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

Published by Pearson

# Chapter 2 - Linear and Quadratic Functions - Section 2.3 Quadratic Functions and Their Zeros - 2.3 Assess Your Understanding - Page 146: 79

#### Answer

Zeros: $\dfrac{5}{2},-\dfrac{3}{5}$ $x$-intercepts: $\dfrac{5}{2},-\dfrac{3}{5}$

#### 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 $G)x)=0$: $$10x^2-19x-15=0$$ Comparing $10x^2-19x-15=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$ $$\therefore a = 10, b=-19 , c =-15$$ Evaluating the discriminant $b^2-4ac$ $$b^2-4ac = (-19)^2-4 \times 10 \times -15 = 961$$ The quadratic formula is given by: $$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$x= \dfrac{19\pm \sqrt{961}}{2\times 10}$$ $$x=\dfrac{19\pm 31}{20}$$ Thus, the zeros, which are also the $x$-intercepts, are $\dfrac{5}{2}$ and $-\dfrac{3}{5}$.

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