\(y < x^2+10x+9\\--------------------\\y=x^2+10x+9=x^2+x+9x+9=x(x+1)+9(x+1)=(x+1)(x+9)\\\\the\ zeros:x=-9\ and\ x=-1\to(-9;\ 0)\ and\ (-1;\ 0)\\\\the\ vertex:\\x_v=\frac{-10}{2\cdot1}=\frac{-10}{2}=-5\\\\y_v=(-5)^2+10(-5)+9=25-50+9=-16\\\\V(-5;-16)\\\\y-intercept\to (0;\ 9)\\============================\\\\look\ at\ the\ picture\)
