Question

PL

Answer 1

Well, as far as the domain goes, this is a polynomial so your domain should be blurred out (correctly) right away.

Answer 2

Is \(\large\color{#000000 }{ \displaystyle f(x) }\) (in our case) ever undefined, for any \(x\) ??

Answer 3

@SolomonZelman no

Answer 4

Yes, right f(x), in this case, is defined for any value of x.
Therefore, what can you conclude about the domain of this function?

Answer 5

@SolomonZelman the vertex is -2, 4?

Answer 6

yup, y=a(x-h)^2+k
h,k is the vertex.
In this case, h and K are -2,4

Answer 7

(2,4) is the vertex, actually, but nice try though...

Answer 8

IM CONFUSED: The vertex is (2, 4), the domain is all real numbers, and the range is y ≤ 4.
The vertex is (2, 4), the domain is all real numbers, and the range is y ≥ 4. WHICH ONE WOULD IT BE? @SolomonZelman

Answer 9

When the parabola opens up the vertex is the minimum point, and when opens down the vertex is the maximum point.

parabola in a form
\(\large\color{#000000 }{ \displaystyle y(x)=a(x-h)^2+k }\)

opens down if \(a<0\).
opens up if \(a>0\).

Answer 10

so it would be the first one!

Answer 11

(1) Does your parabola open down or up?
(2) Is the vertex (2,4) the minimum or maximum of your parabola?
(3) Does that mean that y (the range) is at least 4 (4 or greater), or does that mean that y (the range) is at most 4 (4 or less)?

Answer 12

(1) Opens Up

Answer 13

And from there it should be obvious.

Answer 14

oh ok second one, thanks! @SolomonZelman

Answer 15

@SolomonZelman correct?

Answer 16

@SolomonZelman

Answer 17

should be ionlywantanswers

Answer 18

not the path to stanford...