Answer
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 $x=-4$
Work Step by Step
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:
$x=-b/2a$
$x = -(-4)/2*1$
$x = 4/2 = 2$
$x=2$
$y=(x-2)^2$
$y=(2-2)^2$
$y=0^2$
$y=0$
