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: 31

Answer

Zeros: $2\sqrt{3}-2,-2\sqrt{3}-2$ $x$-intercepts: $2\sqrt{3}-2,-2\sqrt{3}-2$

Work Step by Step

To find the zeros of a function $g$, solve the equation $g(x)=0$ The zeros of the function are also the $x-$intercepts. Let $f(x)=0$: $$x^2+4x-8=0$$ Rearranging the Equation: $$x^2+4x=8$$ The coefficient of $x^2$ is 1 and that of x is 4, complete the square by adding $\left(\dfrac{1 \times 4}{2}\right)^2 = 4$ $$(x^2+4x+4)=8+4$$ $$(x+2)^2 = 12$$ $$(x+2)=\pm \sqrt{12}$$ $\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}$ $\therefore x+2 = 2\sqrt{3} \hspace{20pt}\text{ or } \hspace{20pt} x+2 = -2 \sqrt{3} $ Solve each equation to obtain the zeros of the given function: $x+2 = 2 \sqrt{3} \to x=2\sqrt{3}-2$ $x+2 = -2 \sqrt{3} \to x= -2 \sqrt{3}-2$ Thus, the $x$-intercepts are: $-2\sqrt3-2$ and $2\sqrt3-2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.