## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 1 - Section 1.1 - Functions and Their Graphs - Exercises - Page 12: 12

#### Answer

$P(m)=(\frac{1}{m^2},\frac{1}{m})$

#### Work Step by Step

$\mathrm{See\: the\:figure\:below}$. Let's say $\ P(a,b)\$ is any point on the graph of $\ f(x)=\sqrt{x}\$. If we input $\ x\$ into the function $\ f(x)=\sqrt{x}\$, we get the corresponding value of $\ y\$ to be $\ \sqrt{x}\$. And, since $\ P\$ is on the graph of the function $\ \sqrt{x}\$, we have a relation for its coordinates. $P=(x,\sqrt{x})$ The slope of the line joining the point $\ P\$ and the origin is $\ m=\frac{\sqrt{x}}{x}.$ After rearranging the equation, we have: $\ m=\frac{\sqrt{x}}{x}$ $\Rightarrow\ m=\frac{1}{\sqrt{x}}$ $\Rightarrow\ \sqrt{x}=\frac{1}{m}$ $\Rightarrow\ x=\frac{1}{m^2}$ So, the coordinates of $\ P\$ as a function of the slope $\ m\$ can be written as: $P(m)=(\frac{1}{m^2},\frac{1}{m})$ where $\ m\$ represents the slope of the line.

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