Question

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

Answer 1

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`

Explanation:

`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`