## Algebra and Trigonometry 10th Edition

Standard form: $y=[x-(-5)]^2-11$ Vertex: $(-5,-11)$ Axis of symmetry: $x=-5$ The x-intercepts are: $(-5+\sqrt {11},0)$ and $(-5-\sqrt {11},0)$
Standard form: $y=a(x-h)^2+k$, in which $(h,k)$ is the vertex. $y=x^2+10x+14$ $y=x^2+2(5)x+5^2-5^2+14$ $y=(x+5)^2-11$ $y=[x-(-5)]^2-11$ Vertex: $(-5,-11)$ Axis of symmetry: $x=-5$ $y=0$ $[x-(-5)]^2-11=0$ $[x-(-5)]^2=11$ $x+5=±\sqrt {11}$ $x=-5+\sqrt {11}$ or $x=-5-\sqrt {11}$ The x-intercepts are: $(-5+\sqrt {11},0)$ and $(-5-\sqrt {11},0)$