordinate}$ is, $d=\sqrt{{{x}^{2}}-x+1}$.

Work Step by Step

The distance from point $P\left( x,y \right)$ to point $\left( 1,0 \right)$ is $\begin{align} & d=\sqrt{{{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}} \\ & =\sqrt{{{\left( x-1 \right)}^{2}}+{{y}^{2}}} \end{align}$ Substitute $\sqrt{x}$ for $y$ in the above equation, $d=\sqrt{{{\left( x-1 \right)}^{2}}+{{\left( \sqrt{x} \right)}^{2}}}$ Use the formula ${{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$ in the above equation $\begin{align} & d=\sqrt{{{x}^{2}}-2x+1+x} \\ & =\sqrt{{{x}^{2}}-x+1} \end{align}$ The required distance from a point $P\left( x,y \right)$ from point $\left( 1,0 \right)$ in terms of the point’s $x\text{-coordinate}$ is, $d=\sqrt{{{x}^{2}}-x+1}$.
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