Precalculus (6th Edition) Blitzer

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

Chapter 1 - Mid-Chapter Check Point - Page 229: 38

Answer

See the graph below:

Work Step by Step

We can see that at point $x=0$, the graph passes through $y=-1$. We see that there is no $x$ variable in the first function. Thus, the point is $\left( 0,-1 \right)$. Now, at point $x=0.1$ , the graph passes through the following y-coordinate: $\begin{align} & y=2x+1 \\ & =2\left( 0.1 \right)+1 \\ & =0.2+1 \\ & =1.2 \end{align}$ Thus, the point is $\left( 0.1,1.2 \right)$. And, at point $x=1$, the graph passes through the following y-coordinate: $\begin{align} & y=2x+1 \\ & =2\left( 1 \right)+1 \\ & =3 \end{align}$ Thus, the point is $\left( 1,3 \right)$.
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