Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 2 - Linear and Quadratic Functions - Section 2.3 Quadratic Functions and Their Zeros - 2.3 Assess Your Understanding - Page 146: 84

Answer

Zeros: $\sqrt{2}+2,\sqrt{2}-2$ $x$-intercepts: $\sqrt{2}+2,\sqrt{2}-2$

Work Step by Step

To find the zeros of a function $f$, solve the equation $f(x)=0$ The zeros of the function are also the $x-$intercepts. Let $F(x)=0$: $$\dfrac{1}{2}x^2-\sqrt{2}x-1=0$$ Comparing $\dfrac{1}{2}x^2-\sqrt{2}x-1=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$ $$\therefore a = \dfrac{1}{2}, b=-\sqrt{2}, c =-1$$ Evaluating the discriminant $b^2-4ac$ $$b^2-4ac = (-\sqrt{2})^2-4 \times \dfrac{1}{2} \times -1 = 4$$ The quadratic formula is given by: $$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$x= \dfrac{\sqrt{2}\pm \sqrt{4}}{2\times \dfrac{1}{2}}$$ $$x=\dfrac{\sqrt{2}\pm 2}{1}$$ Thus, the zeros, which are also the $x$-intercepts, are $\sqrt{2}+2$ and $\sqrt{2}-2$.
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