## Algebra and Trigonometry 10th Edition

Standard form: $f(x)=(x-4)^2$ Vertex: $(4,0)$ Axis of symmetry: $x=4$. The x-intercept: $(4,0)$
$f(x)=x^2-8x+16~~$ ($a=1,~b=-8,~c=16$) $f(x)=x^2-2(4)x+4^2$ $f(x)=(x-4)^2$ $-\frac{b}{2a}=-\frac{-8}{2(1)}=4$ $f(4)=(4-4)^2=0$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(4,0)$ So, the axis of symmetry is $x=-\frac{b}{2a}=4$. The x-intercepts: $f(x)=0$ $(x-4)^2=0$ $x-4=0$ $x=4$ So, the x-intercept point is: $(4,0)$