How do you find the vertex and the intercepts for $Y= 3(x – 2)^2 - 4$ ?

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Answer 1 · 2017-03-14T16:32:47

Y has a minimum vertex at (2, -4).
Y-intercept = 8 and
x-intercept = $2(sqrt 3 + 1)/sqrt3$ and $2(sqrt 3 - 1)/sqrt3$

$Y = 3(x - 2)^2 -4$

$Y = a(x-p) -q$ , since
$a > 0$ , it is minimum vertex at (p, -q).

Therefore we conclude that the equation Y has a minimum vertex at (2, -4).

Y-intercept when x =0,
$Y = 3( - 2)^2 -4$
$Y = 3(4) -4 = 8$

x-intercept when y =0,
$0 = 3( x- 2)^2 -4$

$4= 3( x- 2)^2$

$4/3= ( x- 2)^2$
$+-sqrt (4/3)= ( x- 2)$

$2 +- 2/sqrt3= x$

$2(sqrt 3 +- 1)/sqrt3= x$

How do you find the vertex and the intercepts for Y= 3(x – 2)^2 - 4 Y= 3(x – 2)^2 - 4 ?