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

Zeros: $-2+\sqrt{2},-2-\sqrt{2}$ $x$-intercepts: $-2+\sqrt{2},-2-\sqrt{2}$
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 $f(x)=0$: $$x^2+4x+2=0$$ Comparing $x^2+4x+2=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$ $$\therefore a = 1, b=4 , c =2$$ Evaluating the discriminant $b^2-4ac$ $$b^2-4ac = (4)^2-4 \times 1 \times 2 = 8$$ The quadratic formula is given by: $$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$x= \dfrac{-4 \pm \sqrt{8}}{2\times 1}$$ $$x=\dfrac{-4 \pm 2\sqrt{2}}{2}$$ $$x= -2 \pm \sqrt{2}$$ $\therefore x =-2+\sqrt{2} \hspace{20pt} \text{or} \hspace{20pt} x=-2-\sqrt{2}$