University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.1 - Functions and Their Graphs - Exercises - Page 12: 16

Answer

$\mathrm{Domain:}\ \ (-\infty,\infty)$ $\mathrm{See\:the\:graph\:below.}$

Work Step by Step

$\mathrm{Remember:}$ The domain of a polynomial is the entire set of real numbers. The limiting factor on the domain for a rational function is the denominator, which cannot be equal to zero. $\mathrm{Domain:}\ \ (-\infty,\infty)$ To graph the parabola, we need at least three points. First of all, find the vetex of the parabola by using the formula for the $\ \mathrm{x-coordinate}\ $ as: $x=\frac{-b}{2a}$ $f(x)=y=-x^2-2x+1$ Here, $\ a=-1,\quad b=-2,\quad and\quad c=1$ $x=\frac{-(-2)}{2(-1)}=\frac{2}{-2}=-1$ Put this $\ \mathrm{x-coordinate}\ $ in the function to get the corresponding $\ \mathrm{y-coordinate}.$ $y=-(1)^2-2(1)+1=-2$ $\mathrm{Vertex:}\ (-1,2)$ Now apply the quadratic formula to find the roots as: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $x_{1,\:2}=\frac{-(-2)\pm \sqrt{(-2)^2-4(-1)1}}{2(-1)}$ $x_1=-1-\sqrt{2},\quad x_2=\sqrt{2}-1$ So, the other two points are, $\ (-1+\sqrt{2},0) \ \ and \ \ (-1-\sqrt{2},0)$ See the graph below.
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