## Algebra and Trigonometry 10th Edition

Vertex: $(-1,4)$ x-intercept: $(-3,0),(-1,0)$ axis of symmetry: $x=-1$
$f(x)=-(x^2+2x-3)=-x^2-2x+3$ $a=-1,~~b=-2,~~c=3$ Vertex: $\frac{-b}{2a}=\frac{-(-2)}{2(-1)}=-1$ $f(\frac{-b}{2a})=-(-1)^2-2(-1)+3=-1+2+3=4$ x-intercept: $x_1=\frac{-(-2)+\sqrt {(-2)^2-4(-1)3}}{2(-1)}=\frac{2+\sqrt {16}}{-2}=-3$ $x_2=\frac{-(-2)-\sqrt {(-2)^2-4(-1)3}}{2(-1)}=\frac{2-\sqrt {16}}{-2}=1$