## Algebra 1

a) $x=7$ b) $(7,0)$ c) $h=2$; vertex is $(2,0)$ and $x=2$ d) You would expect the vertex to be at $x=-4$
a) $(x-7)^2=0$ $\sqrt {(x-7)^2} = \sqrt 0$ $x-7 = 0$ $x-7+7 = 0+7$ $x = 7$ b) $(x-7)^2 = x^2-7x-7x+49$ $x^2-7x-7x+49 = x^2-14x+49$ vertex: $x=-b/2a$ $x = -(-14)/2*1$ $x = 14/2 = 7$ $x=7$ $y=(x-7)^2$ $y=(7-7)^2$ $y=0^2$ $y=0$ c) $(x-2)^2=0$ $\sqrt {(x-2)^2} = \sqrt 0$ $x-2 = 0$ $x-2+2 = 0+2$ $x = 2$ $(x-2)^2 = x^2-2x-2x+4$ $x^2-2x-2x+4 = x^2-4x+4$ vertex: $x=-b/2a$ $x = -(-4)/2*1$ $x = 4/2 = 2$ $x=2$ $y=(x-2)^2$ $y=(2-2)^2$ $y=0^2$ $y=0$