Answer
Vertex: $(2, -12)$
Opens upward
x-intercepts: 0, 4
y-intercept: 0
Work Step by Step
$f(x)= 3x^2-12x$
$y=3x^2-12x$
$y=3x(x-4)$
$y=3x^2-12x$
$y=3(x^2-4x)$
$y+3*(-4/2)^2=3(x^2-4x)+3*(-4/2)^2$
$y+3*(-2)^2=3(x^2-4x)+3*(-2)^2$
$y+3*4=3(x^2-4x)+3*4$
$y+12=3(x^2-4x+4) $
$y+12=3(x-2)^2$
$y+12-12=3(x-2)^2-12$
$y=3(x-2)^2-12$
Vertex: $(2, -12)$
Opens upward
$x=0$
$y=3x^2-12x$
$y=3*0^2-12*0$
$y=3*0-0$
$y=0-0$
$y=0$
$y=0$
$y=3(x-2)^2-12$
$0=3(x-2)^2-12$
$0+12=3(x-2)^2-12+12$
$12=3(x-2)^2$
$12/3=3(x-2)^2/3$
$4=(x-2)^2$
$\sqrt 4=\sqrt {(x-2)^2}$
$±2 =x-2$
$2=x-2$
$2+2=x-2+2$
$4=x$
$-2=x-2$
$-2+2=x-2+2$
$0=x$
x-intercepts: 0, 4
y-intercept: 0
