Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 44


The required graph is as follows:

Work Step by Step

The given equation $y=x^2-6x+8$ is a parabola $(\cup-\text{shaped})$ curve. It has vertical symmetry about the minimum point. The roots are the points where the y-coordinate is zero. Here, $x=2$ and $x=4$ are the points where $y=0$. Thus, the roots are $(2, 0)$ and $(4, 0)$. The line of symmetry passes from the mid of the roots, i.e., $x=3$ is the line of symmetry. The minimum point lies on the line of symmetry, so, $y=(3)^2-6(3)+8=-1$, i.e., $(3, -1)$ is the minimum point. Now, finally make a $\cup-\text{shaped}$ curve passing through roots and minimum point to obtain the graph as follows:
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