Question

for each of the following quadratic functions:
(i) Find the coordinates for the vertex
(ii) determine whether the vertex is max or min
(iii) find the range for the functions
(a) f(x) = 2x^2 +4

Answer 1

@SolomonZelman

Answer 2

@AaronAndyson

Answer 3

@imqwerty

Answer 4

\(\color{black}{ \displaystyle f(x)=2x^2+4~~~~\Longrightarrow ~~~~f(x)=2(x-0)^2+4 }\)

And this is in the form of \(\color{black}{ \displaystyle f(x)=2(x-h)^2+k }\) (with vertex \((h,k)\) )

Answer 5

ok so the vertex is 0,4

Answer 6

yes, correct.

Answer 7

The vertex is min

Answer 8

If the leading coefficient is POSITIVE, then parabola opens UP, the vertex is the MINIMUM point (or the absolute MINIMUM).

If the leading coefficient is NEGATIVE, then parabola opens DOWN, the vertex is the MAXIMUM point (or the absolute MAXIMUM).

Answer 9

Yes, so the vertex is minimum.

Answer 10

but you said max

Answer 11

When the leading coefficient is negative.

Answer 12

ohh ok

Answer 13

But, as I said, when leading coefficient is positive (in your case it is 2), then the vertex is minimum.

Answer 14

And the range of the opening up parabola will start from the vertex, but has no restrictions on how large the output can be....

Answer 15

so it is (y:y≤4)

Answer 16

Yes, that is exactly right.

Answer 17

??

Answer 18

Or, in interval notation \({\rm y} \in [4,+\infty)\)

Answer 19

we did not learn that yet.

Answer 20

All that means is that the range goes from and including 4, and ends at positive infinity (meaning that it goes up forever endlessly).

Answer 21

don't want to throw in confusion.... in any case you are right y\(\le\)4.

Answer 22

Any questions about what we have done?

Answer 23

nop, thanks a lot really apreciate, really needed it, i have a quiz tmrw.

Answer 24

Good luck on the quiz:)

Answer 25

thanks mate

Answer 26

thanks @imqwerty for seeing my question.

Answer 27

lolol XD

Answer 28

lol