## Calculus with Applications (10th Edition)

$(-\displaystyle \frac{4}{5}$ , $\displaystyle \frac{31}{5})$
$y=-5x^{2}-8x+3$ $=-5(x^{2}+\displaystyle \frac{8}{5}x)+3$ $=-5(x^{2}+2\displaystyle \cdot\frac{4}{5}x+\frac{16}{25}-\frac{16}{25} )+3$ $=-5(x+\displaystyle \frac{4}{5})^{2}+3-(-5)(\frac{16}{25})$ $=-5(x+\displaystyle \frac{4}{5})^{2}+\frac{15+16}{5}$ $=-5(x+\displaystyle \frac{4}{5})^{2}+\frac{31}{5}$ Vertex: $\boxed{\left(-\displaystyle \frac{4}{5} , \displaystyle \frac{31}{5}\right)}$.