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

$(-1,-1)$ and $\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$
To find the points of intersection of $f(x)$ and $g(x)$, solve $f(x)=g(x)$: \begin{align*} -2x^2+1&=3x+2\\ 0&=2x^2-1+3x+2\\ 0&=2x^2+3x+1 \end{align*} By Factoring: $$0=(x+1)(2x+1)$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation: \begin{align*} x+1 &=0 &\text{ or }& &2x+1=0\\ x &=-1 &\text{ or }& &2x =-1\\ x &=-1 &\text{ or }& &x=-\dfrac{1}{2}\\ \end{align*} To find the y-coordinates of the points of intersection, evaluate either of the two functions at $x=-1$ and $x=-\dfrac{1}{2}$ to obtain: $g(-1)=3(-1)+2=-1$ $g(-\dfrac{1}{2})=3(-\dfrac{1}{2})+2=\dfrac{1}{2}$ Therefore, the points of intersection are $(-1,-1)$ and $\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$.