Algebra and Trigonometry 10th Edition

Standard form: $f(x)=-(x-3)^2+9$ Vertex: $(3,9)$ Axis of symmetry: $x=-\frac{b}{2a}=3$. The x-intercepts: $(6,0)$ and $(0,0)$
$f(x)=6x-x^2~~$ ($a=-1,~b=6,~c=0$) $f(x)=-(x^2-6x)=-(x^2-6x+9-9)$ $f(x)=-[x^2-2(3)x+3^2]+9$ $f(x)=-(x-3)^2+9$ $-\frac{b}{2a}=-\frac{6}{2(-1)}=3$ $f(3)=-(3-3)^2+9=9$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(3,9)$ So, the axis of symmetry is $x=-\frac{b}{2a}=3$. The x-intercepts: $f(x)=0$ $-(x-3)^2+9=0$ $0=(x-3)^2$ $x-3=±3$ $x=6$ or $x=0$ So, the x-intercept points are: $(6,0)$ and $(0,0)$