Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 333: 102


The point on the line whose equation is \[2x+y-2=0\] that is closest to the origin is\[\left( \frac{4}{5},\frac{2}{5} \right)\].

Work Step by Step

Consider the equation of the line $2x+y-2=0$: Solve for y, $y=-2x+2$ Apply the distance formula for $y=-2x+2$ , $\begin{align} & d=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & =\sqrt{{{x}^{2}}+{{\left( -2x+2 \right)}^{2}}} \\ & =\sqrt{{{x}^{2}}+{{\left( -2x+2 \right)}^{2}}} \\ & =\sqrt{{{x}^{2}}+4-8x+4{{x}^{2}}} \end{align}$ Simplify further, $d=\sqrt{5{{x}^{2}}-8x+4}$ The minimum value of $5{{x}^{2}}-8x+4$ is the minimum value of the function $d=\sqrt{5{{x}^{2}}-8x+4}$ , Thus minimize $f\left( x \right)=5{{x}^{2}}-8x+4$. Compare it with the standard form $f\left( x \right)=a{{x}^{2}}+bx+c$ The value of $a$ is $5$ , $b$ is $-8$ and $c$ is $4$. Since, $a>0$ , the function has the minimum value at $x=\frac{-b}{2a}$. $\begin{align} & x=\frac{-b}{2a} \\ & =\frac{-\left( -8 \right)}{2\cdot 5} \\ & =\frac{8}{10} \\ & =\frac{4}{5} \end{align}$ Substitute $\frac{4}{5}$ for x in $y=-2x+2$ . $\begin{align} & y=-2\left( \frac{4}{5} \right)+2 \\ & =\frac{2}{5} \end{align}$ Hence, the point is $\left( \frac{4}{5},\frac{2}{5} \right)$.
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